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4 years ago

How many moles of N are in 0.173 g of N2O?

Chemistry

1 answer:

krok68 [10]4 years ago

3 0

Find the RMM of both then cross multiply keeping 0.173 under N2O and you will get the answer

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Approximately how many formula units of NaCl are in 116.88g of table salt (NaCl), knowing that the molar mass of NaCl is 58.44g/

elena-14-01-66 [18.8K]

Answer:

116.88g of table salt (NaCl) contains two formula units

Explanation:

Now,

We know that 1 formula unit of sodium chloride has a molar mass of 58.44g/mol

Hence;

Mass of 1 formula unit = 58.44g

Mass of x formula units = 116.88g

x = 116.88g * 1 formula unit/58.44g

x = 2 formula units

Therefore;

116.88g of table salt (NaCl) contains two formula units

5 0

3 years ago

Read 2 more answers

List the five major pollutants in the United States

lawyer [7]

Fine particles, ground level ozone, sulfur dioxide, nitrogen dioxide, lead

6 0

3 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

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

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

A gas is allowed to expand from a volume of 400 ml to 2000 ml at a constant temperature. If the the initial pressure is 3 atm, c

Lostsunrise [7]

Answer:

$P_{2}) = 0.6 \; atm$

Explanation:

Given the following data;

Initial volume = 400 mL

Final volume = 2000 mL

Initial pressure = 3 atm

To find the final pressure P2, we would use Boyles' law.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

$PV = K$

$P_{1}V_{1} = P_{2}V_{2}$

Substituting into the equation, we have;

$3 * 400 = P_{2}* 2000$

$1200 = 2000P_{2}$

$P_{2}) = \frac {1200}{2000}$

$P_{2}) = 0.6 \; atm$

8 0

3 years ago

How does the law of conservation of mass apply to this reaction C2H4 + O2 > 2H2O + 2CO2

AlexFokin [52]

It shows mass is not created nor lost but re arranged

4 0

3 years ago

Read 2 more answers