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3 years ago

Hey lol hey lol hey lol hey lol

Chemistry

1 answer:

Elanso [62]3 years ago

5 0

Answer:

Hi!!!

Explanation:

Hello hope you are having a great day:)

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If the volume occupied by the gas molecules shown below were doubled, what would happen to the pressure they exert? (Assume cons

AnnZ [28]

there's no picture here but I guess the answer would be:

considering the constant temperature, if you double the volume, the pressure would be halved.

like: volume is 2, pressure is 4

if 2×2, then:4÷2

3 0

3 years ago

2. Which set of symbols represents isotopes of the same element?

MariettaO [177]

Answer:

so you can do the maths your self

Explanation:

isotopes are elements with he same proton number but difference in their neutron number

proton number is also known as the atomic number
mass number is the sum of the proton number and the neutron number
mass number = proton number + neutron number

7 0

3 years ago

(c) O2 gas is transferred from a 3 L vessel containing oxygen at 4 atm to an evacuated 20 L vessel at a constant temperature of

Fudgin [204]

Answer:

After the transfer the pressure inside the 20 L vessel is 0.6 atm.

Explanation:

Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:

$P x V = k$

Therefore, for this problem the step by step explanation is:

$P_{1} xV_{1} = P_{2} xV_{2}$

Clearing P2 and replacing

$P_{2}= \frac{P_{1} xV_{1}}{V_{2} } = \frac{4atmx3L}{20L} = 0.6atm$

7 0

3 years ago

5. How much heat (in calories) is absorbed by a reaction when

Wewaii [24]

Answer:

1.2 × 10⁴ cal

Explanation:

Given data

Mass of water (m): 300 g

Initial temperature: 80 °C

Final temperature: 40°C

We can calculate the heat released by the water ( $Q_w$ ) when it cools using the following expression.

$Q_w = c \times m \times (T_f - T_i)$

where

c is the specific heat capacity of water (1 cal/g.°C)

$Q_w = \frac{1cal}{g.\°C} \times 300g \times (40\°C - 80\°C) = -1.2 \times 10^{4} cal$

According to the law of conservation of energy, the sum of the heat released by the water ( $Q_w$ ) and the heat absorbed by the reaction ( $Q_r$ ) is zero.

$Q_w + Q_r = 0\\Q_r = -Q_w = 1.2 \times 10^{4} cal$

7 0

3 years ago

PLEASE HELP AGAIN lol thank you :))

Annette [7]

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8 0

3 years ago