Answer:
The answer you have selected in the screenshot is correct.
Its tendency to react with oxygen is correct.
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Answer:
0.508 mole
Explanation:
NOTE: Since no hydrogen is attached to the compound given in question above, it means the compound is CCl₄.
The number of mole present in 78.2 g of CCl₄ can be obtained as follow:
Mass of CCl₄ = 78.2 g
Molar mass of CCl₄ = 12 + (35.5×4)
= 12 + 142
= 154 g/mol
Mole of CCl₄ =?
Mole = mass / molar mass
Mole of CCl₄ = 78.2 / 154
Mole of CCl₄ = 0.508 mole
Therefore, 0.508 mole is present in 78.2 g of CCl₄
We know that, M1V1 = M2V2
(Initial) (Final)
where, M1 and M2 are initial and final concentration of soution respectively.
V1 and V2 = initial and final volume of solution respectively
Given: M1 = 12 m, V1 = 35 ml and V2 = 1.2 l = 1200 ml
∴ M2 = M1V1/V2 = (12 × 35)/ 1200 = 0.35 m
Final concentration of solution is 0.35 m
87) A
88) D
89) B
90) E
91) C
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
Learn ore about stoichiometry:brainly.com/question/9743981
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