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3 years ago

How many liters of 3.5 M solution can be made using 23 moles of LiBr must show work to get credit

Chemistry

1 answer:

MrMuchimi3 years ago

5 0

Answer: 6.57 L of solution can be made.

Explanation:

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$ .....(1)

Given values:

Molarity of LiBr = 3.5 M

Moles of LiBr = 23 moles

Putting values in equation 1, we get:

$3.5mol/L=\frac{23mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{23mol}{3.5mol/L}=6.57L$

Hence, 6.57 L of solution can be made.

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Which compound will form from the bonding of CA and N ca3N2, Ca2N3, Ca2N2, caN

Paladinen [302]

Answer:

Ca3N2

Explanation:

Ca+2

N-3

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3 years ago

Standard heats of formation for reactants and products in the reaction below are provided. 2 HA(aq) + MX2(aq) → MA2(aq) + 2 HX(l

Ira Lisetskai [31]

Answer : The standard enthalpy of reaction is, -318.618 kJ

Explanation :

Enthalpy change : It 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:

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

The equilibrium reaction follows:

$2HA(aq)+MX_2(aq)\rightleftharpoons MA_2(aq)+2HX(l)$

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

$\Delta H^o_{rxn}=[(n_{(MA_2)}\times \Delta H^o_f_{(MA_2)})+(n_{(HX)}\times \Delta H^o_f_{(HX)})]-[(n_{(HA)}\times \Delta H^o_f_{(HA)})+(n_{(MX_2)}\times \Delta H^o_f_{(MX_2)})]$

We are given:

$\Delta H^o_f_{(HA(aq))}=-357.05kJ/mol\\\Delta H^o_f_{(MX_2(aq))}=69.602kJ/mol\\\Delta H^o_f_{(MA_2(aq))}=-63.958kJ/mol\\\Delta H^o_f_{(HX(l))}=-449.579kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times -63.958)+(2\times -449.579)]-[(2\times -357.05)+(1\times 69.602)]=-318.618kJ$

Thus, the standard enthalpy of reaction is, -318.618 kJ

6 0

3 years ago

50.0 g N204 (92.02 g/mol) react with 45.0 g N2H4 (32.05 g/mol) forming nitrogen gas, N2

IrinaVladis [17]

Answer:

The excess reactant is N2H4 and the leftover mass is 10.17g.

Explanation:

Step 1:

The balanced equation for the reaction.

N2O4 + 2N2H4 —> 3N2 + 4H2O

Step 2

Determination of the masses of N2O4 and N2H4 that reacted from the balanced equation:

Molar mass of N2O4 = 92.02 g/mol

Mass of N2O4 from the balanced equation = 1 x 92.02 = 92.02g

Molar mass of N2H4 = 32.05 g/mol

Mass of N2H4 from the balanced equation = 2 x 32.05 = 64.1g

From the balanced equation above, 92.02g of N2O4 reacted with 64.1g of N2H4.

Step 3:

Determination of the excess reactant. This is illustrated below:

From the balanced equation above, 92.02g of N2O4 reacted with 64.1g of N2H4.

Therefore, 50g of N2O4 will react with = (50 x 64.1)/92.02 = 34.83g of N2H4.

From the calculations made above, we can see that only 34.83g of N2H4 reacted out of 45g that was given. Therefore, N2H4 is the excess reactant.

Step 4:

Determination of the mass of excess reactant that is leftover.

The excess reactant is N2H4 and the leftover mass can be obtained as follow:

Mass of N2H4 given = 45g

Mass of N2H4 that reacted = 34.83g

Leftover mass of N2H4 =..?

Leftover mass of N2H4 = (Mass of N2H4 given) – (Mass of N2H4 that reacted)

Leftover mass of N2H4 = 45 – 34.83

Leftover mass of N2H4 = = 10.17g.

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3 years ago

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3 years ago

Which profile best describes the reaction C(s) + 2H2(g) → CH4(g), AH = -74.9 kJ?

katrin [286]

Answer:

Option B. A

Explanation:

From the question given above, the following data were obtained:

C(s) + 2H₂ (g) —> CH₄ (g). ΔH = –74.9 kJ

From the reaction above, we can see that the enthalpy change (ΔH) is negative (i.e –74.9 KJ) which implies that the heat content of the reactants is greater than the heat content of the products. Thus, the reaction is exothermic reaction.

For an exothermic reaction, the energy profile diagram is drawn in such a way that the heat content of reactants is higher than the heat content of products because the enthalpy change

(ΔH) is always negative.

Thus, diagram A (i.e option B) gives the correct answer to the question.

8 0

3 years ago

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How many liters of 3.5 M solution can be made using 23 moles of LiBr *must show work to get credit*

How many liters of 3.5 M solution can be made using 23 moles of LiBr must show work to get credit