The solubility of cadmium oxalate, , in 0.150 M ammonia is mol/L. What is the oxalate ion concentration in the saturated solution? If the solubility product constant for cadmium oxalate is , what must be the cadmium ion concentration in the solution? Now, calculate the formation constant for the complex ion

Answer: 125 g

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles[/tex]

The balanced reaction is:

[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]

According to stoichiometry :

1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]

Thus 1.30 moles of [tex]B_2H_6[/tex] will require=[tex]\frac{3}{1}\times 1.30=3.90moles[/tex]  of [tex]O_2[/tex]

Mass of [tex]O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g[/tex]

Thus 125 g of [tex]O_2[/tex] will be needed to burn 36.1 g of [tex]B_2H_6[/tex]

Answer: HA has lowest pH and it is the most acidic as compared to the rest of given acids.

Explanation:

We know that relation between [tex]pK_a[/tex] and [tex]K_a[/tex] is as follows.

       [tex]pK_a = -log K_a[/tex]

This means that more is the value of [tex]K_a[/tex], smaller will be the [tex]pK_a[/tex]. Also, more is the value of [tex]K_a[/tex] smaller will be the pH of a solution.

As, larger is the value of [tex]K_a[/tex] more negative will be the [tex]pK_a[/tex] value. Hence, stronger will be the acid.

In the given options, HA has the smallest [tex]pK_a[/tex] value.

Therefore, we can conclude that HA has lowest pH and it is the most acidic as compared to the rest of given acids.

Answer:

pH = - log [concentration]

pH = - log (0.0043M)

pH = 2.37

Answer: Fossil fuels

Explanation:

Fossil fuels such as petroleum, oil,  and natural gas, are non-renewable energy resources which are formed from the remains of  prehistoric ancient  plants and animals beneath layers of rock of the earth surface.

By analyzing and studying fossil fuels using Radiocarbon analyses by archaeologists, earth scientists  etc, Information about the history of the earth can be obtained from the decomposition of dead organisms present in fossil fuels.

Answer:

  1703 mmHg

Explanation:

Volume and pressure are presumed to be inversely proportional. Hence a change in volume by a factor of 125/325 = 5/13 is expected to change the pressure by a factor of 13/5:

  (13/5)(655 mmHg) = 1703 mmHg

Answer:

The answer is Hg.

Explanation:

Symbol for Mercury is Hg.

Answer:

b. the heating of 10 g of liquid water from 0°C to 100°C.

Explanation:

Hello,

In this case, we must notice a., c. and d. processes are not actually absorbing heat but releasing it since cooling, freezing and condensation are processes with negative heat sign since matter changes from a state of more energy to a state of less energy. We can prove this by realizing that freezing enthalpy of water is -6.00 kJ/mol, condensation enthalpy of eater is -40.8 kJ/mol and a change of temperature from 100 °C to 0 °C is negative.

In such a way, the only process absorbing heat is b. the heating of 10 g of liquid water from 0°C to 100°C since energy must be added to the system, or absorbed by it in order to attain the heating.

Regards.

The process having the greatest amount of heat is:

b. the heating of 10 g of liquid water from 0°C to 100°C.

The options a., c. and d. processes are not actually absorbing heat but releasing it since cooling, freezing and condensation are processes with negative heat sign since matter changes from a state of more energy to a state of less energy.

The freezing enthalpy of water is -6.00 kJ/mol, condensation enthalpy of eater is -40.8 kJ/mol and a change of temperature from 100 °C to 0 °C is negative.

So out of all the options, only process at b is a heating process thus it will absorb greatest amount of heat.

Find more information about Heat here:

brainly.com/question/13439286

Answer:

a) ortho-para isomers predominates

b) 3-nitrobenzoic acid ( meta isomer predominates)

c) 3-bromo nitrobenzene ( meta isomer predominates)

d) the ortho- para isomers predominates

Explanation:

a) Toluene contains -CH3 which is an ortho- para- director hence the major product of the sulphonation of toluene should be the ortho- para isomers.

b) The major product of the nitration of benzoic acid is 3-nitrobenzoic acid. This is an electrophilic substitution in which the meta isomer predominates.

c) The meta isomer predominates giving 3-bromo nitrobenzene as the major product.

d) The isopropyl group is an ortho- para director hence the ortho- para isomers predominates .

