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3 years ago

How many base units exist?

Chemistry

1 answer:

Oksi-84 [34.3K]3 years ago

4 0

Answer:

Explanation:

second

metre

kilogram

ampere

kelvin

mole

candela

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How fast do i have to run 250 m in 30 seconds

elena-14-01-66 [18.8K]

Answer:

S=D/T

250/30= 8.33 miles

Explanation:

7 0

3 years ago

What will be the mass of carbon dioxide

Elza [17]

44.01 g/mol is the mass

4 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

What mass of solid sodium formate (of MW 68.01) must be added to 150 mL of 0.42 mol/L formic acid (HCOOH) to make a buffer solu-

Sergio [31]

Answer:

We need 4.28 grams of sodium formate

Explanation:

Step 1: Data given

MW of sodium formate = 68.01 g/mol

Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L

pH = 3.74

Ka = 0.00018

Step 2: Calculate [base)

3.74 = -log(0.00018) + log [base]/[acid]

0 = log [base]/[acid]

0 = log [base] / 0.42

10^0 = 1 = [base]/0.42 M

[base] = 0.42 M

Step 3: Calculate moles of sodium formate:

Moles sodium formate = molarity * volume

Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles

Step 4: Calculate mass of sodium formate:

Mass sodium formate = moles sodium formate * Molar mass sodium formate

Mass sodium formate = 0.063 mol * 68.01 g/mol

Mass sodium formate = 4.28 grams

We need 4.28 grams of sodium formate

4 0

3 years ago

Ochem question?

insens350 [35]

Answer: 4.) CH3O is the answer, which stands for Methoxide.

Explanation:

3 0

3 years ago