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3 years ago

I am a teacher in a college in a large city in China. One day, some students asked me to join

Chemistry

1 answer:

IrinaVladis [17]3 years ago

8 0

Answer:

c or b

Explanation:

the story is entertaining and it shares the teachers personal experiences with the teachers students

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What type of matter is salt water?

algol13

Salt water is a Homogeneous Mixture......

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3 years ago

How does the arrangement of the particles in a gas compare to that of a solid

blondinia [14]

The particles of the gases are apart from each other and they are movingoing freely, However, in the solid the particles are very close to each other and there is no spaces between them,

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3 years ago

calculate the volume of carbon dioxide of room temperature and pressure obtained from 30 grams of glucose

ivanzaharov [21]

Answer:

Explanation:

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3 years ago

A chemist titrates 220.0 mL of a 0.1917M propionic acid (HC2H5CO2) solution with 0.1787 M KOH solution at 25Â°C. Calculate the p

natima [27]

Answer : The pH at equivalence is, 9.08

Explanation : Given,

Concentration of $HC_2H_5CO_2$ = 0.1917 M

Volume of $HC_2H_5CO_2$ = 220.0 mL = 0.220 L (1 L = 1000 mL)

First we have to calculate the moles of $HC_2H_5CO_2$

$\text{Moles of }HC_2H_5CO_2=\text{Concentration of }HC_2H_5CO_2\times \text{Volume of }HC_2H_5CO_2$

$\text{Moles of }HC_2H_5CO_2=0.1917M\times 0.220L=0.0422$

As we known that at equivalent point, the moles of $HC_2H_5CO_2$ and KOH are equal.

So, Moles of KOH = Moles of $HC_2H_5CO_2$ = 0.0422 mol

Now we have to calculate the volume of KOH.

$\text{Volume of }KOH=\frac{\text{Moles of }KOH}{\text{Concentration of }KOH}$

$\text{Volume of }KOH=\frac{0.0422mol}{0.1787M}$

$\text{Volume of }KOH=0.00754$

Total volume of solution = 0.220 L + 0.00754 L = 0.22754 L

Now we have to calculate the concentration of KCN.

The balanced equilibrium reaction will be:

$HC_2H_5CO_2+KOH\rightleftharpoons C_2H_5CO_2K+H_2O$

Moles of $C_2H_5CO_2K$ = 0.0422 mol

$\text{Concentration of }C_2H_5CO_2K=\frac{0.0422mol}{0.22754L}=0.1855M$

At equivalent point,

$pH=\frac{1}{2}[pK_w+pK_a+\log C]$

Given:

$pK_w=14\\\\pK_a=4.89\\\\C=0.1855M$

Now put all the given values in the above expression, we get:

$pH=\frac{1}{2}[14+4.89+\log (0.1855)]$

$pH=9.08$

Therefore, the pH at equivalence is, 9.08

6 0

3 years ago

When iron(III) oxide reacts with hydrochloric acid, iron(III) chloride and water are formed. How many grams of iron(III) chlorid

Aleksandr [31]

Answer: The mass of iron (III) chloride produced is 14.81 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For iron(III) oxide:

Given mass of iron(III) oxide = 10.0 g

Molar mass of iron(III) oxide = 159.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) oxide}=\frac{10.0g}{159.7g/mol}=0.0626mol$

For hydrochloric acid:

Given mass of hydrochloric acid = 10.0 g

Molar mass of hydrochloric acid = 36.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrochloric acid}=\frac{10.0g}{36.5g/mol}=0.274mol$

The chemical equation for the reaction of iron (III) oxide and hydrochloric acid follows:

$Fe_2O_3+6HCl\rightarrow 2FeCl_3+3H_2O$

By Stoichiometry of the reaction:

6 moles of hydrochloric acid reacts with 1 mole of iron (III) oxide

So, 0.274 moles of hydrochloric acid will react with = $\frac{1}{6}\times 0.274=0.0456mol$ of iron (III) oxide

As, given amount of iron (III) oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

6 moles of hydrochloric acid produces 2 moles of iron (III) chloride

So, 0.274 moles of hydrochloric acid will produce = $\frac{2}{6}\times 0.274=0.0913moles$ of iron (III) chloride

Now, calculating the mass of iron (III) chloride from equation 1, we get:

Molar mass of iron (III) chloride = 162.2 g/mol

Moles of iron (III) chloride = 0.0913 moles

Putting values in equation 1, we get:

$0.0913mol=\frac{\text{Mass of iron (III) chloride}}{162.2g/mol}\\\\\text{Mass of iron (III) chloride}=(0.0913mol\times 162.2g/mol)=14.81g$

Hence, the mass of iron (III) chloride produced is 14.81 grams

7 0

3 years ago