Answer:
3.4 × 10²³ molecules of CBr₄
Explanation:
Given data:
Mass of CBr₄ = 189 g
Number of molecules = ?
Solution:
First of all we will calculate the number of moles.
Number of moles = mass / molar mass
Number of moles = 189 g/ 331.63 g/mol
Number of moles = 0.6 mol
Now the given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
Foe 0.6 moles of CBr₄:
0.6 mol × 6.022 × 10²³ molecules of CBr₄ / 1 mol
3.4 × 10²³ molecules of CBr₄
According to Raoult's low:
We will use this formula: Vp(Solution) = mole fraction of solvent * Vp(solvent)
∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808
when we assume the moles of solute NaCl = X
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
So X (the no.of moles of NaCl) = 0.027 m
Answer:
The concentration of H3O+= 0.15M
Explanation:
From The equation of reaction
HClO2(aq) + H2O(l) ⇌ H3O+(aq) + ClO2−(aq)
0.25mol HClO2(aq) 0.25mol producesClO2−(aq) and x-mol of H3O+
Using Kc = [H3O+][ClO2-]/[HClO2]
0.15= 0.25*x/0.25
Simplify
x=0.15M
Answer:
The limiting reactant is CuSO₄.
Explanation:
The reaction is:
Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s) (1)
To find the limiting reactant we need to find the number of moles of the reactants.
Where:
m: is the mass of iron = 3.26 g
: is the standard atomic weight of iron = 55.845 g/mol
Where:
M: is the concentration of the CuSO₄ = 0.200 mol/dm⁻³
V: is the volume of the solution = 80.0 cm³
First, we need to convert the units of the volume to dm³ knowing that 1 dm = 10 cm and 1 L= 1 dm³.
Now, the number of CuSO₄ moles is:
So, to determine the limiting reactant we need to use the molar ratio from equation (1), Fe:CuSO₄ = 1:1
Since we need 0.016 moles of Fe to react with 0.016 moles of CuSO₄ and initially we have 0.058 moles of Fe, then the limiting reactant is CuSO₄.
Therefore, the limiting reactant is CuSO₄.
I hope it helps you!
Answer:
0.62 L
Explanation:
Step 1:
Data obtained from the question.
Initial Volume (V1) = 1.80 L
Initial pressure (P1) = 785 mmHg
Final pressure (P2) = 3.00 atm
Final volume (V2) =?
Step 2:
Conversion of the pressure in mmHg to atm.
It is important to express the initial and the final pressure in the same unit. Either express both in atm or in mmHg. What ever the case is, we'll still arrive at same answer. Here, we shall be converting from mmHg to atm. This is illustrated below:
760mmHg = 1atm
Therefore, 785 mmHg = 785/760 = 1.03 atm
Step 3:
Determination of the final volume. This is illustrated below.
We shall be applying the Boyle's law equation since the temperature is constant.
P1V1 = P2V2
Initial Volume (V1) = 1.80 L
Initial pressure (P1) = 1.03 atm
Final pressure (P2) = 3.00 atm
Final volume (V2) =?
P1V1 = P2V2
1.03 x 1.8 = 3 x V2
Divide both side by 3
V2 = (1.03 x 1.8) /3
V2 = 0.62 L
Therefore, the new volume of the balloon is 0.62 L
