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2 days ago

What volume in milliliters of concentrated HCl (12 M) is needed to make 1500 mL of a 3.5 M solution?

2 answers:

5 0

This procedure entails diluting the 12 molar HCl. To decrease the concentration, we must create an equation to determine how much of the 12M is needed for the 3.5M solution.

12 moles HCl 3.5 moles HCl
——————— = ———————
1 Liter of Soln ‘x’ Liters of Soln

Note that the ratio of 12 moles over 1 liter corresponds to 12 molar; thus, we maintain the original concentration of the HCl. By equating it to the 3.5 over ‘x’, we are still preserving the concentration.

After computation, we determine ‘x’ to be 0.292. This value indicates that within 0.292 liters of our 12 M HCl solution, there are 3.5 moles of HCl. Yet, we are not finished.

0.292 liters of 12 M HCl can create 1 liter of 3.5 M HCl, but the inquiry demands 1.5 liters. To achieve this, multiply 0.292 liters by 1.5, resulting in 0.4375, which denotes the quantity of 12 M HCl necessary to prepare a 1500 mL 3.5 M HCl solution.

alisha [2.8K]2 days ago

3 0

Response: The amount of 12 M HCl needed for 1500 ml of a 3.5 M solution is 437.5 ml

Explanation:

According to the neutralization principle,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of HCl solution = 12 M

$V_1$ = volume of HCl solution =?

$M_2$ = molarity of the final solution = 3.5 M

$V_2$ = volume of the final solution = 1500 ml

Now substituting all specified values into the above relation, we derive the volume of HCl solution.

$(12M)\times V_1=(3.5M)\times (1500ml)$

$V_1=437.5ml$

Thus, the volume of 12 M HCl necessary to produce 1500 ml of a 3.5 M solution is 437.5 ml

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Response:

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Clarification:

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at 65 degrees celsius, which compound has a vapor pressure of 58 kilopascals? 1. ethanoic acid2. ethanol 3. propanone 4. water

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The vapor pressure of a substance varies with temperature. To identify which compound has a vapor pressure of 58 kilopascals at 65 degrees Celsius, we consult standard tables that list temperature against vapor pressure.

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28 days ago

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Rank the compounds below from slowest to Fastest rate of hydration. 1) formaldehyde 2) 3,3-dimethyIbutan-2-one 3) propanal 4)3-m

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The order of compounds from the slowest to the fastest rate of hydration is 5, 3, 4, 1, 2.

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14 days ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

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Answer: The empirical formula and molecular formula for the analyzed organic compound are $C_9H_{12}O$ and $C_{18}H_{24}O_2$

Explanation:

The combustion chemical equation for a hydrocarbon containing carbon, hydrogen, and oxygen is:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where 'x', 'y', and 'z' represent the subscripts for Carbon, Hydrogen, and Oxygen.

We have the following data:

Mass of $CO_2=39.01g$

Mass of $H_2O=10.65g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For carbon mass calculation:

12 grams of carbon are found in 44 grams of carbon dioxide.

Thus, in 39.01 grams of carbon dioxide, $\frac{12}{44}\times 39.01=10.64g$ grams of carbon will be present.

For hydrogen mass calculation:

In 18 grams of water, 2 grams of hydrogen are contained.

Therefore, in 10.65 grams of water, $\frac{2}{18}\times 10.65=1.18g$ grams of hydrogen will be present.

The oxygen mass in the compound = (13.42) - (10.64 + 1.18) = 1.6 grams.

To derive the empirical formula, you need to perform a few steps:

Step 1: Convert the provided masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.64g}{12g/mole}=0.886moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.18g}{1g/mole}=1.18moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.6g}{16g/mole}=0.1moles$

Step 2: Determine the mole ratio of the elements.

Each mole value is divided by the smallest mole value, which is 0.1 moles, to find the mole ratio.

For Carbon = $\frac{0.886}{0.1}=8.86\approx 9$

For Hydrogen = $\frac{1.18}{0.1}=11.8\approx 12$

For Oxygen = $\frac{0.1}{0.1}=1.99\approx 2$

Step 3: Use the mole ratio values as subscripts.

The ratio of C: H: O = 9: 12: 1

The empirical formula for the compound is $C_9H_{12}O$

To find the molecular formula, it’s necessary to ascertain the valency, which is then multiplied by each elemental subscript of the empirical formula.

The formula used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

Given:

Mass of the molecular formula = 272.38 g/mol

Mass of the empirical formula = 136 g/mol

Substituting values into the equation gives:

$n=\frac{272.38g/mol}{136g/mol}=2$

Multiplying the determined valency with the empirical formula’s element subscripts results in:

$C_{(2\times 9)}H_{(2\times 12)}O_{(2\times 1)}=C_{18}H_{24}O_2$

Therefore, the forms of the organic compound are $C_9H_{12}O$ and $C_{18}H_{24}O_2$

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1 month ago

Shelby measured the volume of a cylinder and determined it to be 54.5 cm3 . The teacher told her that she was 4.25% too high in

alisha [2865]

Answer:

V = 56.816 cm³

Explanation:

It is given that

The recorded value of the cylinder is 54.5 cm³

The measurement error percentage for the volume is 4.25%

Our goal is to determine the actual volume of the cylinder.

First, we calculate the error in the volume determination as follows:

$E=54.5 \times \dfrac{4.25}{100}=\pm2.316$

It mentions in the problem that the teacher indicated that her estimation was 4.25% higher than the actual volume. Therefore,

Total volume = 54.5 + 2.316 = 56.816 cm³

Thus, the true volume of the cylinder amounts to 56.816 cm³.

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