Round to 3 significant figures.
1.4593

Round To 3 Significant Figures.1.4593

Answers

Answer 1

Answer : The correct answer is 1.46

Explanation :

The following rules are used to round off a number to the required number of significant figures:

(1) If the rightmost digit to be removed is more than 5, the preceding number is increased by one.

(2) If the rightmost digit to be removed is less than 5, the preceding number is not changed.

(3) If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number but it is increased by one if it is an odd number.

(4) The same procedure is follow for decimal values.

As we are given, 1.4593

In the given answer, there are 5 significant figures. Now we have to convert it into 3 significant figures.

According to the rules, round off the given measurement in three significant figures as 1.46

Therefore, the correct answer is 1.46


Related Questions

Rubidium is ______ potassium in the periodic table. lodine is ______ bromine in the periodic table. Therefore, the rubidium ion is __________ than the potassium ion, and the iodine ion is___________ than the bromide ion. The _______ the distance between the rubidium ion and the iodide ion is the potassium ion and the bromide ion. Therefore, the energy associated with the interaction between rubidium and iodide is________ atomic radius means that than that between , and the lattice energy of potassium bromide is ________ more exothermidc.

Answers

Answer:

The given blanks can be filled with below, below, larger, larger, larger, larger and smaller.

Explanation:

In the periodic table, rubidium comes below the potassium, and iodine comes below bromine. Therefore, it can be said that the ion of rubidium is larger in comparison to potassium ion, and similarly the ion of iodine is larger in comparison to the ion of bromine.  

When the atomic radius is larger it signifies that the distance in between the ion of iodine and the ion of rubidium is larger in comparison to that between the ion of potassium and the ion of bromine. Thus, smaller energy is associated with the interaction between iodine and rubidium, and potassium bromide's lattice energy is more exothermic.  

Fractionation of Crude Oil Select the correct ranking of the following alkanes according to the height reached in a fractionating column, highest first: butane, heptadecane, dodecane, ethane, decane Select the correct ranking of the following alkanes according to the height reached in a fractionating column, highest first:
butane, heptadecane, dodecane, ethane, decane
A. ethane > butane > decane > dodecane > heptadecane
B. heptadecane > > dodecane > decane butane > ethane
C. ethane > butane > decane> heptadecane >
D. dodecane butane > ethane > decane > dodecane > heptadecane

Answers

Answer:

A. ethane > butane > decane > dodecane > heptadecane

Explanation:

In fractionating column, crude oil is separated by means of fractional distillation due to the wide range of boiling point of the crude products such as ethane, propane, butane pentane etc.

The product with the least weight rises to top height while the product with highest weight will move down.

For the given hydrocarbon products, the ranking according to their molecular weight, starting with the lighter product to heavier product is

ethane (C2), butane (C4), decane(C10), dodecane (C12), heptadecane(C17).

Thus, the correct ranking, starting with the product that will rise highest is ethane > butane > decane > dodecane > heptadecane

An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder is Pf=1.24 atm. What is the enthalpy change for this process? ΔH =

Answers

Answer:

[tex]\Delta H=-11897J[/tex]

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

[tex]\Delta H=\Delta U+V\Delta P[/tex]

Whereas the change in the internal energy is computed by:

[tex]\Delta U=nCv\Delta T[/tex]

So we compute the initial and final temperatures for one mole of the ideal gas:

[tex]T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}[/tex]

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

[tex]\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J[/tex]

Then, the volume-pressure product in Joules:

[tex]V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J[/tex]

Finally, the change in the enthalpy for the process:

[tex]\Delta H=-7140J-4757J\\\\\Delta H=-11897J[/tex]

Best regards.

The change in enthalpy is 70.42J

Data;

Volume of the gas = 4.86LInitial Pressure = 10.90 atmFinal Pressure = 1.24 atmChange in Enthalpy = ?

