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4 years ago

Q1.

Chemistry

1 answer:

valentinak56 [21]4 years ago

4 0

Answer:

4.49dm3

Explanation:

2NH4Cl + Ca(OH)2 —> CaCl2 + 2NH3 + 2H2O

First, we need to convert 10g of ammonium chloride to mole. This is illustrated below:

Molar Mass of NH4Cl = 14 + (4x1) + 35.5 = 53.5g/mol

Mass of NH4Cl = 10g

Number of mole = Mass /Molar Mass

Number of mole of NH4Cl = 10/53.5 = 0.187mol

From the equation,

2moles of NH4Cl produced 2 moles of NH3.

Therefore, 0.187mol of NH4Cl will also produce 0.187mol of NH3

Now we can obtain the volume of NH3 produced by doing the following:

1mole of any gas occupy 24dm3

Therefore, 0.187mol of NH3 will occupy = 0.187 x 24 = 4.49dm3

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In a solution of H2SO4, rank the concentration of H2SO4, H+, SO42- from highest concentration to lowest concentration.

Softa [21]

Answer:

[H⁺] > [SO₄²⁻] > [H₂SO₄]

Explanation:

H₂SO₄ is a strong acid, which means that most of it ionizes in aqueous solution.

Since it is a diprotic acid (two hydrogen ions) its ionization occurs in two steps:

H₂SO₄ (aq) → H⁺(aq) + HSO₄⁻(aq)

HSO₄⁻ (aq) → H⁺(aq) + SO₄²⁻(aq)

Thus, almost all H₂SO₄ has ionized and its final concentration is almost nothing.

After the first ionization, the conentrations of H⁺(aq) and HSO₄⁻ are equal but by the second ionization more H⁺ ions are produced along with SO₄⁻.

You can show it as one step dissociation, assuming 100% dissociation (given this is a strong acid):

By the stequiometry you can build this table:

H₂SO₄ (aq) → 2H⁺(aq) + SO₄²⁻(aq)

Initial A 0 0

Change - x +2x +x

Equilibrium A - x 2x x

As explained, A - x is very low, and 2x is twice x. Thus,

The rank of the concentrations from highest to lowest is:

[H⁺] > [SO₄²⁻] > [H₂SO₄]

5 0

3 years ago

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 their

iris [78.8K]

Answer:

Explanation:

We can obtain the value of x by doing the following:

Mass of hydrated salt (CuSO4.xH2O) = 1.50g

Mass of anhydrous salt (CuSO4) = 0.96g

Mass of water molecule(xH2O) = 1.50 — 0.96 = 0.54g

Molar Mass of CuSO4.xH2O = 63.5 + 32 + (16x4) + x(2 +16) = 63.5 + 32 + 64 + 18x = 159.5 + 18x

Mass of water(xH2O) molecules in the hydrate salt is given by:

xH2O/CuSO4.xH2O = 0.54/1.5

18x/(159.5 + 18x) = 0.36

Cross multiply to express in linear form

18x = 0.36 (159.5 + 18x)

18x = 57.42 + 6.48x

Collect like terms

18x — 6.48x = 57.42

11.52x = 57.42

Divide both side by 11.52

x = 57.42/11.52

x = 5

Therefore, the unknown integer x is 5 and the formula for the hydrated salt is CuSO4.5H2O

8 0

3 years ago

True or false: a mixture is always made up of a combination of elements

FromTheMoon [43]

A mixture has lots of different elements that are not necessarily bonded to each other, like sea water has lots of dirt, animals, and plant parts in it. Compared to a solution (strictly salt and water, which bond and ionize with each other).

8 0

3 years ago

What is the modern atomic model sometimes called?

zavuch27 [327]

Bohr model is the answer

5 0

3 years ago

Read 2 more answers

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

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

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

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

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

Read 2 more answers