1.34 L of HF
Explanation:
We have the following chemical reaction:
Sn (s) + 2 HF (g) → SnF₂ (s) + H₂ (g)
First we calculate the number of moles of SnF₂:
number of moles = mass / molecular weight
number of moles of SnF₂ = 5 / 157 = 0.03 moles
From the chemical reaction we see that 1 mole of SnF₂ are produced from 2 moles of SnF₂. This will mean that 0.03 moles of SnF₂ are produced from 0.06 moles of HF.
Now at standard temperature and pressure (STP) we can use the following formula to calculate the volume of HF:
number of moles = volume / 22.4 (L/mole)
volume of HF = number of moles × 22.4
volume of HF = 0.06 × 22.4 = 1.34 L
Learn more about:
problems with gases at STP
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Answer:
The activation energy for the decomposition = 33813.28 J/mol
Explanation:
Using the expression,
Wherem
is the activation energy
R is Gas constant having value = 8.314 J / K mol
Thus, given that,
= ?
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (5 + 273.15) K = 278.15 K
T = (25 + 273.15) K = 298.15 K
So,




<u>The activation energy for the decomposition = 33813.28 J/mol</u>
The combustion reaction of octane is as follow,
C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O
According to balance equation,
8 moles of CO₂ are released when = 114.23 g (1 mole) Octane is reacted
So,
6.20 moles of CO₂ will release when = X g of Octane is reacted
Solving for X,
X = (114.23 g × 6.20 mol) ÷ 8 mol
X = 88.52 g of Octane
Result:
88.52 g of Octane is needed to release 6.20 mol CO₂.
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