Measure the volume of the gas in the syringe.

CAVA Chemistry 302/303B Unit 2 Lab Report

mafiozo [28]

1 property

2 procedure

Results

5 0

2 years ago

A balloon contains 2.0 L of air at 101.5 kPa . You squeeze the balloon to a volume of 0.25 L.

8_murik_8 [283]

<h3>Answer:</h3>

812 kPa

<h3>Explanation:</h3>

According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.

Mathematically, $P\alpha \frac{1}{V}$

At varying pressure and volume;

P1V1=P2V2

In this case;

Initial volume, V1 = 2.0 L

Initial pressure, P1 = 101.5 kPa

Final volume, V1 = 0.25 L

We are required to determine the new pressure;

$P2=\frac{P1V1}{V2}$

Replacing the known variables with the values;

$P2=\frac{(101.5)(2.0L)}{0.25L}$

= 812 kPa

Thus, the pressure of air inside the balloon after squeezing is 812 kPa

8 0

3 years ago

Nitrogen-13 has a half-life of about 10 minutes. How much of a 320g sample of N-13 would remain after 40 minutes?

lubasha [3.4K]

Answer:

The half-life of a radioisotope describes the amount of time it takes for said isotope to decay to one-half the original amount present in the sample.

Nitrogen-13, because it has a half-life of ten minutes, will experience two half-lives over the course of the twenty minute period. This means that 25% of the isotope will remain after this.

0.25 x 128mg = 32mg

32mg of Nitrogen-13 will remain after 20 minutes.

5 0

3 years ago

Based on the bond angles in CH4, NH3, and H2O, rank the magnitude of these repulsions. Rank from strongest to weakest repulsion.

morpeh [17]

Answer:

H2O> NH3> CH4

Explanation:

According to valence shell electron pair repulsion theory (VSEPR), bond angles and repulsion of electron pairs depends on the nature of electron pairs on the central atom of the molecule. Lone pairs cause more repulsion (and distortion of bond angles) than bond pairs). Lone pair- lone pair repulsion is greater than lone pair bond pair repulsion.

Water contains two lone pairs on oxygen hence it experiences the greatest repulsion. Ammonia has only one lone pair on nitrogen hence there is lesser repulsion between lone pairs and bond pairs. Methane possess only bond pairs of electrons hence it has the least repulsion.

4 0

3 years ago

How many moles of aqueous potassium ions and sulfate ions are formed when 63.7 g of K2SO4 dissolves in water

Viefleur [7K]

Answer:

WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.

Explanation:

Equation for the reaction:

K2SO4 + H20 ------->2 K+ + SO4^2-

When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.

1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-

At STP, 1 mole of K2SO4 will be the molar mass of the substance

Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol

Molar mass = 174 g/mol

So therefore;

1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion

When 63.7 g is used; we have:

174 g = 2 moles of K+

63.7 g = ( 63.7 * 2 / 174) moles of K+

= 0.73 moles of K+

Forr sulfate ion, we have:

174 g = 1 mole ofSO4^2-

63.7 g = (63.7 * 1 / 174) moles of SO4^2-

= 0.366 moles of SO4^2-

In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.

4 0

3 years ago