Answer:
1. 2.28 g
2. 0.82 g
3. 1.35 g
4. 36 %
5. 59 %
6. 2.17 g
7. 95 %
Explanation:
Hello,
1. In this case, the mass of the unknown mixture is obtained by subtracting the mass of the dish plus sample and the mass of the dish:
m = 26.57 g- 24.29 g = 2.28 g
2. In this case, the mass of the NaCl recovered is obtained by subtracting the mass of the dish plus NaCl and the mass of the dish:
m = 68.66 g- 67.84 g = 0.82 g
3. In this case, the mass of the sand recovered is obtained by subtracting the mass of the filter paper plus sand and the mass of the filter paper:
m = 37.69 g- 36.34 g = 1.35 g
4. The percentage of NaCl is computed by considering its mass and the mass of the unknown mixture:
% NaCl = 0.82 g / 2.28 g * 100 % = 36 %
5. The percentage of sand is computed by considering its mass and the mass of the unknown mixture:
% sand = 1.35 g / 2.28 g * 100 % = 59 %
6. Here,we have to add the mass of NaCl and sand:
m = 0.82 g + 1.35 g = 2.17 g
7. Finally, the percent recovery is obtained by diving the total recovered mass by the total obtained mass of the mixture:
% recovery = 2.17 g / 2.28 g * 100 % = 95 %
Best regards.
Which process is a physical change
Answer:
a physical change is something that has not been modified chemically and can possibly be changed back to the state it was once before. A physical change keeps all the same atoms and none of them is modified.
Example:
When a block of clay is morphed into a giraffe statue, it can be morphed back to its original state. If someone burnt the block of clay, the atoms would be modified and it would be unable to go back to its previous state.
-----------------------------------------------------------------------------------------------------------------
(If you're referring to a question with these answers)
A. iron rusting
B. milk turning to curd
C. water boiling
D. paper burning
E. hard water staining pipes
-----------------------------------------------------------------------------------------------------------------
Answer:
C. Water Boiling
(If you are referring to a question with these answers I think this is the correct answer if not I do apologize)
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how many grams of H2 are needed to produce 14.34 g of NH3?
please help, this hw is due in a few hours
Answer:
im pretty sure its 2.54g H2
Explanation:
14.34gNH3 / 17.03gNH3 <-- molar mass
.842g x 3 mol <-- mols of H2
2.52 / 2 mol <-- mols of NH3
1.26 x 2.016gH2= 2.54gH2
What will be the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat? The specific heat of Al is 0.900 J/g·C.
42.1°C is the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat.
What is temperature?The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points or thermometric substances.
The most popular scales include the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the scale of degrees Fahrenheit (°F), or the Kelvin scale (K), with the latter being mostly used for scientific purposes. One of the United Nations System of Units' (SI) seven base units is the kelvin.
Q=725 J
m=55.0 g
c=0.900 J/(°C⋅g)
ΔT=final temperature - initial temperature
ΔT=(x−27.5)°C
725 J=55.0 g⋅0.900 J/(°C⋅g)(x−27.5)
X=42.1°C
Therefore, 42.1°C is the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat.
To know more about temperature, here:
https://brainly.com/question/23411503
#SPJ2
A compound has an empirical formula of CHN. What is the molecular formula, if it’s molar mass is 135.13 g/mol? (C=12.01 amu, H=1.008 amu, N= 14.01)
Answer:
well the MF is 224.78 g/mol
Explanation:
just times them all by the molor mass and divide it by 3
Identify the Lewis acids and Lewis bases in the following reactions:
1. H+ + OH- <-> H2O Lewis acid: Lewis base:
2. Cl- + BCl3 <-> BCl4- Lewis acid: Lewis base:
3. K+ + 6H2O <-> K(H2O)6+ Lewis acid: Lewis base:
Answer: 1. [tex]H^++OH^-\rightarrow H_2O[/tex] Lewis acid : [tex]H^+[/tex], Lewis base : [tex]OH^-[/tex]
2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex] Lewis acid : [tex]BCl_3[/tex], Lewis base : [tex]Cl^-[/tex]
3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex] Lewis acid : [tex]K^+[/tex], Lewis base : [tex]H_2O[/tex]
Explanation:
According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.
