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Mkey [24]
3 years ago
14

A 390 mL solutions of 4.6 M HCl is left out over the weekend and 74 mL of solvent evaporates. What is the new concentration of H

Cl?
Chemistry
1 answer:
storchak [24]3 years ago
8 0

Answer:

5.7 M

Explanation:

You have .390 L of 4.6 M HCl.

.390 L * 4.6 mol/L = 1.794 mol HCl

.390-.074=.316 L solvent remaining

1.794 mol/.316 L=5.7 M HCl

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tresset_1 [31]

Answer:

I think stirring is the right answer

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3 years ago
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A student wants to reclaim the iron from an 18.0-gram sample of iron(III) oxide, which
lilavasa [31]

Answer:

m_{Fe}=12.6gFe

Explanation:

Hello,

In this case, since we have grams of iron (III) oxide whose molar mass is 159.69 g/mol are able to compute the produced grams of iron by using its atomic mass that is 55.845 g/mol and their 2:4 molar ratio in the chemical reaction:

m_{Fe}=18.0gFe_2O_3*\frac{1molFe_2O_3}{159.69gFe_2O_3}*\frac{4molFe_2O_3}{2molFe_2O_3} *\frac{55.845gFe}{1molFe_2O_3} \\\\m_{Fe}=12.6gFe

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3 years ago
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Which scale is being described? Water freezes at 32. Water freezes at 0. Water freezes at 273.
Leya [2.2K]

Scale is defined as the measurement of unit ranges, that has been corresponds to the measurements.

The units for the measurement of temperature are given as:

\rm 0^\circ C\;=\;273\;K\;=\;32\;F

<h3>Which scale is used?</h3>

The scale used for the measurement of the freezing is the temperature scale. The temperature is measured in the units of Kelvin, Fahrenheit , and Celsius.

The unit scale for the measurement of freezing of water are

  • Freezes at 32 -  In the Fahrenheit scale
  • Freezes at 0 - In the Celsius scale
  • Freezes at 273 - In the Kelvin scale

Learn more about measurement scale, here:

brainly.com/question/1152255

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2 years ago
Which option is an example of a chemical change?
MA_775_DIABLO [31]
Firework exploding. Thank you :)
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3 years ago
A student wants to prepare 1.00 L of a 1.00 M solution of NaOH (molar mass 40.00 g/mol). If solid NaOH is available, how would t
Serga [27]

Explanation:

1)

Molarity=\frac{\text{Mass of substance}}{\text{Molar mass of substance}\times \text{Volume of solution(L)}}

Mass of NaOH = m

MOlar mass of NaOH = 40 g/mol

Volume of NaOH solution = 1.00 L

Molarity of the solution= 1.00 M

1.00 M=\frac{m}{40 g/mol\times 1.00 L}

m=1.00 M\times 40 g/mol\times 1.00 L = 40. g

A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.

Upto two significant figures mass should be determined.

2)

M_1V_1=M_2V_2 (dilution equation)

Molarity of the NaOH solution = M_1=2.00 M

Volume of the solution = V_1=?

Molarity of the NaOH solution after dilution = M_2=1.00 M

Volume of NaOH solution after dilution= V_2=1 L

M_1V_1=M_2V_2

V_1=\frac{1.00 M\times 1.00 L}{2.00 M}=0.500 L

A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.

Upto three significant figures volume should be determined.

8 0
3 years ago
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