Answer:
The appropriate answer is "9.225 g".
Explanation:
Given:
Required level,
= 63 ppm
Initial concentration,
= 22 ppm
Now,
The amount of free SO₂ will be:
= 
= 
= 
The amount of free SO₂ to be added will be:
= 
= 
∵ 1000 mg = 1 g
So,
= 
= 
Thus,
"9.225 g" should be added.
Answer:
1.125 moles
Explanation:
2mole of HCl produced 1mole of H2
2.25moles of HCl will produce x moles
cross multiply
2x=™1×2.25
x= 2.25÷2
x=1.125mole
The differences in the properties of diamond and graphite is as a result of how their particles are arranged in space. This space arrangement leads to distinct crystal structures for the two compounds. In diamond, the carbon atoms are arranged in tetrahedral shape while in graphite the carbon atoms arrayed in planes.
Hope this helps :)
1/2 - 3/7 = 7/14 - 6/14 = 1/14
Answer:
1. V₁ = 2.0 mL
2. V₁ = 2.5 mL
Explanation:
<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>
To calculate the volume required (V₁) in each dilution we will use the dilution rule.
C₁ . V₁ = C₂ . V₂
where,
C are the concentrations
V are the volumes
1 refers to the initial state
2 refers to the final state
<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>
C₁ . V₁ = C₂ . V₂
(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL
V₁ = 2.0 mL
<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>
C₁ . V₁ = C₂ . V₂
(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL
V₁ = 2.5 mL
