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Over [174]
3 years ago
8

3. What is the mass of HI used to create 2.50L of a 0.48 M solution of hydroiodic acid?

Chemistry
1 answer:
Daniel [21]3 years ago
5 0

Answer:

Mass = 153.48 g

Explanation:

Given data:

Volume of solution = 2.50 L

Molarity = 0.48 M

Mass required = ?

Solution:

Molarity = number of moles / volume in litter

Number of moles = Molarity × volume in litter

Number of moles = 0.48 M  × 2.50 L

Number of moles =  1.2 mol

Mass of HI:

Number of moles = mass/molar mass

Mass = Number of moles × molar mass

Mass = 1.2 mol × 127.9 g/mol

Mass = 153.48 g

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The temperature of evaporation is much higher for water than for alcohol. Without knowing more about the chemistry of alcohol, w
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6. The Earth’s electric field is directed vertically downward; at the ground level its strength is E = 130 N/C, and at an altitu
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5.31x10⁻⁶ C

Explanation:

The cube is located 100 m altitude from the ground, so the superior face is at 100m and has E = 70 N/C, and the inferior face is at the ground with E = 130 N/C.

The electric field is perpendicular to the bottom and the top of the cube, so the total flux is the flux at the superior face plus the flux at the inferior face:

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Where Ф = E*A*cos(α). α is the angle between the area vector and the field (180° at the topo and 0° at the bottom):

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2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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3 years ago
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