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Nata [24]
3 years ago
7

Calculate the volume in milliliters of a iron(II) bromide solution that contains of iron(II) bromide . Round your answer to sign

ificant digits.
Chemistry
1 answer:
miv72 [106K]3 years ago
5 0

This is an incomplete question, here is a complete question.

Calculate the volume in milliliters of a 1.29 mol/L iron(II) bromide solution that contains 275 mmol of iron(II) bromide . Round your answer to significant 3 digits.

Answer  : The volume of iron(II) bromide solution is, 2.13\times 10^2mL

Explanation : Given,

Concentration of iron(II) bromide = 1.29 mo/L

Moles of iron(II) bromide = 275 mmol = 0.275 mol

conversion used : 1 mmol = 0.001 mol

Now we have to calculate the volume of iron(II) bromide.

\text{Volume of iron(II) bromide}=\frac{Moles of iron(II) bromide}}{\text{Concentration of iron(II) bromide}}

Now put all the given values in this formula, we get:

\text{Volume of iron(II) bromide}=\frac{0.275mol}{1.29mol/L}=0.213L=2.13\times 10^2mL

Thus, the volume of iron(II) bromide solution is, 2.13\times 10^2mL

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Calculate the pH when 60.0 mL of 0.200 M HBr is mixed with 30.0 mL of 0.400 M CH₃NH₂ (Kb = 4.4 × 10⁻⁴).
Mariulka [41]

When a volume of 60.0 mL of 0.200 M HBr is mixed with a volume of 30.0 mL of 0.400 M CH3NH2, The pH value is mathematically given as

pH=10.64

<h3>What is the pH value when 60.0 mL of 0.200 M HBr is mixed with 30.0 mL of 0.400 M CH₃NH₂?</h3>

Question Parameters:

The pH when 60.0 mL of 0.200 M HBr

30.0 mL of 0.400 M CH₃NH₂ (Kb = 4.4 × 10^{-4}).

Generally, the equation for the Chemical Reaction  is mathematically given as

H Br + H_3C NH_2----- > CH_3 NH_3 Br

Therefore

oH=p^{kb}+-log\frac{salt}{base}

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OH=3.36

In conclusion, The equation pH value

pH+OH=14

Therefore

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pH=10.64

Read more about Chemical Reaction

brainly.com/question/11231920

8 0
2 years ago
A sample of an ionic compound that is often used as a dough conditioner is analyzed and found to contain 4.628 g of potassium, 9
Roman55 [17]

Answer:

The empirical formula of the compound is = KBrO_3

The name of the compound is potassium bromate.

Explanation:

Mass of potassium = 4.628 g

Moles of potassium = \frac{4.628 g}{39 g/mol}=0.1187 mol

Mass of bromine = 9.457 g

Moles of bromine = \frac{9.457 g}{80 g/mol}=0.1182 mol

Mass of oxygen = 5.681 g

Moles of oxygen = \frac{5.681 g}{16 g/mol}0.3551

For empirical formula of the compound, divide the least number of moles from all the moles of elements present in the compound:

Potassium :

\frac{0.1187 mol}{0.1182 mol}=1.0

Bromine;

\frac{0.1182 mol}{0.1182 mol}=1.0

Oxygen ;

\frac{0.3551 mol}{0.1182 mol}=3.0

The empirical formula of the compound is = KBrO_3

The name of the compound is potassium bromate.

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3 years ago
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Explanation:

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