The final volume of the methane gas in the container is 6.67 L.
The given parameters;
- <em>initial volume of gas in the container, V₁ = 2.65 L</em>
- <em>initial number of moles of gas, n₁ = 0.12 mol</em>
- <em>additional concentration, n = 0.182 mol</em>
The total number of moles of gas in the container is calculated as follows;
The final volume of gas in the container is calculated as follows;
Thus, the final volume of the methane gas in the container is 6.67 L.
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<span>graduated cylinder is the answer</span>
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!
In this compound (Phosgene) the central atom (carbon is Sp² Hybridized).
Sp, Sp² and Sp³ can be calculated very simply by doing three steps,
Step 1:
Assume triple bond and double bond as one bond and assign s or p to it. In this example carbon double bond oxygen is considered once and let suppose it is s. Now we are having our s.
Step 2:
Count lone pair of electron, each lone pair counts for s and p. In this case there is no lone pair of electron on carbon, so not included.
Step 3:
Count single bonds for s and p. As we have already assigned s to the double bond, now one p for one single bond, and other p for the other single bond.
Result:
So, we counted 1 s for double bond, 1 p for one single and other p for second single bond. As a whole we got,
Sp²
Practice:
You can practice for hybridization of Oxygen in this molecule. Oxygen has 2 lone pair of electrons. (Hint: Sp² Hybridization)
From the stoichiometry of the reaction, 1.4 * 10^-3 g is produced.
<h3>What mass of water is produced?</h3>
The equation of the reaction is written as; CO2 + 2LiOH → Li2CO3 + H2O. This can help us to apply the principle of stoichiometry here.
Thus;
Number of moles of CO2 = 0.00345 g/44 g/mol = 7.8 * 10^-5 moles
If 1 mole of CO2 produced 1 mole of water
7.8 * 10^-5 moles of CO2 produced 7.8 * 10^-5 moles of water
Mass of water produced = 7.8 * 10^-5 moles * 18 g/mol = 1.4 * 10^-3 g
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