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3 years ago

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes.

Chemistry

1 answer:

MAVERICK [17]3 years ago

7 0

Answer:

a. Ca2+ 1M ; NO3- 2M

b. Na+ 4M ; SO4-2 2M

c. NH4+ , Cl- 0,186M

d. K+ 0,056M ; PO4-3 0,0188 M

Explanation:

Here are the ionization equations

Ca(NO3)2 ----> Ca2+ + 2NO3-

Na2SO4 ------> 2Na+ + SO4-2

NH4Cl ------> NH4+ + Cl-

K3PO4 -----> 3K+ + PO4-3

Concentrations must be in Molarity (moles/L)

Ca(NO3)2 0,1 mole in 100mL

100mL = 0,1 L

Molarity Ca(NO3)2 1M (0,1 /0,1)

Ca(NO3)2 ----> Ca2+ + 2NO3-

1 M 1M 2M

Na2SO4 2,5 moles in 1,25L

Molarity Na2SO4 2,5 moles /1,25L = 2M

Na2SO4 ------> 2Na+ + SO4-2

2M 4M 2M

5.00 g of NH4Cl in 500mL

We have to resort to the molar mass

Let's make the rule of three

500mL ______ 5g

1000mL _____ (1000mL . 5g) / 500mL = 10 g

(These are the grams in 1L, so with the molar mass, we get the moles)

Mass / Molar mass = moles --> 10g / 53,5 g/m = 0.186 moles

NH4Cl ------> NH4+ + Cl-

0,186 M 0,186M + 0,186M

1.00 g K3PO4 in 250.0 mL

(The same as before)

250 mL ________1 g

1000 mL _______ (1000 mL . 1 g) /250mL = 4 g

These are the grams in 1L, so with the molar mass, we get the moles

Mass / Molar mass = moles --> 4g /212,26 g/m = 0,0188 moles

K3PO4 -----> 3K+ + PO4-3

0,0188 M 0,056M + 0,0188 M

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Answer:

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Explanation:

Given data:

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3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th

WINSTONCH [101]

Answer: The standard enthalpy change of the reaction is coming out to be -16.3 kJ

Explanation:

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 $\Delta H$

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

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)$

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

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

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