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3 years ago

Please help me with question A and B pleaseeeeee

Chemistry

1 answer:

laiz [17]3 years ago

8 0

Answer:

‍♀️?sorryyyyyyyyyyyyy

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While in Europe, if you drive 109 km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per lite

MrRissso [65]

Answer:

The amount is $x = 113.3 \ dollars$

Explanation:

From the question we are told that

The distance traveled per day is $l = 109\ km = \frac{109}{1.609} = 67.74 \ mi$

The cost of one liter is $c = 1.10 \ euros/liter = 1.10 * 1.26 = 1.36 \ dollars/liter$

The car's gas mileage is $b = 22.0 \ mi/gal$

Generally the amount of distance covered in one week is evaluated as

$z = 67.74 * 7$

$z = 474.18 \ mi$

The amount of gas used in one week by the car is mathematically represented as

$k = \frac{ z}{ b}$

=> $k = \frac{ 474.18}{ 22}$

=> $k = 22 \ gal$

converting to liters

$k = 22 * 3.78541=83.28 \ liters$

Thus the amount spent on gas in one week is

$x = k * c$

=> $x = 83.28 * 1.36$

=> $x = 113.3 \ dollars$

5 0

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:

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

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2 years ago

Photochemical smog results from the interaction of pollutants in the presence of

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Water vapor in the air

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3 years ago

Which of the following is a metal?

kogti [31]

Answer:

A.................. I think the answer

3 0

3 years ago

When two ionic compounds are dissolved in water, a double replacement reaction can... Group of answer choices occur if two of th

sergey [27]

Answer:

occur if two of the ions form an insoluble ionic compound, which precipitates out of solution

Explanation:

When two ionic compounds are dissolved in water, a double replacement reaction takes place if two of the ions form an insoluble ionic compound, which precipitates out of solution. In double displacement reaction ions switch partners. And hence, produce an insoluble precipitate.

5 0

3 years ago