1. [tex]H^++OH^-\rightarrow H_2O[/tex]
As [tex]H^+[/tex] gained electrons to complete its octet. Thus it acts as lewis acid.[tex]OH^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]H^+[/tex].
2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex]
As [tex]BCl_3[/tex] is short of two electrons to complete its octet. Thus it acts as lewis acid. [tex]Cl^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]BCl_3[/tex].
3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex]
As [tex]K^+[/tex] is short of electrons to complete its octet. Thus it acts as lewis acid. [tex]H_2O[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]K^+[/tex].
Alkyl derivatives of mercury are highly toxic and can cause mercury poisoning in humans. Dimethylmercury is one of the strongest known neurotoxins. Although it is said to have a slightly sweet smell, inhaling enough to discern this would be hazardous.
Give the empirical formula of dimethylmercury.
Answer:
The empirical formula of dimethylmercury is C2H6Hg
Explanation:
Dimethylmercury, as it says in the name, presents not only the mercury metal in its structure (Hg) but also two radical groups called methyl, which is why its name begins with the prefix DI, referring to the fact that there are two methyl.
Consider the reaction in a commercial heat pack: 4 Fe (s) + 3 O2(g) ® 2 Fe2O3 (s) DH = -1652 kJ a) How much heat is released when 1.00 g iron is reacted with excess O2? b) What mass of O2 must react with iron in order to generate 2150 kJ of heat?
Answer:
a) -7.395kJ of energy are released.
b) 125g of O₂ must react.
Explanation:
Based on the reaction:
4 Fe (s) + 3 O₂(g) → 2 Fe₂O₃ (s) ΔH = -1652 kJ
4 moles of iron with an excess of oxygen release -1652kJ of energy
a) The heat released is:
1.00g Fe (molar mass: 55.845g/mol)
1.00g × (1mol / 55.845g) = 0.0179 moles de Fe.
As 4 moles release -1652kJ, 0.0179 moles release:
0.0179 mol Fe × (-1652kJ / 4mol Fe) = -7.395kJ of energy are released.
b) As 3 moles of oxygen produce -1652kJ, 2150kJ are released when react:
2150kJ × (3 mol O₂ / 1652kJ) = 3.9 moles of O₂
As molar mass of O₂ is 32g/mol, mass of 3.9 moles of O₂ is:
3.9 mol O₂ × (32g / mol) = 125g of O₂ must react.
What is the limiting reactant if 8 g of Ba reacts with 2.8 g of Al2(SO4)3?
Answer:
Al2(SO4)3 is the limiting reactant
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
3Ba + Al2(SO4)3 → 2Al + 3BaSO4
Next, we shall determine the mass of Ba and the mass of Al2(SO4)3 that reacted from the balanced equation. This is illustrated below:
Molar mass of Ba = 137g/mol
Mass of Ba from the balanced equation = 3 x 137 = 411g
Molar mass of Al2(SO4)3 = 2x27 + 3[32 + (16x4)]
= 54 + 3[32 + 64]
= 54 + 3[96]
= 54 + 288 = 342g/mol
Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g
Summary:
From the balanced equation above,
411g of Ba reacted with 342g of Al2(SO4)3.
Finally, we shall determine the limiting reactant as follow:
From the balanced equation above,
411g of Ba reacted with 342g of Al2(SO4)3.
Therefore, 8g of Ba will react with
= (8 x 342/411 = 6.66g of Al2(SO4)3.
From the calculations made above, we can see that it will take a higher mass of Al2(SO4)3 i.e 6.66g than what was given i.e 2.8g to react completely with 8g of Ba.
Therefore, Al2(SO4)3 is the limiting reactant and Ba is the excess reactant.
If a jet’s cruising altitude is 32,200ft(to three significant figures),the distance in km is :(1 mile=1.61km;1 mile=5280 ft)
Answer:
9.82 km.
