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ValentinkaMS [17]
3 years ago
12

Calculate the amount of grams of water that was used when 5525J of energy was used to boil this amount of water.

Chemistry
1 answer:
djverab [1.8K]3 years ago
4 0

Answer:

2.445 g

Explanation:

Step 1: Given and required data

  • Energy in the form of heat required to boil the water (Q): 5525 J
  • Latent heat of vaporization of water (∆H°vap): 2260 J/g
  • Mass of water (m): ?

Step 2: Calculate the mass of water

We will use the following expression.

Q = ∆H°vap × m

m = Q / ∆H°vap

m = 5525 J / (2260 J/g)

m = 2.445 g

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First we have to find moles of C:
Molar mass of CO2:
12*1+16*2 = 44g/mol
(18.8 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) =0.427 mol C 
Molar mass of H2O:
2*1+16 = 18g/mol
As there is 2 moles of H in H2O,
So,

<span>(6.75 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 0.74mol H </span>

<span>Divide both number of moles by the smaller number of moles: </span>
<span>As Smaaler no moles is 0.427:
So,
Dividing both number os moles by 0.427 :
(0.427 mol C) / 0.427 = 1.000 </span>
<span>(0.74 mol H) / 0.427 = 1.733 </span>

<span>To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula: 
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3 0
3 years ago
Part A
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Answer:

ΔG° = -5.4 kJ/mol

ΔG = 873.2 J/mol = 0.873 kJ /mol

Explanation:

Step 1: Data given

ΔG (NO2) = 51.84 kJ/mol

ΔG (N2O4)  = 98.28 kJ/mol

Step 2:

ΔG = ΔG° + RT ln Q

⇒with Q = the reaction quatient

⇒with T = the temperature = 298 K

⇒with R = 8.314 J / mol*K

⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2 )

⇒ ΔG° = 98.28 kJ/mol - 2* 51.84  kJ/mol

⇒ ΔG° = -5.4 kJ/mol

Part B

ΔG =  ΔG° =RT ln Q

⇒with G° = -5.4 kj/mol = -5400 j/mol

⇒ with R = 8.314 J/K*mol

⇒with T = 298 K

⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577

ΔG = -5400 + 8.314 * 298 * ln(12.577)

ΔG = -5400 + 8.314 * 298 * 2.532

ΔG = 873.2 J/mol = 0.873 kJ/mol

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