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2 years ago

5

A 10% bleach solution for disinfecting countertops can be prepared by: a. weighing the bleach and the water b. weight to volume

percentage c. adding 10 mL bleach to 90 mL water d. adding 10 mL bleach to 990 mL water

1 answer:

Ierofanga [76]2 years ago

3 0

Answer:

The correct answer is - c. adding 10 mL bleach to 90 mL water.

Explanation:

To prepare a effective disinfecting solution to clean the countertops is a bleasce solution of 10% solution of bleach. The ten percent of the bleach solution means there are one volve of bleach and 9 volume of clean water in this solution.

So, preparing the 1/10 or ten percent solution is where adding 10 ml of bleach which is diluted to 90 ml clean water. This diluted solution can be use as disinfecting solution.

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What volume of Co2 (carbon (iv) oxide)

hram777 [196]

Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL

7 0

3 years ago

Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25

Veseljchak [2.6K]

Answer: The molecular formula will be $C_6H_6O_6$

Explanation:

Mass of $CO_2$ = 17.95 g

Mass of $H_2O$ = 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, = $\frac{12}{44}\times 17.95=4.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, = $\frac{2}{18}\times 4.87=0.541g$ of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H = 0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.407}{0.407}=1$

For H = $\frac{0.541}{0.407}=1$

For O = $\frac{0.410}{0.407}=1$

The ratio of C : H : O = 1: 1 : 1

Hence the empirical formula is $CHO$ .

Hence the empirical formula is $CHO$

The empirical weight of $CHO$ = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = $\frac{44.0}{0.25}\times 1=176g$

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6$

The molecular formula will be= $6\times CHO=C_6H_6O_6$

5 0

3 years ago

what is the concentration (in %v/v)) of the following solutions? a. 25 mL of ethanol is diluted to a volume of 150 mL with water

fenix001 [56]

The symbol %v/v means percent by volume. Assuming there is no volume effects when these substances are mixed, we calculate as follows:

% v/v = (25 mL ethanol / 25 mL + 150 mL ) x 100
%v/v = 14.29 mL ethanol / mL solution

Hope this answers the question.

8 0

3 years ago

IMFs are NOT found in what type of<br> substances?

ExtremeBDS [4]

Answer:

False

Explanation:

8 0

3 years ago

Read 2 more answers

Determine the volume of a 0.800 M K2Cr2O7 solution required to completely react with 4.24 g of Cu.

Pachacha [2.7K]

<u>Answer:</u> The volume of solution required is 0.0275 L.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of copper = 4.24 g

Molar mass of copper = 63.55 g/mol

Putting values in above equation, we get:

$\text{Moles of copper}=\frac{4.24g}{63.55g/mol}=0.067mol$

The chemical equation for the reaction of potassium dichromate and copper follows:

$K_2Cr_2O_7+3Cu+7H_2SO_4\rightarrow 3CuSO_4+Cr_2(SO_4)_3+K_2SO_4+7H_2O$

By Stoichiometry of the reaction:

3 moles of copper reacts with 1 mole of potassium dichromate.

So, 0.067 moles of copper will react with = $\frac{1}{3}\times 0.067=0.022mol$ of potassium dichromate

To calculate the volume of potassium dichromate, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Moles of potassium dichromate = 0.022 mol

Molarity of solution = 0.800 M

Putting values in above equation, we get:

$0.08mol/L=\frac{0.022mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=0.0275L$

Hence, the volume of solution required is 0.0275 L.

6 0

3 years ago