Which of the following statement(s) are true about the atoms of any element?

- Ibrahim and Emmanuel have been investigating the amount of copper sulfate that

nignag [31]

The boys can arrive at the conclusion that as temperature increases, a greater mass of copper sulfate is dissolved in water.

<h3>What is an experiment?</h3>

An experiment is a scientific investigation that seeks to obtain cause and effect relationships. There must be a dependent and an independent variable in every experiment.

The average of the temperature readings is; 20 + 30 + 40 + 50 + 60 + 70 + 80/7 = 50

They used an interval of 10 degrees for the temperature readings. The boys should have kept the heating time the same. The independent variable in this experiment is the temperature. The boys can arrive at the conclusion that as temperature increases, a greater mass of copper sulfate is dissolved in water.

Learn more about dependent and independent variables: brainly.com/question/967776

8 0

2 years ago

What is the concentration of a solution of HCl in which a 10.0 mL sample of acid required 50.0 mL of 0.150 M NaOH for neutraliza

puteri [66]

Answer:

The answer is 0.75M HCl

Explanation:

To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:

C1V1 = C2V2

C1 = concentration of acid

C2 = concentration of base

V1 = volume of acid

V2 = volume of base

From the information supplied in the question:

concentration of acid (HCl) is the unknown

volume of acid (HCl) = 10ml

concentration of base (NaOH) = 0.15M

volume of base (NaOH) = 50ml

C1 x 10ml = 0.15M x 50ml

C1 x 10 = 7.5

divide both side by 10

C1 = 0.75M

concentration of acid (HCl) is 0.75M

5 0

3 years ago

If you add heat to a solid, the particles will... (choose all that apply) *

elena55 [62]

Answer: Expand and speeds up

Explanation:

5 0

3 years ago

Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.

Softa [21]

Explanation:

$PbBr_{2}$ will dissociate into ions as follows.

$PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)$

Hence, $K_{sp}$ for this reaction will be as follows.

$K_{sp} = [Pb^{2+}][Br^{-}]^{2}$

We take x as the molar solubility of $PbBr_{2}$ when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

$[Pb^{2+}]$ = $[Pb^{2+}]_{o}$ + x

$[Br^{-}]^{2}$ = $[Br^{-}]_{o}$ + 2x

So, equilibrium solubility expression will be as follows.

$K_{sp}$ = $([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}$

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains $[Br^{-}]_{o}$ = 0.10 and there will be no lead ions. So, $[Pb^{2+}]_{o}$ = 0

So, $[Br^{-}]_{o} + 2x$ will approximately equals to $[Br^{-}]_{o}$ .