Explanation:
Hello,
In this case, given the conversion factors from miles to kilometres and from miles to feet, we can directly compute the jet’s cruising altitude in kilometres as shown below:
[tex]32,200ft\times \frac{1mile}{5280ft}\times \frac{1.61km}{1mile} \\\\=9.82km[/tex]
Best regards.
Which compound has the lowest melting point? KCl CaCl2 Na2O C6H12O6
Consider this reaction:
2Cl2O5 —> 2Cl2 + 5O2
At a certain temperature it obeys this rate law.
rate = (2.7.M^-1•s^-1) [Cl2O5]^2
Suppose a vessel contains Cl2O5 at a concentration of 0.600M. calculate how long it takes for the concentration of Cl2O5 to decrease by 94%. you may assume no other reaction is important. round your answer to two digits
Answer:
[tex]t=9.7s[/tex]
Explanation:
Hello,
In this case, we have a second order kinetics given the second power of the concentration of chlorine (V) oxide in the rate expression, thus, the integrated equation for the concentration decay is:
[tex]\frac{1}{[Cl_2O_5]}=kt+\frac{1}{[Cl_2O_5]_0}[/tex]
Thus, the final concentration for a 94% decrease is:
[tex][Cl_2O_5]=0.600M-0.600M*0.94=0.036M[/tex]
Therefore, we compute the time for such decrease:
[tex]kt=\frac{1}{[Cl_2O_5]}-\frac{1}{[Cl_2O_5]_0}=\frac{1}{0.036M}-\frac{1}{0.60M} =26.1M^{-1}[/tex]
[tex]t=\frac{26.1M^{-1}}{k}= \frac{26.1M^{-1}}{2.7M^{-1}*s^{-1}}\\\\t=9.7s[/tex]
Regards.
how many moles of helium gas occupy 22.4 L at 0 degreeC at 1 atm pressure
Answer:
1 mole of the gas occupies 22. 4L at O0C at 1atm pressure. Hence, the correct option is C.
Explanation:
Why don't we discuss the results during the results section of the project?
Answer: The result section of the project contains your findings while carrying out your research or study.
Explanation:
The Results section of a research or study usually contains only the findings of your study or research.The findings which usually include
1. Data presented in tables, charts, graphs, and other figures.
2. A contextual analysis of this data explaining its meanings. Usually in sentences.
Our result gotten is not discussed in result section because every project or research work has a discussion page where every results or findings are discussed. The result section is expected to carry what you found.
At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=1.9×10−9 M?
Answer:
The correct answer will be "5.26 × 10⁻⁶".
Explanation:
The given values is:
[tex][H^{+}]=1.9\times 10^{-9} M[/tex]
As we know,
⇒ [tex]pH+pOH=14[/tex]
On taking log, we get
⇒ [tex]-log[H^{+}] + -log[OH^{-}] = 14[/tex]
Now,
Taking "log" as common, we get
⇒ [tex]log[H^{+}][OH^{-}]= -14[/tex]
⇒ [tex][H^{+}][OH^{-}]= 10^{-14}[/tex]
⇒ [tex][OH^{-}]=\frac{10^{-14}}{[H^{+}]}[/tex]
On putting the estimated value of "[tex][H^{+}][/tex]", we get
⇒ [tex]=\frac{10^{-14}}{1.9\times 10^{-9}}[/tex]
⇒ [tex]=5.26\times 10^{-6}[/tex]
A researcher placed 25.0 g of silver chloride, AgCl, in sunlight and allowed the substance to decompose completely to form silver, Ag, with the release of chlorine gas, Cl2. The gas was collected in a container during the decomposition. The researcher determined that the mass of the silver formed was 18.8 g, and the mass of the chlorine gas formed was 6.2 g. The equation for the reaction is:
Answer:
A. The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements combines in the same proportion by mass.
B. The law of conservation of mass states that during ordinary chemical reactions, matter can neither be created or destroyed.