Answer:

The correct answer is 15.80 grams.

Explanation:

The reaction taking place in the given question,  

Pb(CH₃COO)₂ + Na₂SO₄ ⇒ PbSO₄ + 2NaCH₃COO

The number of moles can be calculated by using the formula,  

n = weight / molecular mass

Based on the given question, the weight of lead (II) acetate is 23.81 grams and the weight of sodium sulfate is 7.410 grams.  

The number of moles of Pb(CH₃COO)₂ is,  

n = 23.81 g / 325.29 g/mol = 0.0732 moles

The number of moles of Na₂SO₄ is,  

n = 7.410 g / 142.04 g/mol = 0.0521 moles

As one mole of lead (II) acetate needs one mole of sodium sulfate. Therefore, 0.0732 moles of lead (II) acetate needs 0.0732 moles of sodium sulfate.  

However, as sodium sulfate is less, that is, 0.0521, therefore, Na₂SO₄ is a limiting reactant.  

One mole of sodium sulfate produces one mole of PbSO₄. So, 0.0521 moles of Na₂SO₄ produces 0.0521 moles of PbSO₄.  

Now the mass of PbSO₄ is,  

mass = moles × molecular mass

mass = 0.0521 × 303.26 g/mol

mass = 15.80 grams.  

Answer:

Translation energy of 1 mole of H2 molecules = KE x Avogadros number

[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]

Explanation:

Photon wavelength [tex]= 58.4 nm = 58.4 * 10^{-9} m[/tex]

Photon momentum = h/wavelength

[tex]= (6.626 * 10^{-34})/(58.4 * 10^{-9})\\\\ = 1.1346 * 10^{-26} \ kg.m/s[/tex]

Mass of H2 molecule m = molar mass/Avogadros number

[tex]= (2.016)/(6.022 * 10^{23})\\\\= 3.3477 * 10^{-24} \ g = 3.3477 * 10^{-27} \ kg[/tex]

Since momentum is conserved:

Photon momentum = H2 molecule momentum = mass x velocity of H2

[tex]1.1346 * 10^{-26} = 3.3477 * 10^{-27} * v[/tex]

velocity [tex]v = 3.389 m/s = 3.39 m/s[/tex]

Translation energy of 1 H2 molecule = kinectic energy (KE) = (1/2)mv^2

[tex]= 1/2 * 3.3477 * 10^{-27} * 3.389^2\\\\= 1.923 * 10^{-26} J[/tex]

Translation energy of 1 mole of H2 molecules = KE x Avogadros number

[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]

Answer: The solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.

Explanation:

The given data is as follows.

Volume of sample water = 46 ml

Temperature = [tex]21^oC[/tex]

After vaporization, washes and then drying the weight of mineral X = 0.87 g

This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.

          1 ml of water = [tex]\frac{0.87 g}{46.0 ml}[/tex]

                                = [tex]1.891 \times 10^{-2}[/tex] g/ml

Therefore, we can conclude that solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.

Answer:

[CH₂Cl₂] = 7.07x10⁻² M

[CH₄] = 0.319 M

[CCl₄] = 0.164 M  

Explanation:

The equilibrium reaction is the following:

2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)  

The equilibrium constant of the above reaction is:

[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]

When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:

[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]

[tex]C_{CH_{4}} = 0.328 M[/tex]      

Now, the concentrations at the equilibrium are:

2CH₂Cl₂(g)   ⇄   CH₄(g)  +  CCl₄(g)

5.35x10⁻² - 2x   0.328 + x   0.173 + x    

[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]

[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]

Solving the above equation for x:  

x₁ = 0.076 and x₂ = -0.0086

Hence, the concentration of the three gases once equilibrium has been reestablished is:

[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M

[CH₄] = 0.328 + (-0.0086) = 0.319 M

[CCl₄] = 0.173 + (-0.0086) = 0.164 M  

We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.

