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3 years ago

How many molecules are in 25 grams of NH3

Chemistry

1 answer:

N76 [4]3 years ago

7 0

Ok I will 281738 that’s the answer

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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

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Liula [17]

He sponge bob. Kristin craf

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3 years ago

Question 2 of 19

Alborosie

Answer:

Explanation:

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3 years ago

CH3-C_=CH + HCl ( 1:2) -->

svlad2 [7]

CH
3

C≡CH
2mole
HCl

A
CH
3

C(Cl)
2

CH
3

Heat
aq.KOH

B
CH
3

COCH
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3 years ago

A quantity of gas weighing 7.1 g at 741 torr and 44 celius occupies 5.41 L. Find its molar mass.

Aleks04 [339]

1 atmosphere = 760 torr
R = 0.082
44 degrees Celsius = 317 degrees Kelvin

$PV=NRT\\ \frac{744}{760}(5.41)=N(0.082)(317)\\ \frac{4025.04}{760}=25.994N\\\frac{4025.04}{19755.44}=N\\N=\frac{50313}{246943}\\0.203743$

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3 years ago