Note: The full question is as follows;
A researcher placed 25.0 g of silver chloride, AgCl, in sunlight and allowed the substance to decompose completely to form silver, Ag, with the release of chlorine gas, Cl2. The gas was collected in a container during the decomposition. The researcher determined that the mass of the silver formed was 18.8 g, and the mass of the chlorine gas formed was 6.2 g. The equation for the reaction is:
2AgCl ----> 2Ag + Cl2
a. State the law of definite proportions. Then use the researcher's data to confirm the law of definite proportions. Show your calculations.
b. State the law of conservation of matter. Then use the researcher's data to confirm the conservation of matter. Show your calculations.
Explanation:
A. Mass of silver obtained from AgCl = 18.8g.
Percentage mass of silver in the chloride = (18.8/25.0) * 100 = 75.2 %
Mass of chlorine obtained from AgCl = 6.2
Percentage mass of chlorine = (6.2/25) * 100 = 24.8 %
In one mole of AgCl with a molar mass of 143.3 g/mol; mass of silver = 107.8, mass of Cl = 35.5
Percentage mass of Ag = (107.8/143.3) * 100 = 75.2%
Percentage mass of Cl = (35.5/143.3) * 100 = 24.8%
Since the percentages by mass of Ag and AgCl obtained from the sample is the same to that obtained from a mole of AgCl, the law of definite proportions which states that all pure samples of a particular chemical compound contain the same elements combined in the same proportion by mass is verified.
B. Mass of reactant; AgCl sample = 25.0
Mass of products; At = 18.8 g; Cl = 6.2 g
Sum of products masses = 18.8 + 6.2 = 25.0 g
Therefore mass of reactant = mass of products.
This is in accordance with the law of conservation of mass which states that during ordinary chemical reactions, matter is neither created nor destroyed.
Which of the following errors could cause your percent yield to be falsely high, or even over 100%?
Select ALL that apply.
A.) Heating the sample too vigorously.
B.) Handling the crucible directly with your hands.
C.) Failing to completely decompose the sodium bicarbonate sample.
D.) Taking the mass of the empty crucible without the lid, but including the lid in all other mass measurements.
E.) Taking the mass of all samples with the lid included.
Answer:
B.Handling the crucible directly with your hands.
D.Taking the mass of the empty crucible without the lid, but including the lid in all other mass measurements.
E.Taking the mass of all samples with the lid included.
Explanation:
When observed critically , the measures associated with the errors which could cause your percent yield to be falsely high, or even over 100% are those which increase the weight of the substance with the individual neglecting.
Handling the crucible directly with your hands,Taking the mass of the empty crucible without the lid, but including the lid in all other mass measurements and taking the mass of all samples with the lid included will all increase the weight of the substance. Instead the substance should be placed alone without any form of support or contamination.
g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete the mass balance expressions for the sodium and arsenate species in the final solution.
Answer:
The concentration of the sodium and arsenate ions at the end of the reaction in the final solution
[Na⁺] = 0.05512 M
[HAsO₄²⁻] = 0.00185 M
[AsO₄³⁻] = 0.01714 M
Explanation:
Complete Question
A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.
Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O
From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)
Concentration in mol/L = (Number of moles) ÷ (Volume in L)
Number of moles = (Concentration in mol/L) × (Number of moles)
For Na₂HAsO₄
Concentration in mol/L = 0.03798 M
Volume in L = (500/1000) = 0.50 L
Number of moles = 0.03798 × 0.5 = 0.01899 mole
For NaOH
Concentration in mol/L = 0.03428 M
Volume in L = (500/1000) = 0.50 L
Number of moles = 0.03428 × 0.5 = 0.01714 mole
Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.
Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O
0.01899 0.01714 0 0 (At time t=0)
(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)
0.00185 | 0 → 0.01714 0.01714 (end)
Hence, at the end of the reaction, the following compounds have the following number of moles
Na₂HAsO₄ = 0.00185 mole
This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction
NaOH = 0 mole
Na₃AsO₄ = 0.01714 moles
This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction
H₂O = 0.01714 moles
So, at the end of the reaction
Na⁺ has 0.0037 + 0.05142 = 0.05512 mole
(HAsO₄)²⁻ has 0.00185 mole
(AsO₄)³⁻ has 0.01714 mole.
