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3 years ago

11

2. How is the solute being dissolved in the given solvent?

1 answer:

galben [10]3 years ago

8 0

At a definite amount of solute and at a definite tempreture it is dissolved in utane above that it remain unsoluble

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What did the scientists who founded the royal society of London share with Lavoisier

morpeh [17]

They were natural philosophers that used experimentation to establish the cause effect relations and explanations, in an epoque when many "theories' were not supported by experiments but by some authority (antique philosophers like Aristoteles or the Catholic Church).

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3 years ago

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What is the Name of RbC2H3O2

Semenov [28]

Answer:

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Explanation:

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3 years ago

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A volume of 40.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2

Anni [7]

<u>Answer:</u> The concentration of KOH solution is 1.215 M

<u>Explanation:</u>

For the given chemical equation:

$2KOH(aq.)+H_2SO_4(aq.)\rightarrow K_2SO_4(aq.)+2H_2O(l)$

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=1.50M\\V_1=16.2mL\\n_2=1\\M_2=?M\\V_2=40.0mL$

Putting values in above equation, we get:

$2\times 1.50\times 16.2=1\times M_2\times 40.0\\\\M_2=\frac{2\times 1.50\times 16.2}{1\times 40.00}=1.215M$

Hence, the concentration of KOH solution is 1.215 M

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4 years ago

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Explanation:

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