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3 years ago

Balance the following chemical equation by providing the correct coefficients. CO+H2 ———- H2O+CH4

Chemistry

1 answer:

makvit [3.9K]3 years ago

3 0

Answer:

CO + H2 --> CH4 + H2O

Explanation:

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Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /

barxatty [35]

Answer:

$[Pb^{2+}]=3.9 \times 10^{-2}M$

this is the concentration required to initiate precipitation

Explanation:

$PbCl_2$ ⇄ $Pb^{2+}+2Cl^-$

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

$Ip=[Pb^{2}][2Cl^-]^2=Ksp$

$Ksp=2.4\times 10^{-4}$

lets solubility=S

$[Pb^{2+}] = S$

$[Cl^-]=2S$

$Ksp=[Pb^{2+}]\times [Cl^-]^2$

$Ksp=S \times (2S)^2$

$Ksp=4S^3$

$S=\sqrt[3]{\frac{Ksp}{4} }$

$S=3.9\times 10^{-2}$

$[Pb^{2+}]=3.9 \times 10^{-2}M$ this is the concentration required to initiate precipitation

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3 years ago

the melting point of scandium fluoride is 1552°c, explain why scandium fluoride has a high melting point.

Greeley [361]

Answer:

Scandium(III) fluoride, ScF3, is an ionic compound. It is slightly soluble in water but dissolves in the presence of excess fluoride to form the ScF63− anion.

hope it will help you......

5 0

3 years ago

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Amy is in a car moving quickly toward a stationary siren. Just as Amy moves

sineoko [7]

Answer:

Pitch

Explanation:

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3 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

3 years ago