And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L
Hence, the concentration of the sodium and arsenate ions at the end of the reaction is
[Na⁺] = 0.05512 M
[HAsO₄²⁻] = 0.00185 M
[AsO₄³⁻] = 0.01714 M
Hope this Helps!!!
Answer:
[tex]\rm [Na^{+}]= \text{0.055 12 mol/L}[/tex]
[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \text{0.018 99 mol/L}[/tex]
Explanation:
The overall equation for the reaction is
Na₂HAsO₄ + NaOH ⟶ Na₃AsO₄ + H₂O
1. Mass balance for Na
All the Na⁺ comes from the Na₂HAsO₄ and the NaOH.
The mass balance equation for Na is
[tex]\rm c_{Na^{+}} = 2[Na^{+}]_{Na_{2}HAsO_{4}} + [Na^{+}]_{NaOH}[/tex]
At the moment of mixing and before the reaction started, the total volume had doubled, so the concentrations of each component were halved.
[Na₂HAsO₄] = ½ × 0.037 98 =0.018 99 mol·L⁻¹
[NaOH] = ½ × 0.034 28 = 0.017 14 mol·L⁻¹
[tex]\rm c_{Na^{+}} = 2\times 0.01899 + 0.01714 = 0.03798 + 0.01714\\c_{Na^{+}}= \textbf{0.055 12 mol/L}[/tex]
2. Mass balance for arsenate species
All the arsenate species come from the Na₂HAsO₄.
The reactions involved are
HAsO₄²⁻+ OH⁻ ⇌ AsO₄³⁻ + H₂O
HAsO₄²⁻ + H₂O ⇌ H₂AsO₄⁻ + OH⁻
H₂AsO₄⁻ + H₂O ⇌ H₃AsO₄ + OH⁻
The mass balance equation for arsenate species is
[tex]\rm c_{\text{arsenate}} = [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}][/tex]
At the moment of mixing, the concentration of Na₂HAsO₄ had halved.
[Na₂HAsO₄] = ½ × 0.039 78 = 0.018 99 mol·L⁻¹
[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \textbf{0.018 99 mol/L}[/tex]
When 106 g of water at a temperature of 21.4 °C is mixed with 64.3 g of water at an unknown temperature, the final temperature of the resulting mixture is 46.8 °C. What was the initial temperature of the second sample of water? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K.)
Answer:
THE INITIAL TEMPERATURE OF THE SECOND SAMPLE IS 4.93 C OR 277.93 K
Explanation:
Mass of first sample of water = 106 g
Initial temp of first sample = 21.4 °C = 21.4 + 273 K = 294.4 K
Mass of second sample = 64.3 g
Final temp of theresulting mixture = 46.8 °C = 46.8 + 273 K = 319.8 K
Specific heat capacity of water = 4.184 J/g K
It is worthy to note that;
Heat gained by the first sample = Heat lost by the second sample
Since heat = mass * specific heat capacity * change in temperature, we have
Mass * specific heat * change in temp of the first sample = Mass * specific heat * change in temp. of the second sample
MC (T2 - T1) = MC (T2-T1)
106 * 4.184 * ( 319.8 - 294.4) = 64.3 * 4.184 * ( 319.8 - T1)
106 * 4.184 * 25.4 = 269.0312 ( 319.8 - T1)
11 265.0016 = 269.0312 (319.8 - T1)
Since the change in temperature = 319.8 -T1
Change in temperature =11265.0016 / 269.0312
Change in temperature = 41.87
Change in temperature = 319.8 -T1
41.87 = 319.8 - T1
T1 = 319.8 - 41.87
T1 = 277.93 K
T1 = 4.93 °C
So therefore, the initial temperature of the sacond sample is 4.73 °C or 277.93 K
Interpret the following equation for a chemical reaction using the coefficients given: CO(g) Cl2(g) COCl2(g) On the particulate level: _________ of CO(g) reacts with _________ of Cl2(g) to form _________ of COCl2(g). On the molar level: _________ of CO(g) reacts with _________ of Cl2(g) to form _________ of COCl2(g).
