If 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy 25.48 L at STP.
<h3>How to calculate volume?</h3>
The volume of a gas at STP can be calculated using the direct proportion method.
According to this question, 50.75 g of a gas occupies 10.0 L at STP, then 129.3g of the same gas will occupy the following:
= 129.3 × 10/50.75
= 25.48L
Therefore, if 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy 25.48 L at STP.
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NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
Answer:
pH = -log₁₀ [H⁺]
Explanation:
pH is a value in chemistry used in to measure solution trying to determine each quality, purity, risks for health of some products, etc.
As you write in the question, [H⁺] = 10^(-pH)
Using logarithm law (log (m^(p) = p log(m):
log₁₀ [H⁺] = -pH
And
<h3>pH = -log₁₀ [H⁺]</h3>
