Compete Question:
What mass of CDP (403 g mol−1) is in 10 mL of the buffered solution at the beginning of Experiment 1?
Passage: "16 mmol of CDP in 1 L of buffer"
Answer:
6.4 × 10-2 g
Explanation:
we are given from the question that 16 mmol of CDP is in 1 L of buffer
this mean that we have 
moles of CDP in 1 liter of buffer.
so the mass of CDP in one liter of buffer will be calculate as,
mass of CDP = × 403g mol−1
= 
= 6.4 g/L
But because the question
asks us about the mass of CDP in 10 mL of solution, we will go further to calculate it like this:
6.4 g/L × 10 mL
6.4 g/L × 0.01 L = 
Answer:
1. 35 mg of H₃PO₄
2. 27 mol AlF₃; 82 mol F⁻
3. 300 mL of stock solution.
Explanation:
1. Preparing a solution of known molar concentration
Data:
V = 80 mL
c = 4.5 × 10⁻³ mol·L⁻¹
Calculations:
(a) Moles of H₃PO₄
Molar concentration = moles of solute/litres of solution
c = n/V
n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol
(b) Mass of H₃PO₄
moles = mass/molar mass
n = m/MM
m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg
(c) Procedure
Dissolve 35 mg of solid H₃PO₄ in enough water to make 80 mL of solution,
2. Moles of solute.
Data:
V = 4900 mL
c = 5.6 mol·L⁻¹
Calculations:
Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃
Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.
3. Dilution calculation
Data:
V₁= 750 mL; c₁ = 0.80 mol·L⁻¹
V₂ = ? ; c₂ = 2.0 mol·L⁻¹
Calculation:
V₁c₁ = V₂c₂
V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL
Procedure:
Measure out 300 mL of stock solution. Then add 500 mL of water.
Answer:
C)
Explanation:
ozone is a type of oxygen. It's chemical compound is 03. Just 1 more molecule than regualr oxygen.
Glad I was able to help!!
Answer:
7.5 × 10¹⁵ Hz
Explanation:
Given data
- Wavelength of the radio waves (λ): 40 nm = 40 × 10⁻⁹ m = 4.0 × 10⁻⁸ m
- Frequency of the radio waves (ν): ?
- Speed of light (c): 3.00 × 10⁸ m/s
We can determine the frequency of the radio waves using the following expression.
c = λ × ν
ν = c/λ
ν = (3.00 × 10⁸ m/s)/4.0 × 10⁻⁸ m
ν = 7.5 × 10¹⁵ s⁻¹ = 7.5 × 10¹⁵ Hz