Change in Enthalpy

The change of enthalpy is calculated as

[tex]\delta H = \delta V + \delta nRT\\\delta n = 0\\\delta H = \delta U \\[/tex]

The volume change is negligible

The change in enthalpy here is equal to change in internal energy over ΔE

[tex]\delta H = \delta U = nCv\delta T\\\delta H = \frac{3}{2}(nR\delta T)\\\delta H = \frac{3}{2}\{\delta PV)\\ \delta H = \frac{3}{2}[(10.90-1.24)*4.86] \\\delta H = 70.42J[/tex]

The change in enthalpy is 70.42J

Learn more on change in enthalpy here;

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When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation

Answers

All done for you no worries

When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. This reaction describes a non-Markovnikov orientation.

In the reaction between an unsymmetrical alkene (such as propene) and N-bromosuccinimide (NBS) in the presence of aqueous dimethyl sulfoxide (DMSO), the major product is formed with the bromine atom bonded to the less highly substituted carbon atom of the alkene.

In Markovnikov's addition, the major product is formed by adding the electrophile (in this case, the bromine atom) to the carbon atom with more hydrogen atoms bonded to it. However, the given reaction exhibits non-Markovnikov selectivity, as the bromine atom adds to the less substituted carbon atom.

This non-Markovnikov selectivity can be attributed to the presence of DMSO, which acts as a polar solvent and helps generate a bromine radical (Br•). The radical intermediate can then undergo reaction with the alkene, leading to the observed regioselectivity where the bromine atom adds to the less substituted carbon. This process is known as a radical addition reaction.

Hence, the reaction demonstrates a non-Markovnikov orientation due to the addition of the bromine atom to the less highly substituted carbon atom of the propene molecule.

Learn more about the Markovnikov rule here:

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The aluminum in a package containing 75 ft2 of kitchen foil weighs approximately 12 ounces. Aluminum has a density of 2.70 g/cm3 . What is the approximate thickness of the aluminum foil in millimeters?(1 ounce = 28.4g)

Answers

Answer:

18130 mm

Explanation:

Now we have a lot of unit conversions to do in order to correctly answer this question. We shall do these conversions gradually.

First we convert the weight in ounce to grams.

If 1 ounce = 28.4g

12 ounces = 12×28.4 = 340.8 g

Next we convert the area of aluminum from ft2 to m2

1ft2= 0.0929 m2

75 ft2= 75 × 0.0929= 6.9675m2

Now density of aluminum= 2.70 gcm-3

Density= mass/volume

But volume= area× thickness

Density= mass/ area × thickness

Density × area × thickness= mass

Thickness= mass/ density × area

Thickness= 340.8g / 2.70gcm-3 × 6.9675m2

Thickness= 340.8/18.8

Thickness= 18.13 m

Since 1000 milimeters make 1 metre

Thickness= 18130 mm

Which best describes thermal energy? It is the difference between internal energies of two or more substances. It is the sum of internal energies of two or more substances. It is the portion of internal energy that can be transferred from one substance to another. It is the portion of potential energy that can be transferred from one substance to another.

Answers

Answer:

It is the portion of internal energy that can be transferred from one substance to another.

Thermal energy is the portion of internal energy that can be transferred from one substance to another.

What is thermal energy?

Thermal energy is the energy an object posses which is as a result of particles movement within it.

It is also the internal energy system in a state of thermodynamic equilibrium which is as a result of its temperature. Thermal energy cannot be concert to useful work easily.

Therefore, thermal energy is the portion of internal energy that can be transferred from one substance to another.

Learn more about thermal energy from the link below.

https://brainly.com/question/19666326

There are __________ moles of N atoms present in a 2.0 g C8H10O2N4.

