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2 years ago

15

1. How many grams of chromium metal are plated out when a constant current of 8.00 A is passed through an aqueous solution conta

ining Cr3+ ions for 40.0 minutes?

1 answer:

Paul [167]2 years ago

6 0

ion with charge 3⁺, like Cr³⁺ needs 3 Faradays to plate 1 mole

Amount of electric charge per mole (faraday's constant) = 96500 C mol¹⁻

Time = 40min = 2400 sec<span>

you have coulombs = amps x sec = 8 x 2400 coulombs </span>

molar mass of Cr = 51.9961g<span>

96500 x 3 coulombs plates 51.9961g Cr
so 8 x 2400 coulombs gives,</span>

<span>51.9961/(96500 x 3) x 8 x 2400 g Cr = 3.45g Cr </span>

<span>so, 3.45g chromium metal is plated out.</span>

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Gasoline is a mixture of hydrocarbons, a major component of which is octane, CH3CH2CH2CH2CH2CH2CH2CH3. Octane has a vapor pressu

Nitella [24]

Answer:

$\Delta \:H_{vap}=40383.88\ J/mol$

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

$\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c$

Where,

P is the vapor pressure

ΔHvap is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

$\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)$

Given:

$P_1$ = 13.95 torr

$P_2$ = 144.78 torr

$T_1$ = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

$T_1$ = 298.15 K

$T_2$ = 75°C = 348.15 K

So,

$\ln \:\left(\:\frac{13.95}{144.78}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{348.15}-\:\frac{1}{298.15}\:\right)$

$\Delta \:H_{vap}=\ln \left(\frac{13.95}{144.78}\right)\frac{8.314}{\left(\frac{1}{348.15}-\frac{1}{298.15}\right)}$

$\Delta \:H_{vap}=\frac{8.314}{\frac{1}{348.15}-\frac{1}{298.15}}\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)$

$\Delta \:H_{vap}=\left(-\frac{863000.86966\dots }{50}\right)\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)$

$\Delta \:H_{vap}=40383.88\ J/mol$

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2 years ago

Adding one proton to the nucleus of an atom

Sonbull [250]

Adding or removing protons from the nucleus changes the charge of the nucleus and changes that atom's atomic number. So, adding or removing protons from the nucleus changes what element that atom is! For example, adding a proton to the nucleus of an atom of hydrogen creates an atom of helium.

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2 years ago

Olympic cyclist fill their tires with helium to make them lighter. Calculate the mass of air in an air filled tire and the mass

inn [45]

<u>Answer:</u> The mass difference between the two is 7.38 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure = 125 psi = 8.50 atm (Conversion factor: 1 atm = 14.7 psi)

V = Volume = 855 mL = 0.855 L (Conversion factor: 1 L = 1000 mL)

T = Temperature = $25^oC=[25+273]K=298K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles = ?

Putting values in above equation, we get:

$8.50atm\times 0.855L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{8.50\times 0.855}{0.0821\times 298}=0.297mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For air:</u>

Moles of air = 0.297 moles

Average molar mass of air = 28.8 g/mol

Putting values in equation 1, we get:

$0.297mol=\frac{\text{Mass of air}}{28.8g/mol}\\\\\text{Mass of air}=(0.297mol\times 28.8g/mol)=8.56g$

Mass of air, $m_1$ = 8.56 g

<u>For helium gas:</u>

Moles of helium = 0.297 moles

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$0.297mol=\frac{\text{Mass of helium}}{4g/mol}\\\\\text{Mass of helium}=(0.297mol\times 4g/mol)=1.18g$

Mass of helium, $m_2$ = 1.18 g

Calculating the mass difference between the two:

$\Delta m=m_1-m_2$

$\Delta m=(8.56-1.18)g=7.38g$

Hence, the mass difference between the two is 7.38 grams.

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2 years ago

Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2

maxonik [38]

Answer:

Equilibrium constant of the given reaction is $1.3\times 10^{-29}$

Explanation:

$NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr$ .... $(K_{p})_{1}=5.3$

$2NO\rightleftharpoons N_{2}+O_{2}$ .... $(K_{p})_{2}=2.1\times 10^{30}$

The given reaction can be written as summation of the following reaction-

$2NO+Br_{2}\rightleftharpoons 2NOBr$

$N_{2}+O_{2}\rightleftharpoons 2NO$

......................................................................................

$N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr$

Equilibrium constant of this reaction is given as-

$\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}$

$=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})$

$=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}$

$=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}$

7 0

2 years ago

Read 2 more answers

Be sure to answer all parts. There are three different dichloroethylenes (molecular formula C2H2Cl2), which we can designate X,

sertanlavr [38]

Answer:

Explanation:Since the compound X has no net-dipole moment so we can ascertain that this compound is not associated with any polarity.

hence the compound must be overall non-polar. The net dipole moment of compound is zero means that the vector sum of individual dipoles are zero and hence the two individual bond dipoles associated with C-Cl bond must be oriented in the opposite directions with respect to each other.]

So we can propose that compound X must be trans alkene as only in trans compounds the individual bond dipoles cancel each other.

If one isomer of the alkene is trans then the other two isomers may be cis .

Since the two alkenes give the same molecular formula on hydrogenation which means they are quite similar and only slightly different.

The two possibility of cis structures are possible:

in the first way it is possible the one carbon has two chlorine substituents and the carbon has two hydrogens.

Or the other way could be that two chlorine atoms are present on the two carbon atoms in cis manner that is on the same side and two hydrogens are also present on the different carbon atoms in the same manner.

Kindly refer the attachments for the structure of compounds:

7 0

2 years ago