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2 years ago

Giving 30 points!! please help <3

Chemistry

1 answer:

TiliK225 [7]2 years ago

3 0

Answer:

below :)

Explanation:

Bones, droppings, and other dead matter

Energy storage molecules, cellular respiration

Process, energy

Oxygen, energy storage molecules, energy, carbon dioxide

Cellular respiration, carbon

Carbon, nitrogen

Nitrogen

Decomposers, ecosystem

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Use the equation n2 + 3h2=2nh3 if 8.0g N2 react, how many grams of NH3 will be produced

Eduardwww [97]

the balanced equation for the formation of ammonia is

N₂ + 3H₂ ---> 2NH₃

molar ratio of N₂ to NH₃ is 1:2

mass of N₂ reacted is 8.0 g

therefore number of N₂ moles reacted is - 8.0 g / 28 g/mol = 0.286 mol

according to the molar ratio,

1 mol of N₂ will react to give 2 mol of NH₃, assuming nitrogen is the limiting reactant

therefore 0.286 mol of N₂ should give - 2 x 0.286 mol = 0.572 mol of NH₃

therefore mass of NH₃ formed is - 0.572 mol x 17 g/mol = 9.72 g

a mass of 9.72 mol of NH₃ is formed

6 0

3 years ago

What makes noble gases stable?

inn [45]

The correct answer to your question is noble gases are stable <span>due to having the maximum number of valence electrons their outer shell can hold. Meaning their outer shells are stable.

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8 0

3 years ago

Sắp xếp các hợp chất: HF, HI, HBr, HCl theo chiều giảm dần tính axit.

mixer [17]

I don’t know what to answer

6 0

2 years ago

Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h

bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

We know that,

The relation between the $C_p\text{ and }C_v$ for an ideal gas are :

$C_p-C_v=R$

As we are given :

$C_p=28.253J/K.mole$

$28.253J/K.mole-C_v=8.314J/K.mole$

$C_v=19.939J/K.mole$

Now we have to calculate the entropy change of the gas.

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

$\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J$

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. $(w=-pdV)$

(C) Heat during the process will be,

$q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J$

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0

3 years ago

Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:

posledela

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of $HNO_3$ = $\frac{15.5g}{63g/mol\times 9.5L}=0.026M$

Equilibrium concentration of $NO$ = $\frac{16.6g}{30g/mol\times 9.5L}=0.058M$

Equilibrium concentration of $NO_2$ = $\frac{22.5g}{46g/mol\times 9.5L}=0.051M$

Equilibrium concentration of $H_2O$ = $\frac{189.0g}{18g/mol\times 9.5L}=1.10M$

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

For the given chemical reaction:

$2HNO_3(aq)+NO(g)\rightarrow 3NO_2(g)+H_2O(l)$

The expression for $K_c$ is written as:

$K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}$

$K_c=\frac{(0.051)^3\times (1.10)^1}{(0.026)^2\times (0.058)^1}$

$K_c=3.72$

Thus the value of the equilibrium constant Kc for this reaction is 3.72

5 0

3 years ago