Answers

Answer:

[tex]n_N=0.041molN[/tex]

Explanation:

Hello,

In this case, for this mole-mass relationship, we are able to compute the moles of nitrogen atoms by firstly obtaining the moles of the given compound, considering its molar mass that is 194 g/mol:

[tex]n_{C_8H_{10}O_2N_4}=2.0gC_8H_{10}O_2N_4*\frac{1molC_8H_{10}O_2N_4}{194gC_8H_{10}O_2N_4} =0.01molC_8H_{10}O_2N_4[/tex]

Then, by knowing that one mole of the given compound has four moles of nitrogen atoms, we apply the following relationship:

[tex]n_N=0.01molC_8H_{10}O_2N_4*\frac{4molN}{1molC_8H_{10}O_2N_4} \\\\n_N=0.041molN[/tex]

Best regards.

Write a Lewis structure for each atom or ion. Draw the particle by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons and non-bonding electrons. Show the charge of the atom. Particles: S2-, Mg, Mg+2, P.

Answers

Answer:

The Lewis structure to this question can be described as follows:

Explanation:

Structure of Lewis for  [tex]S^{2-}[/tex]:  

The maximum number of electrons from valence in [tex]S^{2-}[/tex]  is 8 (6 from S as well as 2 from negative change).  

The valence electrons in the Lewis structure are placed on four sides of the atom.  

Thus the structure of Lewis for [tex]S^{2-}[/tex] is as follows:

[tex]\left[\begin{array}{ccc} &. .&\\: &S&:\\&. .&\end{array}\right] ^{2-}[/tex]

Lewis Mg Structure:  

Complete valence electrons are 2 in Mg.  

The Lewis structure for Mg, therefore, is as follows:

[tex]\ . \\ Mg\\ \ .[/tex]

The Lewis structure for  [tex]Mg^{2+}[/tex]

The maximum valence of electrons   [tex]Mg^{2+}[/tex] in is=  0.

Thus, the structure for   [tex]Mg^{2+}[/tex] is as follows:

 [tex]Mg^{2+}[/tex]

Lewis structure for P :

The maximum number of valence electrons in P is = 5.

Thus, the structure for P is=

[tex]\ \ \ . \\ : P \ : \\[/tex]

If 196L of air at 1.0 atm is compressed to 26000ml, what is the new pressure

Answers

Answer:

7.5 atm

Explanation:

Initial pressure P1 = 1.0 ATM

Initial volume V1= 196 L

Final pressure P2= the unknown

Final volume V2= 26000ml or 26 L

From Boyle's law we have;

P1V1= P2V2

P2= P1V1/V2

P2= 1.0 × 196/26

P2 = 7.5 atm

Therefore, as the air is compressed, the pressure increases to 7.5 atm.

Hot coffee in a mug cools over time and the mug warms up. Which describes the energy in this system? The average kinetic energy of the particles in the mug decreases. The average kinetic energy of the particles in the coffee increases. Thermal energy from the mug is transferred to the coffee. Thermal energy from the coffee is transferred to the mug.

Answers

Answer:

d

Explanation:

Answer:

D

Explanation:

Edge 2021

When 1.550 gg of liquid hexane (C6H14)(C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 ∘C∘C to 38.13 ∘C∘C. Find ΔErxnΔErxn for the reaction in kJ/molkJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kJ/∘CkJ/∘C.

Answers

Answer:

ΔErxn[tex]= -3.90*10^3KJ[/tex]

Explanation:

Given from the question

T1 = 25.87∘C

T2= 38.13∘C.

C= 5.73Kj/C

CHECK THE ATTACHMENT FOR DETAILED EXPLATION

Ba(OH)2:_______.
A. 1 barium atom, 1 oxygen atom and 1 hydrogen atom.
B. 1 barium atom, 1 oxygen atom and 2 hydrogen atoms.
C. 1 barium atom, 2 oxygen atoms and 2 hydrogen atoms.
D. 1 barium atom, 2 oxygen atoms and 1 hydrogen atom.

Answers

Answer: D

Explanation: Expand this (OH)2 you will get 2O, 2H

Hence 1Ba, 2O, 2H

Answer:

B. 1 barium atom, 1 oxygen atom and 2 hydrogen atoms.

which best describes a mixture.

