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AleksAgata [21]
3 years ago
6

A sample of gas at 300 K has a volume of 2.0 liters. At what temperature will the volume decrease to 1.0 liter if the pressure i

s kept constant?
Chemistry
2 answers:
xxTIMURxx [149]3 years ago
8 0
Pv/t=c 
Therefore 

v/t=c 
2/300=1/t2 
t2=150K 
notsponge [240]3 years ago
6 0

Answer : The temperature will be, 150 K

Explanation :

Charles' Law : This law states that volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T     (At constant pressure and number of moles)

or,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1 = initial volume of gas = 2.0 L

V_2 = final volume of gas = 1.0 L

T_1 = initial temperature of gas = 300 K

T_2 = final temperature of gas = ?

Now put all the given values in the above formula, we get the final temperature of gas.

\frac{2.0L}{300K}=\frac{1.0L}{T_2}

T_2=150K

Therefore, the temperature will be, 150 K

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Markonikov's rule states in the addition of HX to an unsymmetrical alkene, the H atom bonds to the less substituted carbon atom.

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8 0
3 years ago
Mg3N2(s)+6H2O(l)→3Mg(OH)2(s)+2NH3(g) When 36.0 g of H2O react, how many grams of NH3 are produced? When 36.0 g of H2O react, how
Afina-wow [57]

Answer:

11.3 g of NH_{3} are produced from 36.0 g of H_{2}O

Explanation:

1. The balanced chemical equation is the following:

Mg_{3}N_{2}(s)+6H_{2}O(l)=3Mg(OH)_{2}(s)+2NH_{3}(g)

2. Use the molar mass of the H_{2}O, the molar mass of the NH_{3} and the stoichiometry of the balanced chemical reaction to find how many grams of NH_{3} are produced:

Molar mass H_{2}O = 18\frac{g}{mol}

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36.0gH_{2}O*\frac{1molH_{2}O}{18gH_{2}O}*\frac{2molesNH_{3}}{6molesH_{2}O}*\frac{17gNH_{3}}{1molNH_{3}}=11.3gNH_{3}

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3 0
3 years ago
What are the properties of salt
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6 0
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A student pours 800.0 mL of a 3.000 molar solution of sodium hydroxide into a 2.00 liter volumetric flask and fills the flask up
hammer [34]
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5 0
2 years ago
The ka of hypochlorous acid (hclo) is 3.0 x 10-8 at . what is the % ionization of hypochlorous acid in a 0.015 m aqueous solutio
Ahat [919]

Answer is: the % ionization of hypochlorous acid is 0.14.

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Ka = [H⁺] · [ClO⁻] / [HClO].

[H⁺] is equilibrium concentration of hydrogen cations or protons.

[ClO⁻] is equilibrium concentration of hypochlorite anions.

[HClO] is equilibrium concentration  of hypochlorous acid.

Ka is the acid dissociation constant.

Ka(HClO) = 3.0·10⁻⁸.

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Ka(HClO) = α² · c(HClO).

α = √(3.0·10⁻⁸ ÷ 0.015).

α = 0.0014 · 100% = 0.14%.

5 0
3 years ago
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