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2 years ago

What coefficient must be added to the carbon in the reactants to balance the following equation?

Chemistry

1 answer:

Sophie [7]2 years ago

7 0

Answer:

Explanation:

There are 3 carbons on the right side

there are only two on the left side....you need one more ...so add one more... change the '1' coefficient in front of C to '2'

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Klklklkllkllklkklkllkl

Svetradugi [14.3K]

Answer:

klklklkllkllklkklkllkl

Explanation:

Cuz thats what you said.. and why not :)

5 0

3 years ago

A sample of seawater weighs 250 g and has a volume of 173 ml. What is the density of seawater?

kari74 [83]

Answer:

<h2>The answer is 1.45 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass of seawater = 250 g

volume = 173 mL

It's density is

$density = \frac{250}{173} \\ = 1.4450867052...$

We have the final answer as

Hope this helps you

8 0

3 years ago

How many moles are in 20 grams of argon

Sladkaya [172]

Answer:

There are 0.5 mole in 20g of argon.

Explanation:

40 g of argon = 1mole

Then 20g of argon is,

→ 1/40 × 20

→ 0.5 mole

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:

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

6 0

3 years ago

The following is an example of which type of chemical reaction?Na (s) + H2O (l) Imported Asset NaOH (aq) + H2 (g)

lora16 [44]

Answer:

The answer to your question is Single replacement

Explanation:

Data

Chemical reaction

Na + H₂O ⇒ NaOH + H₂

Single replacement is a chemical reaction in which a metal replaces the cation of a compound.

Decomposition is a chemical reaction in which a compound forms 2 or more products.

Double replacement in this chemical reaction two compounds react interchanging their cations and anions.

Combustion in this chemical reaction the reactants must be a compound with carbon and the other oxygen and the products must be carbon dioxide and water.

4 0

3 years ago

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