Ask question

Ask question

All categories

2 years ago

14

Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate ∙ x-hydrate (CuSO4 ∙ xH2O), where x is an integer. Part of t

heir practical exam is to determine this integer x. They are working in pairs, though Hermione is doing most of the work. This should be discouraged! After dehydration they find that they are left with 0.96 g of the an-hydrate CuSO4. What is the unknown integer x? Round the answer to the nearest integer.

1 answer:

Gwar [14]2 years ago

7 0

Answer

5

Explanation:

We can go about this using the percentage compositions.

First, we calculate the percentage composition of the copper sulphate. This is obtainable by using the mass.

0.96/1.5 * 100 = 64%

Hence the percentage by mass of the water present is 36%

The molar mass of the anhydrous sulphate is 64 + 32 +4(16) = 160g/mol

The molar mass of the water is 2(1) + 16 = 18g/mol

Not forgetting that it is in multiples of x, the total molar mass of the water is 18x moles

The total mass of the copper sulphate hydrate is 160+ 18x

Now how do we get x? Like it is said earlier, the percentage composition is constant.

Hence, 64/100 * (160 + 18x) = 160

16000 = 64(160 + 18x)

16000 = 10,240 + 1152x

16,000 - 10,240 = 1152x

1152x = 5760

x = 5760/1152

x = 5

You might be interested in

A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this p

xxTIMURxx [149]

Answer:

Limiting reagent = lead(II) nitrate

Theoretical yield = 3.75435 g

% yield = 65.26 %

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium chloride :

Molarity = 1.20 M

Volume = 25.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of potassium chloride :

$Moles=1.20 \times {25.0\times 10^{-3}}\ moles$

<u>Moles of potassium chloride = 0.03 moles</u>

For lead(II) nitrate :

Molarity = 0.900 M

Volume = 15.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=0.900 \times {15.0\times 10^{-3}}\ moles$

<u>Moles of lead(II) nitrate = 0.0135 moles</u>

According to the given reaction:

$2KCl_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbCl_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium chloride react with 1 mole of lead(II) nitrate

1 mole of potassium chloride react with 1/2 mole of lead(II) nitrate

0.03 moles potassium chloride react with 0.03/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.015 moles

<u>Limiting reagent is the one which is present in small amount. Thus, lead(II) nitrate is limiting reagent. (0.0135 < 0.015)</u>

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) nitrate gives 1 mole of lead(II) chloride

0.0135 mole of lead(II) nitrate gives 0.0135 mole of lead(II) chloride

Molar mass of lead(II) chloride = 278.1 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.0135 × 278.1 g = 3.75435 g

<u>Theoretical yield = 3.75435 g</u>

Given experimental yield = 2.45 g

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (2.45/3.75435) × 100 = 65.26 %</u>

6 0

2 years ago

For double-helix formation, change in Gibbs free energy, ΔG, can be measured to be −54 kJ⋅mol−1 (−13 kcal⋅mol−1) at pH 7.0 in 1

77julia77 [94]

Answer:

Explanation:

Entropy change in the system : --

ΔG = −54 kJ⋅mol−1 (−13 kcal⋅mol−1) = −54 kJ⋅mol−1 (−13 x 4.2 kJ⋅mol−1)

= - 108.6 KJ / mol

ΔH = -251 kJ/mol (-60 kcal/mol) = -251 kJ/mol (-60 x 4.2 kJ/mol)

= - 503 KJ / mol

ΔG = ΔH - TΔS

ΔS = ( ΔH - ΔG ) / T

= - 503 + 108.6 / ( 273 + 25 ) KJ / mol k⁻¹

= - 1323.48 J / mol k⁻¹

Entropy change in the surrounding

+ 1323.48 J / mol k⁻¹

7 0

2 years ago

What mass of oxygen reacts when 84.9 g of iron is consumed in the following reaction: Fe+O2= Fe2O3

konstantin123 [22]

First we will calculate the number of moles of Iron:
$n = \frac{m}{M}
$ , where n is the number of moles, m is the mass of iron in the reaction and M is the Atomic weight.
$n= \frac{84.9}{55.845} = 1,52$ moles of Iron.
The same number of moles of Oxygen will take part in the reaction.
So $1,52= \frac{mOxygen}{32}$ where 32 is the Atomical Weight of Oxygen (16 x 2).
=> $mOxygen=32*1,52=48,64$ g

6 0

2 years ago

Read 2 more answers

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

Kisachek [45]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $C_9H_{12}O$ and $C_{18}H_{24}O_2$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.01g$

Mass of $H_2O=10.65g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.01 g of carbon dioxide, $\frac{12}{44}\times 39.01=10.64g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 10.65 g of water, $\frac{2}{18}\times 10.65=1.18g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.64 + 1.18) = 1.6 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.64g}{12g/mole}=0.886moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.18g}{1g/mole}=1.18moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.6g}{16g/mole}=0.1moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.1 moles.

For Carbon = $\frac{0.886}{0.1}=8.86\approx 9$

For Hydrogen = $\frac{1.18}{0.1}=11.8\approx 12$

For Oxygen = $\frac{0.1}{0.1}=1.99\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 12 : 1

The empirical formula for the given compound is $C_9H_{12}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 272.38 g/mol

Mass of empirical formula = 136 g/mol

Putting values in above equation, we get:

$n=\frac{272.38g/mol}{136g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 9)}H_{(2\times 12)}O_{(2\times 1)}=C_{18}H_{24}O_2$

Hence, the empirical and molecular formula for the given organic compound is $C_9H_{12}O$ and $C_{18}H_{24}O_2$

6 0

2 years ago

Why is it important to have regular supervision of the weight and measurements in the market

Sergio [31]

Answer:

Supervision of weights and measures promotes accurate measurements of goods and services to ensure that everybody gets a fair trade in the marketplace. Not so coincidentally it also is a deterrent to ensure that traders are being honest in their trade practises.

Explanation:

4 0

2 years ago