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mr Goodwill [35]
2 years ago
15

Calculate the mass of nitric acid required to make a 250mL solution with 2.40 ph

Chemistry
1 answer:
meriva2 years ago
3 0

The mass of nitric acid required to make the given solution is 0.0627 g.

The given parameters:

  • <em>Volume of the acid, V = 250 mL</em>
  • <em>pH of the acid, = 2.4</em>

The hydrogen ion (H⁺) concentration of the nitric acid is calculated as follows;

H^+ = 10^{-pH}\\\\H^+ = 10^{-2.4}\\\\H^+ = 0.00398

The molarity of the nitric acid is calculated as follows;

=  0.00398 \ H^+ \times \frac{1 \ M \ HNO_3}{1 \ H^+} \\\\= 0.00398 \ M

The number of moles of the nitric acid is calculated as follows;

moles = M\times L\\\\moles = 0.00398\ M \ \times \ \frac{250 \ mL}{1000} \\\\moles = 9.95 \times 10^{-4} \ mol.

The molar mass of nitric acid is calculated as;

HNO_3 = (1) \ + (14) \ + (16 \times 3) = 63 \ g/mol

The mass of the nitric acid contained in the calculated number of moles is calculated as;

mass = moles\  \times \ molar \ mass\\\\mass = 9.95\times 10^{-4} \ mol. \ \times \ 63 \ g/mol\\\\mass = 0.0627 \ g

Thus, the mass of nitric acid required to make the given solution is 0.0627 g.

Learn more about molarity of acids here: brainly.com/question/13864682

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Svet_ta [14]

Answer:

2.40 M

Explanation:

The molarity of a solution tells you how many moles of solute you get per liter of solution.

Notice that the problem provides you with the volume of the solution expressed in milliliters,

mL

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1 L

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10

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Now, in order to get the number of moles of solute, you must use its molar mass. Now, molar masses are listed in grams per mol,

g mol

−

1

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1 g

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10

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Sodium chloride,

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−

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, which means that your sample will contain

unit conversion



280.0

mg

⋅

1

g

10

3

mg

⋅

molar mass



1 mole NaCl

58.44

g

=

0.004791 moles NaCl

This means that the molarity of the solution will be

c

=

n

solute

V

solution

c

=

0.004791 moles

2.00

⋅

10

−

3

L

=

2.40 M

The answer is rounded to three sig figs, the number of sig figs you have for the volume of the solution.

6 0
3 years ago
Assuming you have 6.24 x 1014 electrons and the surface area of the pail is 0.2 m2, what is the charge density (C/m2)?
pentagon [3]

Answer:

σ = 4.998 E-4 C/m²

Explanation:

  • 1 Coulomb (C) ≡ 6.241509 E18 electrons (e)

∴ # elect = 6.24 E14 elect

charge (Q):

⇒ Q = (6.24 E14 elect)/( 1 C /6.241509 E18 elect) = 9.998 E-5 C

charge density (σ):

  • σ = Q/S

∴ surface area (S) = 0.2 m²

⇒ σ = ( 9.998 E-5 C ) / ( 0.2 m²)

⇒ σ = 4.998 E-4 C/m²

4 0
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A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and place
kakasveta [241]

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8  cm³

mass of gold cube =  20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

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Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

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