Answer:
1210
Explanation:
Because the 121 cm times 10 is 1210
The percent concentration of a solution can be calculated from; mass of solute /mass of solution * 100. The mass of the solute here is 8.1 g.
<h3>What is concentration?</h3>
The term concentration refers to the amount of solute presnt in a solution. There are many ways of expressing concentration such as molarity, molality and percentage.
Here;
mass of solution = 230.5 g
Percent of solute = 3.5 %
3.5 = x/ 230.5 * 100
3.5 = 100x/230.5
230.5(3.5) = 100x
x = 230.5(3.5) /100
x = 8.1 g
Learn more about percent concentration: brainly.com/question/202460?
Explanation:
In order to be able to calculate the volume of oxygen gas produced by this reaction, you need to know the conditions for pressure and temperature.
Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature and Pressure.
STP conditions are defined as a pressure of
100 kPa
and a temperature of
0
∘
C
. Under these conditions for pressure and temperature, one mole of any ideal gas occupies
22.7 L
- this is known as the molar volume of a gas at STP.
So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.
The balanced chemical equation for this decomposition reaction looks like this
2
KClO
3(s]
heat
×
−−−→
2
KCl
(s]
+
3
O
2(g]
↑
⏐
⏐
Notice that you have a
2
:
3
mole ratio between potassium chlorate and oxygen gas.
This tells you that the reaction will always produce
3
2
times more moles of oxygen gas than the number of moles of potassium chlorate that underwent decomposition.
Use potassium chlorate's molar mass to determine how many moles you have in that
231-g
sample
231
g
⋅
1 mole KClO
3
122.55
g
=
1.885 moles KClO
3
Use the aforementioned mole ratio to determine how many moles of oxygen would be produced from this many moles of potassium chlorate
1.885
moles KClO
3
⋅
3
moles O
2
2
moles KClO
3
=
2.8275 moles O
2
So, what volume would this many moles occupy at STP?
2.8275
moles
⋅
22.7 L
1
mol
=
64.2 L
A. 1/2
Explanation- There is a 5/10 chance of choosing on of the numbers which simplifies to 1/2
Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
