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3 years ago

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A gas has a volume of 100 mL and a pressure of 6 atm. What volume will the gas occupy when the pressure is reduced to 1 atm (tot

al)?

1 answer:

serious [3.7K]3 years ago

8 0

<u>We are given:</u>

V1 = 100 mL P1 = 6 atm

V2 = x mL P2 = 1 atm

<u>Solving for 'x' : </u>

According to the Boyle's law:

P ∝ 1/V (pressure and volume are inversely proportional)

PV = k (where k is a constant)

since the constant k will be the same:

P1V1 = P2V2

replacing the variables

6 * 100 = 1 * x

x = 600 mL OR 0.6L

Therefore, the gas will have a volume of 600mL or 0.6L

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After he finished his work, Mendeleev predicted _____.

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Answer:

the properties of three undiscovered elements is the correct answer.

Explanation:

After he completed his work Mendeleev predicted the properties of the undiscovered elements.

In the year 1871, Dmitri Mendeleev predicted the detailed existence and the properties of the three elements that are undiscovered and he gained tremendous fame.

He left the gap in the table to put the elements which are not identified at that time and by glancing at the physical and chemical properties of an element next to the gap, Mendeleev could predict the properties of the not discovered elements.

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2 years ago

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2 years ago

Read 2 more answers

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

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3 years ago

To determine the freezing point depression of a LiCl solution, Toni adds 0.411 g of LiCl to the sample test tube along with 19.7

Cerrena [4.2K]

Answer:

LiCl = 0.492 m

Explanation:

Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.

Our solute is lithium chloride, LiCl.

Our solvent is distilled water.

We do not have the mass of water, but we know the volume, so we should apply density to determine mass.

Density = mass / volume

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We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg

Let's determine the moles of LiCl

0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles

Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m

7 0

2 years ago