All categories

3 years ago

If water is added to a 0.70 molar solution of CuSO4 what will change

Chemistry

1 answer:

lana [24]3 years ago

7 0

Answer:

B. Molarity will decrease

Explanation:

Molarity is one of the measures of the molar concentration of a solution. It is calculated by dividing the number of moles of the solute by the volume of the solvent. This means that the higher the amount of solute in relation to the volume of solvent, the higher the molarity of that solution.

In essence, adding water to a solution dilutes it i.e it increases the solvent's volume in relation to the solute, causing the molarity to decrease. In a nutshell, diluting a solution (by adding water or more solvent) causes the molarity of such solution to decrease. For example, if water is added to a 0.70 molar solution of CuSO4, the molarity of the solution will DECREASE.

You might be interested in

which of the following describes a similarity between a liquid and gas A. both are forms of heat are used to warm objects. C. bo

Leona [35]

Answer:

It's A

Explanation:

Liquid can heat things and so can steam.

8 0

3 years ago

Calculate the molar mass for S206[g).

KiRa [710]

Answer:

6605.3900 g/

Explanation:

3 0

3 years ago

What is the change in density if a sample goes from 3.21 g/L to 5.43 g/mL?

Step2247 [10]

Answer:

$\Delta \rho =2.22 g/mL$

Explanation:

Hello,

In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:

$\Delta \rho=\rho _f-\rho _i\\ \\\Delta \rho=5.43g/mL-3.21g/mL\\\\\Delta \rho =2.22 g/mL$

Best regards.

3 0

3 years ago

In Part B the given conditions were 1.00 mol of argon in a 0.500-L container at 18.0°C. You identified that the ideal pressure (

Natasha2012 [34]

Answer:

4,38%

small molecular volumes

Decrease

Explanation:

The percent difference between the ideal and real gas is:

(47,8atm - 45,7 atm) / 47,8 atm × 100 = 4,39% ≈ 4,38%

This difference is considered significant, and is best explained because argon atoms have relatively small molecular volumes. That produce an increasing in intermolecular forces deviating the system of ideal gas behavior.

Therefore, an increasing in volume will produce an ideal gas behavior. Thus:

If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to decrease

I hope it helps!

5 0

3 years ago

Calculate the entropy change for the reaction: Fe2O3(s) +3C(s) -> 2Fe(s) + 3CO(g)Entropy data:Fe2O3(s): 90 J/K molC(s): 5.7 J

Annette [7]

Explanation:

We are given: entropy of Fe2O3 = 90J/K.mol

: entropy of C = 5.7J/K.mol

: entropy of Fe = 27.2J/K.mol

: entropy of CO = 198J/K.mol

$\begin{gathered} \Delta S\text{ = S}_{products}-S_{reactants} \\ \\ \text{ = \lparen3}\times198+2\times27.2)-(3\times5.7+90) \\ \\ \text{ = 541.3J/K.mol} \end{gathered}$

Answer:

The correct answer is C.

5 0

1 year ago