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Soloha48 [4]
3 years ago
15

Chemistry questions below! (●'◡'●)

Chemistry
1 answer:
bearhunter [10]3 years ago
4 0

(1) Islands are always surrounded by bodies of water, in this case oceans. These islands tend to have more 'stable climates' and 'smaller temperature fluctuations' due to the specific heat of water, which is quite high, 4180 J/kg C. The higher the specific heat, the more the energy needed to raise the temperature of the substance. Land requires less energy to raise it's temperature than water due to it's lower specific heat. Thus, islands tend to have more stable temperatures than regions surrounded by land.

(2) Based on the conditions provided, the thermochemical equation for the reaction should be:  

<u>2NaHCO₃ (s) ---> Na₂CO₃ (s) + H₂O (g) + O₂ (g)</u>

This is a decomposition reaction, as it explains that the baking soda breaks down into the products. All we had left to do is balance the equation, adding 2 as the coefficient of the reactant.

(3) Let's rewrite the given equations:

First chemical reaction: H₂(g) + F₂(g)  ---> 2HF(g), ΔH₁ = –537 kJ.

Second chemical reaction: C(s) + 2F₂(g) ---> CF₄(g), ΔH₂ = –680 kJ

Third chemical reaction: 2C(s) + 2H₂(g) ---> C₂H₄(g), ΔH₃ = +52.3 kJ.

Fourth chemical reaction: C₂H₄(g) + 6F₂(g) ---> 2CF₄(g) + 4HF(g), ΔH₄ = ?

Now the 'fourth chemical reaction' is the one in which we need to determine the enthalpy:

ΔH₄ = 2(ΔH₁) + 2(ΔH₂) - ΔH₃,

ΔH₄ = 2(–537) + 2(–680) - 52.3 =<u> -2486.3 kJ</u>

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A balloon has a volume of 10,500 liters and the tempature is 15 C . If the temp tire were -25C what would the volume of the ball
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1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

5 0
3 years ago
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