The new volume of the air bubble that has an initial volume of 5.0 ml released at the bottom of a lake where the pressure is 3.0 atm is 15mL.
<h3>How to calculate volume?</h3>
The volume of a given gas can be calculated by using the following formula:
P1V1 = P2V2
Where;
- P1 = initial pressure
- V1 = initial volume
- P2 = final pressure
- V2 = final volume
5 × 3 = 1 × V2
15 = V2
V2 = 15mL
Therefore, the new volume of the air bubble that has an initial volume of 5.0 ml released at the bottom of a lake where the pressure is 3.0 atm is 15mL.
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The relative molecular mass of acid A : 50 g/mol
<h3>Further explanation</h3>
Given
40.0 cm³(40 ml) of 0.2M sodium hydroxide
0.2g of a dibasic acid
Required
the relative molecular mass of acid A
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence(number of H⁺/OH⁻)
NaOH ⇒ n = 1
Dibasic acid = diprotic acid (such as H₂SO₄)⇒ n = 2
mol = M x V
Input the value in the formula :(1 = NaOH, 2=dibasic acid)
0.2 x 40 x 1 = M₂ x V₂ x 2
M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A
The relative molecular mass of acid A (M) :
The smallest mass among the choices is Choice (ii):2.3 x 10-3 μg
This is so because, the prefix μ signifies a factor of 10^-6.
For choice I:
- 23 cg means 23 centigrams
For choice II:
- 2.3 × 10-3 μg means 2.3 × 10-9 grams
For choice III:
- 0.23 mg means 0.23 × 10-3 grams
For choice IV:
- 0.23 grams simply means 0.23grams
For choice V:
- 2.3 × 10-2 kg means 23 grams
Ultimately, Choice II represents the smallest mass.
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To calculate the moles of AgNO3 in a solution, we need to know the volume and concentration of the solution.
Moles of AgNO3 = Volume of AgNO3 solution (L) * concentration of AgNO3 solution (M or mole/L) = 1.50 L * 0.050 M = 0.075 mole.
So 0.075 moles of AgNO3 are present in 1.50 L of a 0.050 M solution.