A it has a single composition and it has a set of characteristics

B it can have different compositions but it has a set of charactaristics that does not change

C it has a single composition but it has a set of characteristics that does change

D it can have different compositions and it has a set of characteristics that does change​

Answers

Answer:

B) It can have different compositions, but it has a set of characteristics that does not change.

Explanation:

On e d g e n u i t y

I believe the answer is d lmk if  im  wrong or right

reasons for good care on computer​

Answers

answer

1)maximise your software efficiency

2)Prevention against viruses and malware

3)Early detection of problematic issues

4)prevent data loss

5)Speed up your computer

What would form a solution?
O A. Mixing two insoluble substances
O B. Mixing a solute and a solvent
O C. Mixing a solute and a precipitate
O D. Mixing two solutes together

Answers

The correct answer is B

One of the reagents below gives predominantly 1,2 addition (direct addition) while the other gives predominantly 1,4 addition (conjugate addition). a) Which major organic product is the result of 1,2 addition? ---Select--- b) Draw the skeletal structure of major organic product A

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The correct option is reagent B

b

The  skeletal structure of major organic product A is shown on the third uploaded image

Explanation:

The mechanism of the reaction for A and  B  are shown on the second the second reaction and looking at this we can see that the reagent that  predominately gives 1,2 addition is reagent B  

Ni
Express your answer in condensed form in the order of orbital filling as a string without blank space between orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2.

Answers

Answer:

[Ar]3d^84s^2

Explanation:

From the question given, we are asked to write the condensed form of electronic configuration of nickel, Ni.

To do this, we simply write the symbol of the noble gas element before Ni in a squared bracket followed by the remaining electrons to make up the atomic number of Ni.

This is illustrated below:

The atomic number of Ni is 28.

The noble gas before Ni is Argon, Ar.

Therefore, the condensed electronic configuration of Ni is written as:

Ni(28) => [Ar]3d^84s^2

Answer:

[Ar] 4s^23d^8

Explanation:

The osmotic pressure exerted by a solution is equal to the molarity multiplied by the absolute temperature and the gas constant . Suppose the osmotic pressure of a certain solution is measured to be at an absolute temperature o of 312. K. Write an equation that will let you calculate the molarity c of this solution.

Answers

Answer:

Explanation:

From the question, osmotic pressure exerted by a solution is equal to the MOLARITY multiplied by the absolute TEMPERATURE and the GAS CONSTANT r.

Let P = osmotic pressure,

C = molarity, then

T = absolute temperature

r=gas constant

The Osmotic pressure Equation exerted by a solution [tex]P=C*T*r[/tex]

[tex]P=CTr[/tex]

Then it was required in the question to write an equation that will let you calculate the molarity c of this solution, and this equation should contain ONLY symbols

C= molarity of the solution

P=osmotic pressure

r = gas constant

T= absolute temperature

[tex]C=P/(rT)[/tex]

The equation that will let us calculate the molarity c of this solution = [tex]C=P/(rT)[/tex]

Which of the compounds below are amines?
1. H4C-NH-CH3
2. H3C-NH-C-CH3
H2C-CH3
1 +
H3C-CH2-N-CH3
CH3
N
3. H
4.

Answers

Answer:

1. H4C-NH-CH3

2. H3C-NH-C-CH3

H2C-CH3

1 +

H3C-CH2-N-CH3

CH3

N

3. H

4.

.

.

.

.

.

.

A water tank measures 24in.×48in.×12in. Find the capacity of the water tank in cubic feet. Do not include units in your answer.

Answers

Answer: 8 (feet)

Explanation:

24 inches = 2 feet

48 inches = 4 feet

12 inches = 1 foot

To find volume you do Base * Width * Height

2*4*1 = 8

Hope this helps!

The correct answer is  8 (feet).

How to calculate ?

