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1 year ago

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A solution has a concentration of 0.6mol/dm³. If a container of this solution holds 3 moles of solute, what volume of solution i

s in the container?

1 answer:

FinnZ [79.3K]1 year ago

6 0

Answer: 3 moles solute x 1 dm^3/0.60 moles solute = 5 dm^3

Explanation:

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A sample of a compound of Cl and O reacts with an excess of H2 to give 0.233g of HCl and 0.403g of H2O. Determine the empirical

Vsevolod [243]

Answer:

Cl₂O₇

Explanation:

For the reaction:

ClₓOₙ + H₂ → HCl + H₂O

Moles of HCl and moles of H₂O are:

HCl: 0.233g HCl ₓ (1mol / 36.46g) = 6.39x10⁻³ mol HCl

H₂O: 0.403g H₂O ₓ (1mol / 18.02g) = 2.236x10⁻² mol H₂O

As you can see, moles of HCl are equivalent to moles of Cl in the compound and moles of H₂O are equivalent to moles of O in the compound, that means:

6.39x10⁻³ mol Cl

2.236x10⁻² mol O

Empirical formula is the simplest ratio of atoms presents in a molecule. If Cl is <em>1</em>, Oxygen will be:

2.236x10⁻² mol / 6.39x10⁻³ = <em>3.5</em>

As empirical formula must be given in natural numbers, the empirical formula is:

<em>Cl₂O₇</em>

<em></em>

4 0

2 years ago

Read 2 more answers

Which of the following is not a nanoscale object?

KiRa [710]

If I am correct I would believe that it would be a muscle cell.

6 0

1 year ago

Read 2 more answers

Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.

Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$C(s)+2H_2(g)\rightarrow CH_4(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ$

Hence, the enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

3 0

1 year ago

You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain

IrinaVladis [17]

Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of $O_2$ :-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{49.8\ g}{32\ g/mol}$

$Moles_{O_2}= 1.55625\ mol$

Since pressure and volume are constant, we can use the Avogadro's law as:-

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

Given ,

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

$\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}$

n₁ = 0.778125 moles

Moles of carbon dioxide = 0.778125 moles

Molar mass of $CO_2$ = 44.0 g/mol

Mass of $CO_2$ = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>

5 0

1 year ago

The density of silver metal bar is 10.5 g/cm3 and it occupies a volume space of 0.5l. What is its mass in kilograms (kg).

mojhsa [17]

Density is equal to the ratio of mass to the volume.

The mathematical expression is given as:

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

Density of silver metal bar= $10.5 g/cm^{3}$

Convert $g/cm^{3}$ into g/L

$1 cm^{3}$ = 0.001 L

Thus, density = $\frac{10.5}{10^{-3}} g/L$

= $10.5\times 10^{3}g/L$

Volume = 0.5 L

Put the values,

$10.5\times 10^{3}g/L= \frac{mass}{0.5 L}$

$m= 10.5\times 10^{3}g/L\times 0.5 L$

= $5.25 \times 10^{3}g$

Now, convert gram into kg

1 g = 0.001 kg

Therefore, mass in kg= $\frac{5.25 \times 10^{3}}{10^{-3}}$

= 5.25 kg

Thus, mass of silver metal bar in kg=5.25 kg

4 0

1 year ago