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3 years ago

How many milliliters of a 12-molar HCl must be diluted to obtain 2.0 liters of 3.0-molar HCl?

Chemistry

1 answer:

Vladimir [108]3 years ago

7 0

Answer: 500ml

Explanation:

Using the Dilution formula that

C₁V₁=C₂V₂

Where

C₁=the molarity of concentrated solution

= 12 M

V₁ = Volume of the concentrated solution

C₂=the molarity of the diluted solution

=3.0M

V₂= Volume of the diluted solution= 2 liters

Changing 2 liters to millilitres

1000ml= 1 Liters

2Liters=2000ml

Bringing the formulae and inputting the given values

C₁V₁=C₂V₂

12 x V₁= 3.0 x 2000ml

V₁= 3.0 x 2000 / 12

V₁=500ml

Therefore 500ml of 12-molar HCl mst be diluted to obtain 2 Liters of 3.0 Molar HCl.

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To prepare 250mL of calcium chloride solution with a molar concentration of 1.20mol/L, what mass of calcium chloride would be re

e-lub [12.9K]

Answer:

33.30 grams of CaCl2 will be required

Explanation:

Given,

Volume of solution, V= 250 ml

Molarity of solution, M= 1.20 mol/L

Molecular mass of CaCL2, S= 40+(35.5 X 2)= 111

We know,

Required mass, W= SVM/1000

Now,

W = (111 X 250 X 1.20)/1000

= 33300/1000

= 33.30

Therefore, 33.30 grams of Calcium Chloride will be required.

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3 years ago

How is the periodic table useful today?

S_A_V [24]

Answer:

The periodic table is useful today for finding out all the elements that exists on Earth. The elements there can be used to write down chemical formulas, calculate things like molar mass/atomic mass of each element, atomic number for each, number of valence electrons each element has, the oxidation number for each, etc. Lastly, it can be used to predict the properties of elements yet to be discovered.

Explanation:

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3 0

3 years ago

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Marcia is given an incomplete chemical equation that includes the number of nitrogen atoms present in the products of the reacti

Elden [556K]

If the products contain 3 nitrogen atoms, then so did the reactants since overall mass is conserved in a chemical reaction.

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3 years ago

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¿Cuál es el % m/m de una disolución en que hay disueltos 22 g de soluto en 44 g de disolvente?

Archy [21]

The question is as follows: What is the% m / m of a solution in which 22 g of solute are dissolved in 44 g of solvent?

Answer: The% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.

Explanation:

Given: Mass of solute = 22 g

Mass of solvent = 44 g

The percentage m/m is calculated using the following formula.

$Mass percentage = \frac{mass of solute}{mass of solvent} \times 100$

Substitute the values into above formula as follows.

$Mass percentage = \frac{mass of solute}{mass of solvent} \times 100\\= \frac{22 g}{44 g} \times 100\\= 50 percent$

Thus, we can conclude that the% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.

5 0

3 years ago

A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this

NeTakaya

Answer:

a) molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane $CH_4}= 16 g/mol$

Molar mass of ethane $C_2H_6= 30 g/mol$

Molar mass of ethylene $C_2H_4 = 28 g/mol$

Molar mass of water $H_2O=18g/mol$

number of moles = $\frac{mass}{molar mass}$

Their molar composition can be calculated as follows:

$n_{CH_4}= \frac{75}{16}$

$n_{CH_4}=$ 4.69 moles

$n_{C_2H_6} = \frac{10}{30}$

$n_{C_2H_6} =$ 0.33 moles

$n_{C_2H_4} = \frac{5}{28}$

$n_{C_2H_4} =$ 0.18 moles

$n_{H_2O}= \frac{10}{18}$

$n_{H_2O}=$ 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas = $\frac{n_{H_2O}}{n_{drygas}}$

= $\frac{0.56}{5.2}$

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

$CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O$

4.69 2× 4.69

moles moles

$C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O$

0.33 3.5 × 0.33

moles moles

$C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O$

0.18 3× 0.18

moles moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

$CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h$

Total moles of $O_2$ required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles $O_2$

Thus, moles of air required = $\frac{1}{0.21}*11.075$

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

5 0

4 years ago