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2 years ago

Calculate the mass percent of oxygen in CO2 (Work shown)

Chemistry

1 answer:

Genrish500 [490]2 years ago

4 0

Answer:

$mass \: of \: CO _{2} = (12 + (16 \times 2)) = 44 \: g \\ mass \: of \: oxygen \: in \: CO _{2} = (16 \times 2) = 32 \\ \% \: of \: oxygen = \frac{32}{44} \times 100\% \\ = 72.7\%$

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Water at the bottom of a narrow metal tube is held at a constant temperature of 293 K. The total pressure of air (assumed dry) i

defon

Answer:

1.595 x 10-7kmol/m2.s Explanation:

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3 years ago

If 8.00 g NH4NO3 is dissolved in 1000 g of water, the water decreases in temperature from 21.00 degrees Celsius to 20.39 degrees

telo118 [61]

Answer:

25.7 kJ/mol

Explanation:

There are two heats involved.

heat of solution of NH₄NO₃ + heat from water = 0

q₁ + q₂ = 0

n = moles of NH₄NO₃ = 8.00 g NH₄NO₃ × 1 mol NH₄NO₃/80.0 g NH₄NO₃

∴ n = 0.100 mol NH₄NO₃

q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln

m = mass of solution = 1000.0 g + 8.00 g = 1008.0 g

q₂ = mcΔT = 58.0 g × 4.184 J°C⁻¹ g⁻¹ × ((20.39-21)°C) = -2570.19 J

q₁ + q₂ = 0.100 mol ×ΔHsoln – 2570.19 J = 0

ΔHsoln = +2570.19 J /0.100 mol = +25702 J/mol = +25.7 kJ/mol

7 0

3 years ago

Help with this question

boyakko [2]

True. Our earth Is getting hotter due to global warming.

6 0

2 years ago

2C4H10(g) +13O2(g) → 8CO2(g) +10H2O(g) Suppose a grill lighter contains 50.0 g of butane. How many grams of butane in the lighte

MatroZZZ [7]

Answer is: mass of butane is D)11.6 g.
m(butane) = 50,0 g.
V(CO₂) = 17,9 L.
n(CO₂) = V(CO₂) ÷ Vm.
n(CO₂) = 17,9 L ÷ 22,4 L/mol.
n(CO₂) = 0,8 mol.
From chemical reaction n(CO₂) : n(C₄H₁₀) = 8 : 2.
n(C₄H₁₀) = 0,8 mol ÷ 4.
n(C₄H₁₀) = 0,2 mol.
m(C₄H₁₀) = n(C₄H₁₀) · M(C₄H₁₀).
m(C₄H₁₀) = 0,2 mol · 58 g/mol.
m(C₄H₁₀) = 11,6 g.

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2 years ago

A scooter maintains a pace of 1.45 minutes per mile, with fuel economy of 85 miles per gallon. How many seconds can it travel on

vlada-n [284]

Answer:

281s

Explanation:

Given parameters:

Speed of the scooter = 1.45min per mile

Fuel economy of engine = 85miles per gallon

Quantity of fuel = 145mL

Unknown:

Time of travel with the volume of fuel given = ?

Solution:

To solve this problem, we need to find the distance the fuel will last.

Rate of fuel consumption by the engine = 85miles per gallon

Convert 145mL to gallons;

3785.41mL = 1 gallon

145mL will therefore give $\frac{145}{3785.41}$ = 0.038gallons

So;

Distance covered = 85miles per gallon x 0.038gallons = 3.23miles

From;

Rate of travel = $\frac{time }{distance}$

Time = rate of travel x distance = 1.45 minutes per mile x 3.23miles

Time = 4.7min

1 min = 60s

4.7min = 4.7 x 60 = 281s

6 0

2 years ago