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3 years ago

A gas mixture contains 88 grams of oxygen gas, 88 grams of nitrogen gas, and 12 grams of nitrogen dioxide gas,

Chemistry

2 answers:

emmainna [20.7K]3 years ago

7 0

BROO DEE YOU ALREADY KNOW WHAT IS STOP DA CAP, o.81

bixtya [17]3 years ago

6 0

88 g O2 x 1 mol O2/32 g = 2.75 moles O2
88 g N2 x 1 mol N2/28 g = 3.14moles N2
12 g NO2 x 1 mol NO2/46 g = 0.261 moles NO2
TOTAL moles = 6.15 moles
mole fraction N2 = 3.14/6.15 = 0.51

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Consider 2H2 + O2 → 2H2O. To produce 1.2 g water, how many grams of H2 are required? Report to the correct number of significant

Elden [556K]

Answer:

0.133 mol (corrected to 3 sig.fig)

Explanation:

Take the atomic mass of H=1.0, and O=16.0,

no. of moles = mass / molar mass

so no. of moles of H2O produced = 1.2 / (1.0x2+16.0)

= 0.0666666 mol

From the equation, the mole ratio of H2:H2O = 2:2 = 1:1,

meaning every 1 mole of H2 reacted gives out 1 mole of water.

So, the no, of moles of H2 required should equal to the no, of moles of H2O produced, which is also 0.0666666 moles.

mass = no. of moles x molar mass

hence,

mass of H2 required = 0.066666666 x (1.0x2)

= 0.133 mol (corrected to 3 sig.fig)

3 0

3 years ago

Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for t

ratelena [41]

Answer:

$\Delta G^0 _{rxn} = 207.6\ kJ/mol$

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

$2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}$

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

$\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol$

$\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol$

$\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol$

$\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})$

$\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)$

$\Delta G^0 _{rxn} = 207.6\ kJ/mol$

The equilibrium constant determined from the partial pressure denoted as $K_p$ can be expressed as :

$K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}$

$K_p = \dfrac{1}{ (22.20)}$

$K_p$ = 0.045

$\Delta G = \Delta G^0 _{rxn} + RT \ lnK$

where;

R = gas constant = 8.314 × 10⁻³ kJ

$\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)$

$\Delta G =207.6 + 2.4788191 \times \ ln(0.045)$

$\Delta G =207.6+ (-7.687048037)$

$\Delta G =$ 199.912952 kJ

ΔG ≅ 199.91 kJ

7 0

4 years ago

Name something made of aluminum and explain why malleability is a useful property

avanturin [10]

Soft drink and beer cans. If aluminum were not malleable, these couldn't be made

3 0

4 years ago

Read 2 more answers

Convert 98.7 °F to Celsius Formula for Fahrenheit to Celsius

dmitriy555 [2]

Answer:

$\boxed {\tt 37.0555556 \textdegree C}$

Explanation:

The formula to convert Fahrenheit to Celsius, use the following formula.

$C=(\textdegree F -32)*\frac{5}{9}$

We want to convert 98.7 degrees Fahrenheit to Celsius, so we must substitute 98.7 in for degrees Fahrenheit.

$C=(98.7 -32)*\frac{5}{9}$

Solve inside the parentheses. Subtract 32 from 98.7.

$C=(66.7)*\frac{5}{9}$

Multiply 66.7 by 5/9.

$C=37.0555556$

98.7 degrees Fahrenheit is equal to 37.0555556 degrees Celsius

3 0

4 years ago

Read 2 more answers

An element has an atomic number of 30. How many protons and electrons are in a neutral atom of the element?

anygoal [31]

30 protons, 30 electrons is your answer!:)

7 0

3 years ago