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3 years ago

For a titration of 25 ml of naoh solution, 23.30 ml of hcl solution if used. what is the molarity of naoh

Chemistry

1 answer:

Eduardwww [97]3 years ago

3 0

HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

As you can see here, one mole of acid neutralizes one mole of base.

We use the concentration equation, which states that,

c=nv

n is the number of moles v is the volume of solution

Rearranging for moles, we get,

n=c⋅v

So, we have:

nNaOH=0.1 M⋅0.05 L

=0.005 mol

Since one mole of acid neutralizes one mole of base, then we must have: nHCl=nNaOH.

And so,

cHCl=nHClvHCl

=0.005 mol0.03 L

≈0.17 M

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Balance the following reaction in KOH (under basic conditions). What are the coefficients in for C3H8O2 and KMnO4 in the balance

GrogVix [38]

Answer:

Coefficient of $C_3H_8O_2=3$

Coefficient of $KMnO_4$ =8

Explanation:

We are given that a reaction in which $C_3H_8O_2$ reacts with $KMnO_4$

We have to find the coefficient of each reactants in balanced reaction

$3C_3H_8O_2(aq)+8KMnO_4(aq)\rightarrow 3C_3H_2O_4K_2(aq)+8MnO_2(aq)+2KOH+8H_2O$

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Coefficient of $C_3H_8O_2$ =3

Coefficient of $KMnO_4=8$

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Coefficient of $MnO_2=8$

Coefficient of $H_2O=8$

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3 years ago

How does Hanukkah relate to science?

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2 years ago

Please help !!

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Answer:

For 2: The % yield of the product is 92.34 %

For 3: 12.208 L of carbon dioxide will be formed.

Explanation:

For 2:

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(1)

Given values:

Actual value of the product = 78.4 g

Theoretical value of the product = 84.9 g

Plugging values in equation 1:

$\% \text{yield}=\frac{78.4 g}{84.9g}\times 100\\\\\% \text{yield}=92.34\%$

Hence, the % yield of the product is 92.34 %

For 3:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(2)

Given mass of carbon dioxide = 24 g

Molar mass of carbon dioxide = 44 g/mol

Plugging values in equation 1:

$\text{Moles of carbon dioxide}=\frac{24g}{44g/mol}=0.545 mol$

At STP conditions:

1 mole of a gas occupies 22.4 L of volume

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2 years ago

In the reaction 2H2(g) + O2(g) ® 2H2O(g), what is the volume ratio of H2 to H2O?

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3 years ago