24 inches = 2 feet48 inches = 4 feet12 inches = 1 footTo find volume the method is  Base * Width * Height

Therefore, 2*4*1 = 8

Hence, the capacity of the water tank in cubic feet is 8 feet.

learn more about capacity below,

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A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Litre volumetric flask. What is the molarity (M) of the resulting solution relative to the nitrate ion, NO3 1-

Answers

Answer:

[tex]M=0.213M[/tex]

Explanation:

Hello,

In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:

[tex]n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-[/tex]

[tex]n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-[/tex]

[tex]n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-[/tex]

We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:

[tex]n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-[/tex]

Finally, we compute the molarity:

[tex]M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M[/tex]

Regards.

The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that also contains 300mM imidazole. Describe how the instructor made the 0.25L elution buffer for all the students this week given 500ml of 1M of Tris (121.1 g/mole) (pH8.0), 750ml of 5M NaCl (MW

Answers

Answer:

Explanation:

From the given information ;the objective is to determine how the instructor made the 0.25L elution buffer

0.25 L elution buffer = 250 mL elution butter

The breaking buffer that we use this week contains

10mM Tris    =   0.01 M

150mM NaCl  =   0.15 M

300mM imidazole.  = 0.3 M

The stock concentration  of Tris in 1M

Therefore ; by using the formula: [tex]M_1V_1 = M_2 V_2[/tex]; we can determine the volume in the preparation; so;

[tex]1*V_1 = 0.0 1 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0.0 1 \ M * 250 \ mL}{1 }[/tex]

[tex]V_1 = 2.5 \ mL[/tex]

In NaCl, The amount of stock concentration is 5 M

so; using the same formula; we have:

[tex]5*V_1 = 0.15 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0. 15 \ M * 250 \ mL}{5 }[/tex]

[tex]V_1 = 7.5 \ mL[/tex]

From Imidazole ; the amount of stock concentration is

[tex]1*V_1 = 0.3 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0. 3 \ M * 250 \ mL}{1 }[/tex]

[tex]V_1 = 75 \ mL[/tex]

Thus; we can have a table as shown as :

Stock concentration        volume to be added        Final concentration

1 M of Tris                              2.5 mL                            10 mM

5 M of  NaCl                          7.5 mL                             150 mM

1 M of Imidazole                    75  mL                            300  mM

In conclusion. the addition of all the volume make up the 250 mL elution buffer that is equivalent to 0.25 L.

which statements describe how chemical formulas, such as H2O, represent compounds?

Answers

Answer:

2 Hydrogen One oxygen

Explanation:

Calculate the percentage of the void space out of the total volume occupied by 1 mole of water molecules. The density of water is assumed to be 1.0 g/mL that is 1.0 g/cm3. The molar mass of water is 18.0 g/mol. The atomic radius of hydrogen is 37 pm and of oxygen is 73 pm. The formula for the volume of a sphere is 4/3(r3

Answers

Answer:

The percentage of the void space out of the total volume occupied is 93.11%

Explanation:

A mole of water contains 2 atoms of hydrogen and 1 atom of oxygen.

To calculate the volume of a mole of water, we calculate 2 times the volume of the hydrogen atom and 1 times the volume of the oxygen atom

Let's calculate this one after the other.

For the hydrogen, formula for the volume will be

[tex]V_{hydrogen[/tex] = 2 × 4/3 × π × [tex]r_{H}^{3}[/tex]

where [tex]r_{H}^{3}[/tex] = 37 pm which is read as 37 picometer (1 picometer = 10^-12 m) = 37 × [tex]10^{-12}[/tex] meters

Volume of the hydrogen = 8/3 × (37 × 10^-12)^3 = 4.05 * 10^-31

we multiply this by the avogadro's number = 6.02 * 10^23

= 4.05 * 10^-31 * 6.02 * 10^23 = 2.6 * 10^-8 m^3

We do same for thr oxygen, but this time we do not multiply the volume of the oxygen by 2 as we have only one atom of oxygen

