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4 years ago

How many grams of solute are present in 275 ml of a 12.5% (m/v) solution of methanol?

Chemistry

1 answer:

butalik [34]4 years ago

5 0

Answer:

34.375 grams of solute are present in 275ml of a 12.5%(m/v) solution of methanol.

Explanation:

12.5% methanol solution which means

100ml solution contain 12.5 grams methanol

1 ml solution contain 12.5/100 grams methanol

275 ml solution contain 12.5/100*275

= 34.375 grams of methanol .

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Describe the relationship between volume and temperature of an ideal gas

kvv77 [185]

Answer:

Explanation:

Here, we want to describe the relationship between the volume and temperature of an ideal gas

This relationship is defined by Charles' law

From this law, we know that the volume of a given mass of gas is directly proportional to its temperature at a fixed pressure

What this means is that as long as the pressure remains unchanged, when the volume increases, the temperature increases, and when the volume decreases, the temperature decreases

These can be represented by the mathematical formula below:

$\frac{V_1}{T_1}\text{ = }\frac{V_2}{T_2}$

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2 years ago

Which noble gas is closer to magnesium

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3 years ago

Which polymer is a biopolymer?

kodGreya [7K]

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3 years ago

Read 2 more answers

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h

statuscvo [17]

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

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3 years ago

2NaOH+CO2----->Na2CO3+H2O

kondaur [170]

Answer:

13.2 g Na2CO3

Explanation:

Convert 10.0 g NaOH to mol.

10.0 g x 1 mol/39.997 g = 0.250 mol

Use mol ration given by the equation: 2 mol NaOH to 1 mol Na2CO3

0.250 mol NaOH x 1 mol Na2CO3/2 mol NaOH = 0.125 mol Na2CO3

Finally, convert the moles of Na2CO3 to grams.

0.125 mol Na2CO3 x 105.99 g/1 mol = 13.2 g

8 0

3 years ago