Answer:
D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Explanation:
Step 1: Detemine the mass of O in SO₂
There are 2 atoms of O in 1 molecule of SO₂. Then,
m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g
Step 2: Determine the mass of SO₂
m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g
Step 3: Detemine the mass percent of oxygen in SO₂
We will use the following expression.
m(O)/m(SO₂) × 100%
(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Answer:
pH = 6.999
The solution is acidic.
Explanation:
HBr is a strong acid, a very strong one.
In water, this acid is totally dissociated.
HBr + H₂O → H₃O⁺ + Br⁻
We can think pH, as - log 7.75×10⁻¹² but this is 11.1
acid pH can't never be higher than 7.
We apply the charge balance:
[H⁺] = [Br⁻] + [OH⁻]
All the protons come from the bromide and the OH⁻ that come from water.
We can also think [OH⁻] = Kw / [H⁺] so:
[H⁺] = [Br⁻] + Kw / [H⁺]
Now, our unknown is [H⁺]
[H⁺] = 7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]
[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) / [H⁺]
This is quadratic equation: [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴
a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴
(-b +- √(b² - 4ac) / (2a)
[H⁺] = 1.000038751×10⁻⁷
- log [H⁺] = pH → 6.999
A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.
Answer:
B
Explanation:
B is the best showing of a chemical reaction out of the choices
Answer:
The concentration of chloride ion is 
Explanation:
We know that 1 ppm is equal to 1 mg/L.
So, the 
content 100 ppm suggests the presence of 100 mg of in 1 L of solution.
The molar mass of is equal to the molar mass of Cl atom as the mass of the excess electron in is negligible as compared to the mass of Cl atom.
So, the molar mass of is 35.453 g/mol.
Number of moles = (Mass)/(Molar mass)
Hence, the number of moles (N) of present in 100 mg (0.100 g) of is calculated as shown below:
So, there is 
of present in 1 L of solution.
Answer:
'See Explanation
Explanation:
Determine the [OH−] , pH, and pOH of a solution with a [H+] of 9.5×10−13 M at 25 °C.
Given [H⁺] = 9.5 x 10⁻¹³M => [H⁺][OH⁻] = 1.0 x 10⁻¹⁴ => [OH⁻] = 1.0 x 10⁻¹⁴/9.5 x 10⁻¹³ = 0.0105M
pH = -log[H⁺] = -log(9.5 x 10⁻¹³) = - (-1202) = 12.02.
pOH = -log[OH⁻] = -log(0.0105) = -(-1.98) = 1.98
Now you use the same sequence in the remaining problems.
