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3 years ago

An extremely soluble salt is added to water, and all of it dissolves. Is it at equilibrium? Explain.

1 answer:

5 0

Once the substance stops dissolving, the system is at equlibrium with the water and the undissolved salt now, if it is in the process of dissolving because it is completely soluble but has not been able to completely dissolve, it is not at equilibrium

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PLEASE HELP! 25 POINTS!!!! I got the #1, just not #2 and #3. An industrial chemical company has opened a new plant that will pro

podryga [215]

Answer :

Part 1 : Balanced reaction, $3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

Part 2 : The theoretical yield of $NH_3$ gas = 440.96 g

Part 3 : The % yield of ammonia is 90.03 %

Solution : Given,

Mass of $N_2$ = 475 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $NH_3$ = 17 g/mole

Experimental yield of $NH_3$ = 397 g

Answer for Part (1) :

The balanced chemical reaction is,

$3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

Answer for Part (2) :

First we have to calculate the moles of $N_2$ .

$\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles$

From the given reaction, we conclude that

1 moles of $N_2$ gas react to give 2 moles of $NH_3$ gas

16.96 moles of $N_2$ gas react to give $\frac{2}{1}\times 16.96=33.92$ moles of $NH_3$ gas

Now we have to calculate the mass of $NH_3$ gas.

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g$

Therefore, the theoretical yield of $NH_3$ gas = 440.96 g

Answer for Part (3) :

Formula used for percent yield :

$\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}\times 100$

$\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%$

Therefore, the % yield of ammonia is 90.03 %

3 0

3 years ago

The whiye pigment TiO2 is prepared by the reaction of titanium tetrachloride, TiCl4, with water vapor in the gas phase:

omeli [17]

Answer:

$\boxed{\text{62.1 kJ}}$

Explanation:

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

$\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})$

TiCl₄(g) + 2H₂O(g) ⟶ TiO₂(s) + 4HCl(g)

ΔH°f/kJ·mol⁻¹: -763.2 -241.828 -939.7 -92.307

$\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [-939.7 + 4(-92.307)] - [-763.2 + 2(-241.828)\\& = & [-939.7 - 369.228] - [-763.2 - 483.656]\\& = & -1308.928 + 1246.856\\& = & \mathbf{-62.1}\\\end{array}\\\text{The amount of heat evolved is } \boxed{\textbf{62.1 kJ}}$

5 0

3 years ago

Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ΔH° for the r

PolarNik [594]

Answer: -105 kJ

Explanation:-

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]$

$\Delta H=[(n_{N_2}\times B.E_{N_2})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{NH_3}\times B.E_{NH_3})]$

$\Delta H=[(n_{N_2}\times B.E_{N\equiv N})+(n_{H_2}\times B.E_{H-H}) ]-[(n_{NH_3}\times 3\times B.E_{N-H})]$

where,

n = number of moles

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

$\Delta H=[(1\times 945)+(3\times 432)]-[(2\times 3\times 391)]$

$Delta H=-105kJ$

Therefore, the enthalpy change for this reaction is, -105 kJ

8 0

3 years ago

What can you infer from the fact that metals are good conductors of electricity?

choli [55]

Answer:

Explanation:

makes sense

3 0

2 years ago

Read 2 more answers

Be sure to answer all parts. For each of the following pairs of elements, state whether the binary compound they form is likely

andrew-mc [135]

Answer:

(a) Covalent bond. NF₃ (nitrogen trifluoride)

(b) Ionic bond. LiCl (lithium chloride)

Explanation:

(a) N and F

Nitrogen and fluorine are nonmetals, with high and similar electronegativities, so they form covalent bonds, in which they share pairs of electrons to complete the octet in their valence shell. N has 5 valence electrons so it will form 3 covalent bonds while each Cl has 7 valence electrons so it will form 1 covalent bond. As a result, the empirical formula is NF₃ (nitrogen trifluoride).

(b) Li and Cl

Lithium is a metal and Chlorine is a nonmetal. They have different electronegativities so they form an ionic bond, in which Cl gains 1 electron (7 valence e⁻) and Li loses 1 electron (1 valence e⁻). The empirical formula is LiCl (lithium chloride).

5 0

3 years ago