All categories

3 years ago

Air trapped in a cylinder fitted with a piston occupies 145 mL at 1.08 atm

Chemistry

1 answer:

pav-90 [236]3 years ago

7 0

Answer: 0.0014 atm

Explanation:

Given that,

Original pressure of air (P1) = 1.08 atm

Original volume of air (T1) = 145mL

[Convert 145mL to liters

If 1000mL = 1l

145mL = 145/1000 = 0.145L]

New volume of air (V2) = 111L

New pressure of air (P2) = ?

Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

P1V1 = P2V2

1.08 atm x 0.145L = P2 x 111L

0.1566 atm•L = 111L•P2

Divide both sides by 111L

0.1566 atm•L/111L = 111L•P2/111L

0.0014 atm = P2

Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm

You might be interested in

Suppose you have 1000 lb of waste copper (II) oxide sitting in the basement of your factory. Do you see a problem with sending i

Free_Kalibri [48]

Answer:

Yes.

Explanation:

Yes, we have a problem with sending it to a landfill of copper oxide because it has harmful effect on the health of humans as well as more weight of the copper oxide. Copper oxide usually found in powder form which can easily be inhaled that causes death of the cell due to toxic effect on the mitochondria and lysosomes of the cell. It makes problem of health in carrying the copper oxide from the basement of the factory to the landfill region due to its power form so we can say that it can do problems to human health while carrying from one place to another.

8 0

2 years ago

Many tests to distinguish aldehydes and ketones involve the addition of an oxidant. Only __________ can be easily oxidized becau

tatiyna

Answer:

Aldehydes
A hydrogen atom
Oxygen

Explanation:

Many tests to distinguish aldehydes and ketones involve the addition of an oxidant. Only aldehydes can be easily oxidized because there is a hydrogen atom next to the carbonyl and oxidation does not require oxygen

4 0

2 years ago

A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is

Charra [1.4K]

Answer:

$[H^{+}] = 0.761 \frac{mol}{L}$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

$pH = 0.119$

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

$n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)$

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

$n_{H^{+} } from HCl = (5.00)(0.093)$

$n_{H^{+} } from HCl = 0.465 mol$

$n_{H^{+} } from HNO_{3} = (8.00)(0.037)$

$n_{H^{+} } from HNO_{3} = 0.296 mol$

$n_{H^{+}(total) } = 0.296 + 0.465$

$n_{H^{+}(total) } = 0.761 mol$

For molar concentration of hydrogen ions:

$[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}$

$[H^{+}] = \frac{0.761}{1.00}$

$[H^{+}] = 0.761 \frac{mol}{L}$

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

$K_{w} = [H^{+} ][OH^{-} ]$

$[OH^{-}]=\frac{Kw}{[H^{+}] }$

$[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

The pH of the solution can be measured by the following formula:

$pH = -log[H^{+} ]$

$pH = -log(0.761)$

$pH = 0.119$

5 0

3 years ago

What is the density of a cube of water measuring 2cmX4cmX1cm, with a mass of 8g?

Vinil7 [7]

The actual formula for volume for a cube is the length multiplied by the width and then multiplied by the height. Since all three measurements are the same, the formula results in the measurement of one side cubed. For the example, 5^3 is 125 cm^3. Multiply the volume by the known density, which is the mass per volume.

7 0

2 years ago

The ideal gas constant, R has several different values that could be used. Which quantity causes these differences?

Vera_Pavlovna [14]

Answer:

The answer is A

Explanation:

6 0

2 years ago