When the reaction equation is:
CaSO3(s) → CaO(s) + SO2(g)
we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.
to get the moles of SO2 we are going to use the ideal gas equation:
PV = nRT
when P is the pressure = 1.1 atm
and V is the volume = 14.5 L
n is the moles' number (which we need to calculate)
R ideal gas constant = 0.0821
and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K
so, by substitution:
1.1 * 14.5 L = n * 0.0821 * 285.5
∴ n = 1.1 * 14.5 / (0.0821*285.5)
= 0.68 moles SO2
∴ moles CaSO3 = 0.68 moles
so we can easily get the mass of CaSO3:
when mass = moles * molar mass
and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol
∴ mass = 0.68 moles* 120 g/mol = 81.6 g
Answer:
HBr(aq) + LiOH(aq) = LiBr(aq) + H2O(l)
Explanation:
For this reaction, the reactants are the hydrobomic acid and the lithium hydroxide which produces the products lithium bromide and water.
Answer:
119 kCal per serving.
Explanation:
The heat energy necessary to elevates water's temperature from 23.4°C to 37.9°C can be calculated by the equation below:
Q = mcΔT
Q: heat energy
m: mass in g
c: specific heat capacity in cal/g°C
ΔT = temperature variation in °C
m is the mass of water, considering the density of water to be 1g/mL, 100 mL of water weights 100g. Therefore:
Q = 100 g x 1.00 cal/g°C x (37.9 - 23.4)°C
Q = 1450 cal
1450 cal ____ 0.341 g peanuts
x ____ 28 g peanuts
x = 119061.58 cal
This means that the cal from fat per serving of peanuts is at least 119 kCal.
Answer:
44.01 g/mol
Explanation:
Add each elements atomic mass. For oxygen you will do that twice because their is two oxygens.
- Hope that helps! Please let me know if you need further explanation.
Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol
