Make sure all units are consistent. Convert ml to L first. (500ml = 0.5L). Then divide the mole by the volume which would result to 1.6*10^-4mol/L. This can also be expressed as interms of molarity (M). 1 M = 1 mol/L hence the final answer is 1.6*10^-4 M.

8 0

4 years ago

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A lead ball is added to a graduated cylinder containing 30.8 ml of water, causing the level of the water to increase to 89.9 ml.

enot [183]

A lead ball added to a graduated cylinder containing 30.8 ml of water, causing the level of the water to increase to 89.9 ml, has a volume of 59.1 ml.

Volume is defined as the amount of space that matter occupies. There are three different ways to find the volume:

1) Volume by Space - by measuring its physical dimensions

2) Volume by Density and Mass - by dividing its weight by its density

3) Volume by Displacement - by measuring the volume displaced when immersed in a liquid or gas

To solve for the volume of the lead ball, we can use the displacement method by subtracting the initial volume from the final volume.

If the level of the water increased from 30.8 ml to 89.9 ml upon adding the lead ball, the displaced volume is:

volume = final volume - initial volume

volume = 89.9 - 30.8

volume = 59.1 ml

Hence, the volume of the lead ball, using displacement method, is 59.1 ml.

To learn more about volume by displacement: brainly.com/question/1945909

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3 0

2 years ago

All the following involve an increase in entropy except O most endothermic reactions O sublimation O vaporization O combustion

daser333 [38]

Answer:

most endothermic reactions

Explanation:

In endothermic reactions, the entropy of the environment decreases.

The entropy is the measure of the disorder of a specific system.

An endothermic reaction is every chemical reaction in which energy is absorbed.

In a comparison of an endodermic reaction in which energy is released in light or heat form.

5 0

4 years ago

If you complete and balance the following oxidation-reduction reaction in basic solution NO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(a

andreev551 [17]

Answer:

NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

There is 1 hydroxide ion, on the reactant side

Explanation:

NO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(aq)

Step 1: The half reactions

NO2- (aq) → NH3(g)

Al(s) → Al(OH)4-

Step 2: Balancing electrons

NO2- → NH3

On the left side N has an oxidation number of +3 and on the right side -3.

NO2- +6e-→ NH3

Al(s) → Al(OH)4-

On the left side, Al has an oxidation number of 0 and on the right side +3.

Al(s) → Al(OH)4- +3e-

To have the same amount of electrons transfered, we have to multiply the second reaction by 2

NO2- +6e-→ NH3

2(Al(s) → Al(OH)4- +6e-)

Step 3: Balance with OH/H2O

NO2- +6e +5H2O → NH3 +7OH-

2Al +8OH- → 2Al(OH)4- + 6e-

Step 4: The netto reaction

NO2- + 5H2O + 2Al + 8OH- → NH3 +7OH- + 2Al(OH)4-

NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

There is 1 hydroxide ion, on the reactant side

6 0

4 years ago