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:
0.052mL
Explanation:
1mole of a gas occupy 22.4L.
Therefore, 1 mole of CO2 will also occupy 22.4L.
If 1mole of CO2 occupies 22.4L,
Then 2.3moles of CO2 will occupy = 2.3 x 22.4 = 51.52L
coverting this volume to mL, we simply divide by 1000 as shown below:
51.52/1000 = 0.05152mL = 0.052mL
Answer:
For H3O concentration you do 10^-pH so if pH is 5 then H3O+ is 10^-5= 1*10^-5 H3O+ ions
For OH is one extra step. First find H3o+ ions using equation above then you have to use that to divide 1*10^-14
So if pH is 5....the H3O+ is 1*10^-5 then OH- = (1*10^-14)/(1*10^-5) = 1*10^-9 OH ions
as far as acid/base pH 0-6 is Acid 8-14 is Base. pH of 7 is neutral. Recheck your work *hint* *hint* water is neutral. Spit is above 7 so is base.
