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2 years ago

5

If 45.0 mL of a 4.00 M sodium sulfate (Na2SO4) solution is used for a reaction, how many moles of solute were present in the sol

ution?
____ moles

1 answer:

mixas84 [53]2 years ago

6 0

797 jenehsbsehshehehbehshehhebwbbss

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a. Calculate the pH of a solution that has [H3O+]= 4.3 x 10-5 M. b. Is the solution acidic or basic? How do you know?

Luda [366]

Answer:

* $pH=4.37$

* The solution is acidic since the pH is below 7.

Explanation:

Hello,

In this case, we can mathematically define the pH by:

$pH=-log([H_3O^+])$

Thus, for the given hydronium concentration we simply compute the pH:

$pH=-log(4.3x10^{-5})=4.37$

Thereby, we conclude the solution is acidic due to the fact that the pH is below 7 which is the neutral point and above it the solutions are basic.

Regards.

3 0

3 years ago

Mario uses a hot plate to heat a beaker of 50mL of water. He used a thermometer to measure the

mojhsa [17]

Answer:

Mario uses a hot plate to heat a beaker of 50mL of water. He used a thermometer to measure the

temperature of the water. The water in the beaker began to boil when it reached the temperature of

100'C. If Mario completes the same experiment with 25mL of water, what would happen to the boiling

point?

a) The water will not reach a boil.

b) The boiling point of water will increase.

c) The boiling point of water will decrease.

d) The boiling point of water will stay the same.

Explanation:

6 0

3 years ago

A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following

marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

$\frac{V_1}{n_1} =\frac{V_2}{n_2}$

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}$

<em>V₂ = 4,00L</em>

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}$

<em>V₂ = 16,00L</em>

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

$\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}$

<em>V₂ = 12,31L</em>

I hope it helps!

6 0

4 years ago

Balance each of the following redox reactions occurring in basic solution.MnO−4(aq)+Br−(aq)→MnO2(s)+BrO−3(aq)Express your answer

Ahat [919]

Answer : The balanced chemical equation is,

$2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)$

Explanation :

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.
Now balance the main atoms in the reaction.
Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion $(OH^-)$ at that side where the less number of hydrogen are present.
Now balance the charge.

The half reactions in the basic solution are :

Reduction : $MnO_4^-(aq)+2H_2O(l)+3e^-\rightarrow MnO_2(s)+4OH^-(aq)$ ......(1)

Oxidation : $Br^-(aq)+6OH^-(aq)\rightarrow BrO_3^-(aq)+3H_2O(l)+6e^-$ .......(2)

Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)$

8 0

3 years ago

When an acid reacts with a strong base which product always forms

patriot [66]

Answer:طيزي

Explanation:

4 0

3 years ago