<u>Answer:</u> The mass of iron (III) nitrate is 11.16 g/mol
<u>Explanation:</u>
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
We are given:
Molarity of solution = 0.3556 M
Molar mass of Iron (III) nitrate = 241.86 g/mol
Volume of solution = 129.8 mL
Putting values in above equation, we get:
Hence, the mass of iron (III) nitrate is 11.16 g/mol
Answer:
¹/3 C3H8(g) + ⁵/3 O2(g)
Explanation:
The coefficient before every molecule is representative of the number of moles. We can represent it in ration form so as to calculate the question;
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l) means;
For every 1 mole of C₃H₈(g) and 5 moles of O₂(g) produces 3 moles of CO₂(g) and 4 moles of H₂O(l).
Therefore to produce 1.00 mole of CO₂(g);
We represent it in ratio;
C₃H₈(g) : CO₂(g)
1 : 3
For more on evaluating moles in chemical reactions check out;
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To determine the volume of the gas mixture, we first need to determine the total pressure of the mixture. To do this, we use the definition of the partial pressure of a component in the gas mixture. The partial pressure is the pressure of a component as if it were alone in the container. It is equal to the mole fraction of the component times the total pressure of the system. From this, we determine total pressure.
Pneon = xneonP
P = Pneon / xneon
P = 8.87 kPa / (225 / (225 + 320 + 175))
P = 8.87 kPa / 0.3125 = 28.384 kPa
Assuming ideal gas, we use PV=nRT to calculate for the volume,
PV = nRT
V = nRT / P
n = 225 mg ( 1 mmol / 20.18 mg) + 320 mg ( 1 mmol / 16.05 mg ) + 175 mg ( 1 mmol / 39.95 mg ) = 35.47 mg = 35467.0 g
V = 35467.0 (8.314) (300) / (28384) = 3116.68 m^3
<span>A. Commercial cooking
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