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Arturiano [62]
3 years ago
6

The pressure of a gas is 1.0 atm, the volume is 3.0 L, and the temperature is 200 K. A chemist changes one factor while keeping

another constant so that the new pressure is 2.0 atm. Which of the following could be the new conditions?
The final volume is 6.0 L, while temperature is kept constant.
The final temperature is 100 K, while volume is kept constant.
The final volume is 4.0 L, while temperature is kept constant.
The final temperature is 400 K, while volume is kept constant.
Chemistry
1 answer:
anygoal [31]3 years ago
4 0

Answer: fourth statement is correct. The final temperature is 400 K, while volume is kept constant.

Explanation: At constant temperature, volume is inversely proportional to the pressure and,

At constant volume, pressure is directly proportional to the kelvin temperature.

Initially the gas pressure is 1.0 atm, volume is 3.0 L and the temperature is 200 K.  The new pressure is 2.0 atm.

If we consider the temperature is constant, then new volume must be 1.5 L since the pressure is doubled and we know that volume is inversely proportional to the pressure.

So, as the pressure is doubled, the volume will be halved means it will decrease from 3.0 L to 1.5 L.

First statement is not correct since it says the new volume will be 6.0 L while temperature is kept constant.

If the volume is kept constant, the temperature must be doubled as the pressure has also doubled. So, new temperature will be 400 K.

second statement is also not correct as it says, the final temperature is 100 K, while volume is kept constant.

Third statement is also not correct as we already have seen that the new volume will be 1.5 L, while the temperature is kept constant.

Hence, the only and only correct statement is fourth as the final temperature will be 400 K(double of initial 200 K), while the volume is kept constant.

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Calculate the empirical formula for a compound containing 67.6% Hg 10.8% S 21.6% O
pashok25 [27]

Answer:

HgSO₄

Explanation:

% => g => moles => ratio => reduce => empirical ratio

%Hg = 67.6% => 67.6g/201g/mol = 0.34mol

%S    = 10.8% => 10.8g/32g/mol = 0.34mol

%O   = 21.6% => 21.6g/16g/mol = 1.35mol

Hg:S:O => 0.34:0.34:1.35

Reduce to whole number ratio by dividing by the smaller mole value...

Hg:S:O => 0.34/.34:0.34/.34:1.35/.34 => Empirical Ratio = 1:1:4

∴ Empirical Formula is HgSO₄

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3 years ago
What is true about asteroids
Igoryamba

Answer:

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Which laws can be combined to form the ideal gas law?
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Option 3- Avogadro's, Charles's and Boyle's


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3 years ago
When one mole of a solute is present in 500 cm3 of solution, then the concentration of the solution is:
Schach [20]

Answer:

0.5M is the answer.

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Using unitary method,

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7 0
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What is the oxidation number of cl− in the hypochlorite ion clo−?
mariarad [96]

Answer : The oxidation number of chlorine (Cl) is, (+1)

Explanation :

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Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

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The oxidation number of a monatomic ion equals the charge of the ion.

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The oxidation number of  oxygen (O)  in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

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The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

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Therefore, the oxidation number of chlorine (Cl) is, (+1)

5 0
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