I hope it helps you!

Answer:

[tex]M=0.0637M[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)[/tex]

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

[tex]n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2} *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}[/tex]

Finally, the resulting molarity in 30.8 mL (0.0308 L):

[tex]M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M[/tex]

Regards.

Answer: The oxygen's oxidation number is -2.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

In [tex]Fe_2O_3[/tex], Fe is having an oxidation state of +3 called as  cation and oxygen  is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]Fe_2O_3[/tex]

The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.

Answer:

The correct IUPAC name for the organic reactant is :

d) 3-methylbutene

Explanation:

Firstly the  missing diagram is attached in the diagram below.

The objective of this question is to draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.

From the image attached below; we would see the reaction that occurs between the alkene and the HBr (hydrobromic acid). What really occur in the reaction is that; in the presence of HBr with an alkene compound a secondary 2° carbocation is usually formed. This secondary 2° carbocation formed is usually unstable, so what we called an hydride shift occurs (Markovnikov's product) here to form a stable tertiary 3° carbocation.

The correct IUPAC name for the organic reactant is : 3-methylbutene

Answer:

10 g of CO2

Explanation:

Equation of the reaction:

CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2

Fom the above balanced equation,

1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2

Molar mass of Octane = 114 g/mol

Molar mass of oxygen gas = 32 g/mol

Molar mass of CO2 = 44 g/mol

Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.

From the given mass of reactants;

3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.

Therefore oxygen is the limiting reactant.

15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.

Mass of CO2 produced will be

(352 * 15.6)/544 = 10 g of CO2

Answer:

0.031 m

HgO(s) ⇒ Hg(l) + 1/2 O₂(g)

Chemical change

Element

Explanation:

A 75 gram solid cube of mercury (II) oxide has a density of 2.4 × 10³ kg/m³.  What is the length of one side of the cube in cm?

Step 1: Convert the mass to kilograms

We will use the relationship 1 kg = 1,000 g.

[tex]75g \times \frac{1kg}{1,000g} = 0.075kg[/tex]

Step 2: Calculate the volume (V) of the cube

[tex]0.075kg \times \frac{1m^{3} }{2.4 \times 10^{3} kg} = 3.1 \times 10^{-5} m^{3}[/tex]

Step 3: Calculate the length (l) of one side of the cube

We will use the following expression.

[tex]V = l^{3} \\l = \sqrt[3]{V} = \sqrt[3]{3.1 \times 10^{-5} m^{3} }=0.031m[/tex]

The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?

The balanced chemical equation is:

HgO(s) ⇒ Hg(l) + 1/2 O₂(g)

This is a chemical change because new substances are formed.

The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?

The liquid gray substance is Hg(l), which is an element because it is formed by just one kind of atoms.

Answer:

[tex]2.11\ * 10^{-2}[/tex]  is the correct answer to the given question.

Explanation:

Given k=6.40 x 10-3 min-1.

According to the first order reaction .

The concentration of time can be written as

[tex][\ A\ ]\ = \ [\ A_{0}\ ] * e \ ^\ {-kt}[/tex]

Here [tex][\ A\ ]_{0}[/tex] = Initial concentration.

So  [tex][\ A\ ]_{0}= 0.0314 M[/tex]

Putting this value into the above equation.

[tex]0.0314 \ *\ e^{6.40 x 10^{-3} \ * \ 62.0 }[/tex]

=0.211 M

This can be written as

[tex]=\ 2.11 *\ 10^{-2}[/tex]

Answer:

The correct answer is 66.35 kilograms.

Explanation:

Based on the data given in the question, the energy consumed by the body of a human being is 50%. Based on the given data, the energy consumed in a day is 8000 kJ, 50 percent is the energy transfer efficiency. Thus, the consumption of total energy is 4000 kJ, and for the transformation of ADP to ATP, the energy involved is 30.6 kJ per mole.  

Hence, the total ATP produced in the process is,  

ATP = 4000 kJ / 30.6 kJ/mol

= 130.7189 mol.  