Answer:
On the particulate level: 6.02 * 10²³ particles of CO(g) reacts with 6.02 * 10²³ particles of Cl₂(g) to form 6.02 * 10²³ particles of COCl2(g).
On the molar level: 1 mole of CO(g) reacts with 1 mole of Cl2(g) to form 1 mole of COCl₂(g).
Explanation:
The particulate level refers to the microscopic or atomic level of substances. It also involves the ions, protons, neutrons and molecules present in substances.
The molar level refers to the quantitative measure of substances in terms of the mole, where a mole represents the amount of substances containing the Avogadro number of particles which is equal to 6.02 * 10³ particles.
Equation of the reaction: CO(g) + Cl₂(g) ----> COCl₂(g)
From the equation above, I mole of CO gas reacts with 1 mole of Cl₂ gas to produce 1 mole of COCl₂ gas.
Since 1 mole of a substance contains 6.02 * 10²³ particles, on a particulate level, 6.02 * 10²³ particles of CO gas reacts with 6.02 * 10²³ particles of Cl₂ gas to produce 6.02 * 10²³ particles of COCl₂ gas.
A student sets up the following equation to convert a measurement. (The ? stands for a number the student is going to calculate.) Fill in the missing part of this equation. Note: your answer should be in the form of one or more fractions multiplied together. (23. Pa cm^3)____?kPa . m^3
Answer:
The correct answer will be "-6.7 × 10¹⁰ kg.m/s".
Explanation:
The required conversions are:
⇒ [tex]1 \ kg=1000 \ g[/tex]
⇒ [tex]1 \ m=100 \ cm[/tex]
Now,
The complete conversion will be:
= [tex][-6.7\times 10^5 \ \frac{kg \ m}{s} ]\times [\frac{10^3 \ g}{kg}\times \frac{10^2 \ cm}{1 \ m}][/tex]
On cancelling the terms, we get
= [tex]-6.7\times 10^{10} \frac{kg \ m}{s}[/tex]
So that the missing terms will be [tex][\frac{10^3 \ g}{kg}\times \frac{10^2 \ cm}{1 \ m}][/tex] and [tex][-6.7\times 10^{10}\frac{kg \ m}{s}][/tex]
A gas has a volume of 6.6 L at a temperature of 40 C. What is the volume of
the gas if the temperature changes to 15 C?
Answer:
6.07 L
Explanation:
It appears that the reading has been made at constant pressure .
At constant pressure , the gas law formula is
V/T = constant V is volume and T is temperature of the gas.
V₁ / T₁ = V₂ / T₂
V₁ = 6.6 L ,
T₁ = 40°C
= 273 + 40
= 313 K
T₂ = 15+ 273
= 288K
V₂ = ?
Putting the values in the formula above
6.6 / 313 = V₂ / 288
V₂ = 6.07 L.
AHP for the formation of rust (Fe2O3) is -826 kJ/mol. How much energy is
involved in the formation of 5.00 grams of rust?
A 25.9 kJ
B 25.9 J
C 66.0 kJ
D 66.0)
Answer:
A- 25.9 kJ
Explanation:
ΔH of formation is defined as the amount of energy that is involved in the formation of 1 mole of substance.
ΔH of rust is -826kJ/mol, that means when 1 mole of rust is formed, there are released -826kJ.
Moles of 5.00g of Fe₂O₃ (Molar mass: 159.69g/mol) are:
5.00g ₓ (1 mole / 159.69g) = 0.0313 moles of Fe₂O₃.
If 1 mole release -826kJ, 0.0313 moles release:
0.0313 moles ₓ (-826kJ / 1 mole) = -25.9kJ
Thus, heat involved is:
A- 25.9 kJHow many moles of CO2 are produced when 84 0 mol O2 completely react?
