Answer:
4.87g
Explanation:
Step 1:
Data obtained from the question. This includes the following:
Mass of solution = 0.35kg
Molality = 0.238 m
Mass of NaCl =..?
Step 2:
Determination of the number of mole of NaCl in the solution.
Molality of a solution is simply defined as the mole of solute per unit kg of the solvent. It is given as:
Molality = mol of solute /mass of solvent (kg)
With the above formula, we calculate the mole of NaCl present in the solution as follow:
Molality = mol of solute /mass of solvent (kg)
0.238 = mol of NaCl /0.35
Cross multiply
mol of NaCl = 0.238 x 0.35
mol of NaCl = 0.0833 mol
Step 3:
Determination of the mass of NaCl in 0.0833 mol of NaCl.
This is illustrated below:
Number of mole NaCl = 0.0833 mol
Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol
Mass of NaCl =..?
Mass = number of mole x molar Mass
Mass of NaCl = 0.0833 x 58.5
Mass of NaCl = 4.87g
Therefore, 4.87g of NaCl is contained in the solution.
Add all weighs together
calculate carbons weigh
(carb weigh/total)×100
Amount of CH4 is excess, so no need to worry about it
<span>but the limiting factor is the Oxygen </span>
<span>according to stranded equation, </span>
<span>CH4 + 2 O2 --> CO2 + 2 H2O ΔH = -889 kJ/mol </span>
<span>just by taking proportions </span>
<span>(-889 kJ/mol) / 2 x 0.8 mol = - 355.6 kJ </span>
<span>so i think the answer is (a)</span>
<u>Ans: Mass of NaNO3 = 574 g</u>
<u>Given:</u>
Volume of NaNO3, V = 4.50 L
Molarity of NaNO3, M = 1.50 M i.e. 1.50 moles/L
Molar mass of NaNO3 = 85.00 g/mol
<u>To determine:</u>
The mass of NaNO3
<u>Explanation:</u>
Molarity of a solution can be expressed as the number of moles of the solute in a liter volume of solution
Molarity of NaNO3 = moles of NaNO3/volume of solution in L
1.50 moles/L = moles/4.50 L
moles NaNO3 = 1.50 *4.50 = 6.75 moles
Now, moles of NaNO3 = mass of NaNO3/molar mass NaNO3
mass of NaNO3 = moles * molar mass =
= 6.75 moles * 85.00 g/mole
= 573.75 g = 574 g
<u>Answer: </u>The amount of heat released is 84 calories.
<u>Explanation:
</u>
The equation used to calculate the amount of heat released or absorbed, we use the equation:

where,
Q = heat gained or released = ? Cal
m = mass of the substance = 10g
c = specific heat of aluminium = 0.21 Cal/g ° C
Putting values in above equation, we get:
Q = -84 Calories
Hence, the amount of heat released is 84 calories.