Answer: 1.
moles
2. 90 mg
Explanation:

According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus
moles of ozone is removed by =
moles of sodium iodide.
Thus
moles of sodium iodide are needed to remove
moles of 
2. 
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 0.0003 moles of ozone is removed by =
moles of sodium iodide.
Mass of sodium iodide=
(1g=1000mg)
Thus 90 mg of sodium iodide are needed to remove 13.31 mg of
.
Hey there!
Oxygen has a molar mass of 16. That means 16g of oxygen is 1 mole.
32.6 ÷ 16 = 2.0375 moles
We have 2.0375 moles.
There are 6.022 x 10²³ atoms in one mole.
2.0375 x 6.022 x 10²³
1.3 x 10²⁴
There are 1.3 x 10²⁴ atoms in 32.6 grams of oxygen.
Hope this helps!
Answer:It’s C on edge 2020
(Combustion of car engines producing pollutants in the air)
Explanation:
I got it right :))
Use the clapeyron equation:
T in kelvin : 6.80 + 273 => 279.8 K
R = 0.082
n = 71.5 moles
P = 5.03 atm
Therefore:
P x V = n x R x T
5.03 x V = 71.5 x 0.082 x 279.8
5.03 x V = 1640.4674
V = 1640.4674 / 5.03
V = 326.13 L
hope ths helps!
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.