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3 years ago

A sample of oxygen occupies a volume of 250mL at 114 kPa. What pressure is required to change the volume to 593 mL?

Chemistry

1 answer:

rewona [7]3 years ago

6 0

Answer:

48.1 kPa

Explanation:

Data Given

initial volume V1 of oxygen gas = 250 mL

initial pressure P1 of oxygen gas = 114 kPa

final volume V2 of oxygen gas = 593 mL

final pressure P2 oxygen gas = ?

Solution:

This problem will be solved by using Boyle's law equation at constant Temperature.

The formula used

P1V1 = P2V2

As we have to find out Volume, so rearrange the above equation

P2 = P1V1 / V2

Put value from the data given

P2 = 114 kPa x 250 mL / 593 mL

P2 = 48.1 kPa

So the final pressure required = 48.1 kPa

You might be interested in

If 10.57 g of magnesium reacts completely with 6.96 g of oxygen, what is the percent by mass of oxygen in magnesium oxide? Round

Deffense [45]

Answer:

39.7 %

Explanation:

magnesium + oxygen ⟶ magnesium oxide

10.57 g 6.96 g 17.53 g

According to the Law of Conservation of Mass, the mass of the product must equal the total mass of the reactants.

Mass of MgO = 10.57 + 6.96

Mass of MgO = 17.53 g

The formula for mass percent is

% by mass = Mass of component/Total mass × 100 %

In this case,

% O = mass of O/mass of MgO × 100 %

Mass of O = 6.96 g

Mass of MgO = 17.53 g

% O = 6.96/17.53 × 100

% O = 0.3970 × 100

% O = 39.7 %

5 0

4 years ago

If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did

user100 [1]

Answer:

4.5 L water we have in litres (L).

Explanation:

$Q=m\times c \times \Delta T$

where

$\Delta T$ = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0 cal/(g℃))

m is the mass of water

Plugging in the values

$\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$$