Answer:
Explanation:boom
what is the color of benzene and bromine
Explanation:
Benzene is colorless, with a sweet odour.
Color of Bromine is reddish brown .
Hope this helps.
WILL GIVE BRAINLIEST!!!!! will give brainliest!!!! ******According to the reaction in question I above, how many grams of solid copper will theoretically be
produced when 14.4 g of aluminum are reacted with 14.4 g of copper (II) sulfate? Which reactant is the
limiting reactant? Show your work. Be sure to include units!
Answer:
5.73 g Cu
Explanation:
M(CuSO4) = 159.6 g/mol
M(Al) = 27.0 g/mol
M(Cu) = 63.5 g/mol
14.4 g Al * 1 mol/27.0 g = 0.5333 mol Al
14.4 g CuSO4 * 1 mol/159.6 g = 0.0902 mol CuSO4
2 Al + 3CuSO4 -------> 3Cu + Al2(SO4)3
from reaction 2 mol 3 mol
given (0.5333 mol )x 0.0902 mol
needed 0.0601 mol
x= 2*0.0902/3 = 0.0601 mol Al
Al is excess, CuSO4 is a limiting reactant.
2 Al + 3CuSO4 -------> 3Cu + Al2(SO4)3
from reaction 3 mol 3 mol
given 0.0902 mol x mol
x = 0.0902 mol Cu
0.0902 mol Cu * 1 mol/63.5 g Cu = 5.73 g Cu
The height of the sun in the sky at noontime is called
Answer:
It is called 'solar noon'
Explanation:
It's when the sun is at its highest
And also the moment the sun crosses the meridian.
A semipermeable sac containing 4% NaCl, 9% glucose, and 10% albumin is suspended in a solution with the following composition: 10% NaCl, 10% glucose, and 40% albumin. Assume that the sac is permeable to all substances except albumin. State whether each of the following will (a) move into the sac, (b) move out of the sac, or (c) not move.
glucose: a. moves into sac
water: b. moves out of sac
albumin: c; does not move
NaCl: a; moves into sac
Answer:
Glucose: (a) moves into sac
Water: (b) moves out of sac
Albumin: (c) does not move
NaCl: (a) moves into sac
Explanation:
A semi-permeable membrane is a membrane that allow certain molecules or ions to pass through by diffusion.
Diffusion is the movement of molecules from a region of higher concentration to a region of lower concentration until equilibrium is attained. A special form of diffusion known as osmosis transports water molecules across a semi-permeable membrane.
Osmosis is the movement of water molecules from a region lower solute concentration (high water molecules concentration) to a region of higher solute concentration (low water molecules concentration) until equilibrium is attained.
From the above definitions and the given data;
Glucose concentration is higher in solution outside the sac, thus, glucose molecules will move into the sac.
Water molecules are higher in the sac as a result of the lower concentration of solutes, therefore, water molecules will move out of the sac into the solution outside.
Since the sac is impermeable to albumin, it does not move.
NaCl concentration is lower in the sac, therefore, it will move ro the solution outside into the sac.
Glucose will move into the sac, water will move out of the sac, albumin will neither move in nor out, and NaCl will move into the sac.
The molecules of each of the substances will move by diffusion from the region of higher concentration to the region of lower concentration of each substance as long as there is a permeable membrane. Water, on the other hand, will move by osmosis from the region of high to low water potential through a permeable membrane. Regions with higher concentrations of substances usually have low water potentials and vice versa.Thus, both glucose and NaCl molecules will diffuse from the solution into the sac, and water molecules will move from the sac into the surrounding solution. Since the sac is not permeable to albumin, then the movement in or out is inhibited.
More on osmosis and diffusion can be found here: https://brainly.com/question/19867503?referrer=searchResults
1. Reaccionan 9.7 Kg de un mineral de níquel al 70% con 8L de una solución de ácido fosfórico al 60% y con una densidad de 1.36g/ml.