Volume of oxygen = 4/3 * π * (73 * 10^-12) ^3 * avogadro's number = 9.81 * 10^-7 m^3

adding both volumes together, we have 1.24 * 10^-6 m^3 or simply 1.24 ml ( 0.01 m = 1 ml)

Dividing the molar mass of one mole of water by its density, we can get the volume of 1 mole of water

= (18g/mol)/(1 g/ml) = 18 ml/mol

Now we proceed to calculate the volume of void = Total volume - volume of molecule = 18 - 1.24 = 16.76 ml

Now, the percentage of void = volume of void/total volume * 100%

= 16.76/18 * 100% = 93.11%

What is the conjugate acid in the following equation hbr + H2O yields h30 positive + BR negative

Answers

Answer:

HBr + H2O = H3O+ + Br-

So our conjugate acid is the H3O+ to H2O

Explanation:

A conjugate acid of a base results when the base accepts a proton.

Consider ammonia reacting with water to form an equilibrium with ammonium ions and hydroxide ions:

NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)

Ammonium, NH4+, acts as a conjugate acid to ammonia, NH3.

At 25.0 °C the Henry's Law constant for sulfur hexafluoride (SP) gas in water is 2.4x 10 M/atm Calculate the mass in grams of SFo, gas that can be dissolved in S25. ml. of water at 25.0 C and a SF, partial pressure of 1.90 atm Be sure your answer has the correct number of significant digits.

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The mass is [tex]m = 0.0349 \ g[/tex]

Explanation:

From the question we are told that

   The Henry's Law constant is  [tex]k = 2.4 *10^{10} M/atm[/tex]

   The volume of water is [tex]V = 525 \ ml = 0.525 \ L[/tex]

   The partial pressure is  [tex]P = 1.90 \ atm[/tex]

   The temperature is [tex]T = 25 ^oC[/tex]

Henry's  law is mathematically represented as

       [tex]C = P * k[/tex]

Where C is  the concentration of sulfur hexafluoride(SP)

substituting value

        [tex]C = 1.90 * 2.4*10^{-4}[/tex]

        [tex]C = 4.56*10^{-4} \ M[/tex]

The number of moles of  SP is mathematically represented as

        [tex]n = C * V[/tex]

substituting value

       [tex]n = 0.525 * 4.56*10^{-4}[/tex]

       [tex]n = 2.39 *10^{-4} \ moles[/tex]

The mass of SP that dissolved is

          [tex]m = n * Z[/tex]

Where Z is the molar mass of SP which has a constant value of

           [tex]Z = 146 g/mole[/tex]

So

         [tex]m = 2.394*10^{-4} * 146[/tex]

         [tex]m = 0.0349 \ g[/tex]

Show work plzzz
Unknown Metal Bar #8
Mass of Unknown Metal bar 11.3g
Length of bar 13.90cm
Width of bar 2.9cm
Thickness of bar 0.081cm

1. Calculate the volume of the bar:

2. Calculate the (experimental) density of the bar:

3. Based on the provided list of (true) densities, what is the possible identity of the Unknown metal?

4. What is the percent difference between the true density of your metal and the calculated density?
= | − | ∗ 100%

Answers

Answer:

1= Volume

= Length x breath x height

= 13.90 x 2.9 x 0.081

=3.26511

2= Density = Mass ÷ volume

= 11.3 ÷ 3.26511

= 3.461 (3d.p)

idk the rest because you haven't shown a picture of the rest

Answer:

1. 3.3 cm³; 2. 3.5 g/cm³; 3. barium; 4. 4%

Explanation:

Experimental data:

Mass          = 11.3    g

Length      = 13.90 cm

Width        =  2.9    cm

Thickness = 0.081 cm

Calculations:

1. Volume of bar

V = lwh = 13.90 cm × 2.9 cm × 0.081 cm = 3.3 cm³

2. Experimental density

[tex]\text{Density} = \dfrac{m}{V} = \dfrac{\text{11.3 g}}{\text{3.27 cm}^{3}} = \textbf{3.5 cm}^{\mathbf{3}}[/tex]

3. Identity of metal

The three most likely metals are scandium (3.00 g/cm³), barium (3.59 g/cm³), and yttrium (4.47 g/cm³)

The metal is probably barium.

