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3 years ago

A) Balance the equation below.

Chemistry

1 answer:

Advocard [28]3 years ago

3 0

Answer:

254.7 g

Explanation:

Hg(ONC)2 -----------> Hg + N2 + 2CO

In every problem of stoichiometry, we must commence with writing the balanced reaction equation as a guide to our calculation. Once we have written the reaction equation, we can now proceed with the solution.

Amount of CO produced= mass of CO produced/ molar mass of CO

Molar mass of CO = 28 gmol-1

Mass of CO produced= 50.2g

Amount of CO produced= 50.2g/28gmol-1

Amount of CO produced= 1.79 moles of CO

From the reaction equation:

1 mole of Hg(ONC)2 produced 2 moles of CO

x moles of Hg(ONC)2 will produce 1.79 moles of CO

x= 1.79 × 1 / 2

x= 0.895 moles of Hg(ONC)2

Molar mass of Hg(ONC)2 = 284.624 g/mol

Mass of Hg(ONC)2 = number of moles × molar mass = 0.895 moles × 284.624 g/mol

Mass of Hg(ONC)2 = 254.7 g

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3 years ago

The vapor pressure of the liquid NH3 is measured at different temperatures. The following vapor pressure data are obtained.

S_A_V [24]

Answer:

$\Delta \:H_{vap}=23.3054\ kJ/mol$

Explanation:

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

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

Where,

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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:

$P_1$ = 234.2 mmHg

$P_2$ = 522.6 mmHg

$T_1$ = 217.9 K

$T_2$ = 232.4 K

So,

$\ln \:\left(\:\frac{234.2\ mmHg}{522.6\ mmHg}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314\times 10^{-3}\ kJ /mol K}\:\left(\:\frac{1}{232.4\ K}-\:\frac{1}{217.9\ K}\:\right)$

$\Delta \:H_{vap}=\frac{\left\{\ln \left(\:\frac{234.2}{522.6}\right)\:\times \:8.314\times 10^{-3}\right\}}{\left(\:\frac{1}{232.4}-\:\frac{1}{217.9}\:\right)}\ kJ/mol$

$\Delta \:H_{vap}=-\frac{10^{-3}\times \:8.314\ln \left(\frac{234.2}{522.6}\right)}{\frac{14.5}{50639.96}}\ kJ/mol$

$\Delta \:H_{vap}=-\frac{421020.62744\ln \left(\frac{234.2}{522.6}\right)}{14500}\ kJ/mol$

$\Delta \:H_{vap}=23.3054\ kJ/mol$

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3 years ago

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beks73 [17]

<u>Answer:</u> The 4 nuclear equations for the given series of emissions are written below.

<u>Explanation:</u>

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In this decay process, the atomic number of the parent nucleus decreases by a factor of 2 and the mass number also decreases by a factor of 4

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$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$