Answer:
The reaction produces 201.4 g of hydrogen gas and 12.2 kg of Nickel Phosphate.
Explanation:
English Translation
9.7 Kg of a 70% nickel mineral react with 8L of a 60% phosphoric acid solution and with a density of 1.36g / ml.
Solution
The problem doesn't seen to be complete as it doesn't ask a question in the end. But, we will just calculate the amount of each product expected to cover the grounds.
The balanced chemical reaction between Nickel and Phosphoric acid is given as
3Ni + 2H₃PO₄ → 3H₂ + Ni₃(PO₄)₂
We need to first obtain the limiting reagent, that is, the reagent that is used up during the reaction and is in short supply. This reagent determines the amount of products that will be formed.
Mass of nickel that is present at the start = 70% of 9.7 kg = 6.79 kg
Mass of Phosphoric acid present at the start of the reaction = 60% of (8000 mL × 1.36 g/mL) = 6528 g = 6.528 kg
Converting both of these to number of moles
Number of moles = (mass)/(Molar mass)
For nickel,
Mass = 6.79 kg = 6790 g
Molar mass = 58.6934 g/mol
Number of moles at the start = (6790/58.6934) = 115.7 moles
For Phosphoric acid
Mass = 6528 g
Molar mass = 97.994 g/mol
Number of moles = (6528/97.994) = 66.6 moles
3 moles of Ni reacts with 2 moles of H₃PO₄
From the number of moles present initially, shows that Phosphoric acid is in limited supply and is the limiting reagent.
From the stoichiometric balance of the reaction
2 moles of H₃PO₄ gives 3 moles of H₂
66.6 moles of H₃PO₄ will give (66.6×3/2) of H₂, that is, 99.9 moles of H₂.
Mass of H₂ liberated from the reaction = (Number of moles) × (molar mass) = 99.9 × 2.016 = 201.3984 g = 201.4 g
2 moles of H₃PO₄ gives 1 mole of Ni₃(PO₄)₂
66.6 moles of H₃PO₄ will give (66.6×1/2) of Ni₃(PO₄)₂, that is, 33.3 moles of Ni₃(PO₄)₂.
Mass of Ni₃(PO₄)₂ produced from the reaction = (Number of moles) × (molar mass) = 33.3 × 366.02 = 12,188.466 g = 12.2 kg
Hope this Helps!!!
Now construct a different electrochemical cell. You put a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of the electric potential at the moment the reaction begins
Answer:
[tex]1.66~V[/tex]
Explanation:
We have to start with the half-reactions for both ions:
[tex]Zn^+^2~+2e^-~->Zn[/tex] V= -0.76
[tex]Ag^+~e^-~->~Ag[/tex] V= +0.80
If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:
[tex]Zn~->~Zn^+^2~+2e^-~[/tex] V= +0.76
[tex]Ag^+~+~e^-~->~Ag[/tex] V= +0.80
If we want to calculate ºE we have to add the two values, so:
ºE=0.76+0.80 = 1.56 V
Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:
[tex]E=ºE~+~\frac{0.059}{n}LogQ[/tex]
On this case, Q is equal to:
[tex]Q=\frac{[Zn^+^2]}{[Ag^+]^2}[/tex]
Because the total reaction is:
[tex]Zn~+~2Ag^+~->~Zn^+^2~+~2Ag[/tex]
So, the value of "Q" is:
[tex]Q=\frac{[0.052 M]}{[0.0042]^2}=2947.84[/tex]
Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:
[tex]E=1.56~V~+~\frac{0.059}{2}Log(2947.84)=1.66~V[/tex]
I hope it helps!
When you look at an ant up close, using a convex lens, what do you see?
Answer:
You would be able to see the ants clearly with the unique body parts.
Explanation:
Convex lens is also known as Converging lens. It helps in converging rays of light to become a principal focus which is usually very clear and visible to the eyes. The converging lens when used to view the ant makes the ant appear very visible and the individual is able to see all the unique body parts of the ant. This type of lens is used in individuals who are longsighted.