4. Percent difference

[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{ True - Calculated}\lvert}{ \text{True}} \times 100 \,\%\\\\& = & \dfrac{\lvert 3.59 - 3.5\lvert}{3.59} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.1\lvert}{3.59} \times 100 \, \%\\ \\& = & 0.04 \times 100 \, \%\\& = & \mathbf{4 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{4 \, \%} }$}[/tex]

Compare the conjugate bases of these three acids. Acid 1: hypochlorous acid , HClO Acid 2: phosphoric acid , H3PO4 Acid 3: hydrogen sulfide , HS- What is the formula for the weakest conjugate base ?

Answers

Answer:

The weakest conjugate is HClO-.

Explanation:

As a general rule, the stronger the Bronsted-Lowry acid, the weaker its conjugate base, and vice versa.  

Acid 1: HClO is a strong acid, hence its conjugate base would be weak

Acid 2: H3PO4 is a weak acid, hence its conjugate base would be strong

Acid 3: hydrogen sulphide is also a moderately weak acid with a moderately strong conjugate base.

In order of increasing strengths:

HClO < H2S < H3PO4

what is the correct ionic equation, including all coefficients, charges, and phases for the following sets of reactants? Assume that the contribution of protons from H2SO4 is near 100%.

Ba(OH)2(aq)+H2SO4(aq) —>


help, I have no clue

Answers

Answer:

Ba(OH)2(aq)+H2SO4(aq) gives us 2BaH+H2O

Explanation:

Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a copper(II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of . Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to significant digits.

Answers

Answer:

Concentration of Copper (II) Sulfate in the original sample in mol/L = 0.0035 M

Concentration of Copper (II) Sulfate in the original sample in g/L = 0.56 g/L

Explanation:

Complete Question

Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)

Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.

Solution

Noting that the precipitate is Copper as it is the only solid by-product of this reaction.

89 mg of Copper is produced from this reaction.

We convert this into number of moles for further stoichiometric calculations

Mass of Copper = 89 mg = 0.089 g

Molar mass of Copper = 63.546 amu

Number of moles of Copper produced from the reaction = (0.089/63.546) = 0.0014005602 = 0.001401 mole

From the stoichiometric balance of the reaction,

1 mole of Copper is produced from 1 mole of Copper (II) Sulfate

0.001401 mole of Copper will be produced similarly from 0.001401 mole of Copper (II) Sulfate.

Number of moles of Copper (II) Sulfate in the original sample = 0.001401 mole

Concentration of Copper (II) Sulfate in the original sample in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = 0.001401 mole

Volume in L = (400/1000) = 0.4 L

Concentration of Copper (II) Sulfate in the original sample in mol/L = (0.001401/0.4) = 0.0035025 mol/L = 0.0035 mol/L to 2 s.f.

Concentration in g/L = (Concentration in mol/L) × (Molar Mass)

Concentration in mol/L = 0.0035025 M

Molar mass of Copper (II) Sulfate = 159.609 g/mol

Concentration of Copper (II) Sulfate in the original sample in g/L = 0.0035025 × 159.609 = 0.559 g/L = 0.56 g/L to 2 s.f

Hope this Helps!!!!

The concentration of the original copper solution is 0.035 M.

The equation of the reaction is;

Fe(s) + CuSO4(aq) -------> FeSO4(aq) + Cu(s)

Number of moles of copper obtained = 89 × 10^-3g/63.5 = 0.0014 moles

Since the reaction is 1:1, the number of moles of copper sulfate that reacted is c.

From the question, we are told that the volume of solution is 400.mL or 0.04L.

Hence, the concentration of the solution is; number of moles /volume

=  0.0014 moles/0.04L = 0.035 M

Learn more: https://brainly.com/question/9352088

Missing parts;

Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.

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