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In-s
1 month ago
8

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

Chemistry
2 answers:
lorasvet [2.7K]1 month ago
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 [2.9K]1 month 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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1 atomic mass unit (amu) represents the mass of an atom or is used to measure mass on an atomic scale. It is also referred to as a dalton, abbreviated as Da, while atomic mass unit is indicated as amu.

1 amu can be translated into grams as follows:

1 amu = 1.6 * 10^-2^4 g

Mass of Te = 127.6 amu

For conversion into grams:

M = (127.6 ) * 1.6 * 10^-2^4 g

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2 months ago
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Answer:

1.5

Explanation:

It is given that:

Compound A and B originate from Sulfur + Oxygen.

Compound A:

6g sulfur + 5.99g Oxygen

Compound B:

8.6g sulfur + 12.88g oxygen

By comparing the ratios:

Compound A:

S: O = 6.00: 5.99

S/0 = 6.0g S / 5.99g O

Compound B:

S: O = 8.60: 12.88

S / O = 8.60g S / 12.88g O

The mass ratio of A and that of B

(6.0g S / 5.99g O) ÷ (8.60g S / 12.88g O)

(6.0 g S / 5.99g O) × (12.88g O / 8.60g S)

(6 × 12.88) / (5.99 × 8.60)

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Answer:

Refer to the explanation.

Explanation:

Formation reactions involve the creation of one mole of a compound from its elements in their standard states.

NaBr (s)

The equation for the standard formation is

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As per appendix C, the standard heat of formation for NaBr(s) is

ΔH∘f = -359.8 kJ/mol.

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The equation for the standard formation is

S (s) + (3/2) O₂ (g) → SO₃ (g)

<paccording to="" appendix="" c="" the="" standard="" heat="" of="" formation="" for="" so="" is="">

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Pb(NO₃)₂ (s)

The equation for the standard formation is

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Answer:

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We derive the molar mass of XCl2 and YCl2 by recalling the molar mass formula when both mass and the number of moles are known.

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