Thus, with the energy transfer efficiency of 50 percent, the total moles of ATP produced is 130.7 mol.  

The mass of ATP can be calculated by using the formula,  

moles = mass/molecular mass

The molecular mass of ATP is 507.18 g per mol

Now by putting the values we get,  

mass of ATP = 130.7189 mol * 507.18 g/mol

= 66298.011 g or 66.298 kg

It is mentioned that human comprise 50 g of ATP or 0.05 kg of ATP. Therefore, the sum of the available ATP will be.  

= Total production of ATP + Total ATP available

= 66.298 kg + 0.05 kg

= 66.348 kg

Hence, the sum of the ATP that is turned over by a resting human in a day is 66.35 kg.  

Answer:

Mass of Ga = 0.73694 gram

Explanation:

Given:

Current = 0.850 A

Time = 60 minutes

Find:

Amount of gas deposit.

Computation:

Total charge = Current × Time in second

Total charge = 0.850 × 60 × 60

Total charge = 3,060 C

Mole of electron = Total charge / Faraday constant         [Faraday constant = 96,485.3329]

Mole of electron = 3,060 / 96,485.3329

Mole of electron = 0.0317146

Moles of Ga = 1/3 [Mole of electron]

Moles of Ga = 1/3 [0.0317146]

Moles of Ga = 0.01057

Mass of Ga = molar mass × Moles of Ga

Mass of Ga = 69.72 × 0.01057

Mass of Ga = 0.73694 gram

Answer:

I think that the fire will continue burning, because the oil and water don't mix and the water is heavier (denser) than oil, so the oil will go up and the fire with it. That's why because the gas station have sand instead of water

Water is heavier than oil. Because oil is lighter and immiscible with water, it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire. As a result, the fire won't be put out.

A small amount of water will instantly sink to the bottom of a pan or deep fryer filled with hot, burning oil and explode there. The Scientific American claims that the characteristics of oils explain why they do not mix with water.

Oil or petroleum-related fires cannot be put out with water. Water sinks below the oil because it is heavier than oil and does not float, allowing the fire to continue to burn. Oil and petroleum fires can be put out with fire extinguishers or sand.

The temperature of the burning substance is lowered by water. The fire goes out when the temperature drops below the burning substance's ignition temperature. Here, the water serves as an acclimatizer.

Thus,  it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire.

To learn more about the oil fire, follow the link;

https://brainly.com/question/15173100

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Answer:

Explanation:

In cobalt - 65 ,

no of protons is 27 ( p )

no of neutron = 65 - 27 ( n )

= 38

n / p ratio

=  38 / 27

= 1.41

If case of emission of alpha particle

no of proton p = 27 - 2 = 25

no of neutrons = 38 - 2 = 36

n / p ratio = 36 / 25

= 1.44

So it increases

In case of emission of beta particle

No of neutron n = 38 - 1 = 37

No of proton = 27 + 1 = 28

n / p ratio = 37 / 28

= 1.32

Hence ratio decreases.

Hence beta ray decay will result in decrease in n / p ratio.

Answer:

pH = 9.08

Explanation:

Quinine, C₂₀H₂₄O₂N₂, Q, is a weak base that, in water, has as equilibrium:

Q + H₂O ⇄ QH⁺ + OH⁻

Where pKb is 5.10

Using H-H equation for weak bases:

pOH = pKb + log₁₀ [QH⁺] / [Q]

The reaction of quinine with HCl is:

Q + HCl → QH⁺ + Cl⁻

Initial moles of quinine are 0.125 moles and moles added of HCl are:

0.05000L × (1.00mol / L) = 0.05000moles.

That means after the addition of 50.00mL of the HCl solution, moles of Q and QH⁺ are:

Q = 0.125mol - 0.050mol = 0.075 moles

QH⁺ = 0.050 moles

Replacing in H-H equation:

pOH = 5.10 + log₁₀ [0.050] / [0.075]

pOH = 4.92

As pH = 14 - pOJ

Answer: D

Explanation: The elements in two compunds switch places

Answer:

D

Explanation:

trust me its correct i think