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vampirchik [111]
3 years ago
12

How are a substance and a solution similar? how are they different?

Chemistry
1 answer:
inn [45]3 years ago
5 0
A substance and a solution have both uniform composition throughout. A substance is a matter that is uniform in composition and can be an element or compound. A solution is another name for homogeneous which is a mixture. The difference of substances and solution, a substance has a definite composition and a solution has a variable composition. It is also uniform in composition, but a solution is made by combining two or more substances.
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What volume of 0.0748 M perchloric acid can be neutralized with 115 mL of 0.244 M sodium hydroxide?
Irina18 [472]

Answer:

0.375 L

Explanation:

We know that at neutralization, the number of mol  of acid must equal the number of equivalents of base.

This is a reaction 1:1 acid to base:

HClO₄ + NaOH ⇒ NaClO₄ + H₂O

We re given the  moles  of the base indirectly since we know the volume and molarity. From there we can calculate the volume of HClO₄.

Moles NaOH = 0.115 L x 0.244 M = 0.115 L x 0.244 mol/L =0.028 mol

Thus we require 0.028 mol of HClO₄ in the pechloric acid solution:

Molarity = # moles / V ⇒ V = # moles / M

V = 0.028 mol / 0.0748 mol/L = 0.375 L

Note  that this problem can be solved in just one step since

M(HClO₄) x V(HClO₄) = M(NaOH) x V(NaOH)  ⇒

V(HClO₄) = M(NaOH) x V(NaOH) / M(HClO₄)

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3 years ago
True or False.There is a chemical difference between the gaseous and liquid state of the same substance.
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Read 2 more answers
Consider the hypothetical reaction 4A + 3B → C + 2D Over an interval of 3.00 s the average rate of change of the concentration o
pogonyaev

Answer:

Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M

Explanation:

4A + 3B ------> C + 2D

In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s

The amount of A that has reacted at the end of 3 seconds will be

0.08 × 3 = 0.24 M

Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.

From the chemical reaction,

4 moles of A gives 1 mole of C

0.24 M of reacted A will form (0.24 × 1)/4 M of C

Amount of C formed at the end of the 3s interval = 0.06 M

If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.

If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M

3 0
3 years ago
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