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2 years ago

If a chemist has a stock solution of HBr that is 10.0 M and would like to make 450.0 mL of 3.0 M HBr, how would he do it?

1 answer:

4 0

This is a dilution that requires a certain volume from the stock solution to be diluted with distilled water to make a solution of HBr with a lesser concentration than the stock solution

Following dilution formula can be used

c1v1 = c2v2

Where c1 is concentration and v1 is the volume of the stock solution

c2 is concentration and v2 is volume of the diluted solution to be prepared

Substituting these values

10.0 M x v1 = 3.0 x 450.0 mL

v1 = 135.0 mL

A volume of 135.0 mL from HBr stock solution needs to be taken and diluted with distilled water upto 450.0 mL. The resulting solution will have a concentration of 3.0 M

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A student was given four unknown solutions. Each solution was checked for. conductivity and tested with. phenolphthalein. The re

777dan777 [17]

Answer: (3) C

Explanation:

4 0

2 years ago

PLEASE HELP I NEED IT!!!!!

galina1969 [7]

Answer:
Molarity = 2.3 M

Explanation:
Molarity can be calculated using the following rule:
Molarity = number of moles of solute / volume of solution

1- getting the number of moles:
We are given that:
mass of solute = 105.96 grams
From the periodic table:
atomic mass of carbon = 12 grams
atomic mass of hydrogen = 1 gram
atomic mass of oxygen = 16 grams
Therefore:
molar mass of C2H6O = 2(12) + 6(1) + 16 = 46 grams
Now, we can get the number of moles as follows:
number of moles = mass / molar mass = 105.96 / 46 = 2.3 moles

2- The volume of solution is given = 1 liter

3- getting the molarity:
molarity = number of moles of solute / volume of solution
molarity = 2.3 / 1
molarity = 2.3 M

Hope this helps :)

8 0

2 years ago

Rank the following fertilizers in decreasing order of mass percentage of nitrogen:

charle [14.2K]

<h3>Answer:</h3>

NH₃ > NH₄NO₃ > (NH₄)₂HPO₄ > (NH₄)₂SO₄ > KNO₃ > (NH₄)H₂PO₄

<h3>Soution:</h3>

In (NH₄)₂HPO₄:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of (NH₄)₂HPO₄ = 132.06 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)₂HPO₄ × 100

Mass %age = 28 g.mol⁻¹ / 132.06 g.mol⁻¹ × 100

Mass %age = 21.20 %

In (NH₄)₂SO₄:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of (NH₄)₂SO₄ = 132.14 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)₂SO₄ × 100

Mass %age = 28 g.mol⁻¹ / 132.14 g.mol⁻¹ × 100

Mass %age = 21.18 %

In KNO₃:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of KNO₃ = 101.10 g.mol⁻¹

Mass %age = Mass of N / M.Mass of KNO₃ × 100

Mass %age = 14 g.mol⁻¹ / 101.10 g.mol⁻¹ × 100

Mass %age = 13.84 %

In (NH₄)H₂PO₄:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of (NH₄)H₂PO₄ = 115.03 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)H₂PO₄ × 100

Mass %age = 14 g.mol⁻¹ / 115.03 g.mol⁻¹ × 100

Mass %age = 12.17 %

In NH₃:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of NH₃ = 132.14 g.mol⁻¹

Mass %age = Mass of N / M.Mass of NH₃ × 100

Mass %age = 14 g.mol⁻¹ / 17.03 g.mol⁻¹ × 100

Mass %age = 82.20 %

In NH₄NO₃:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of NH₄NO₃ = 80.04 g.mol⁻¹

Mass %age = Mass of N / M.Mass of NH₄NO₃ × 100

Mass %age = 28 g.mol⁻¹ / 80.04 g.mol⁻¹ × 100

Mass %age = 34.98 %

5 0

2 years ago

Help me please, will give brainly

geniusboy [140]

I don’t know sorry :(

6 0

2 years ago

Read 2 more answers

A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom

Taya2010 [7]

Answer:

28/95 = 29.,5 % of Arsine decomposed

Explanation:A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom- poses to solid arsenic and hydrogen gas. The flask is then cooled to 273 K, at which tem- perature the pressure in the flask is 488 Torr. What percentage of arsine molecules have de- composed?

Answer in units of %.

initial pressure 332 Torr initial volume 0.46 L initial temperature 223K

final pressure 488 Torr final volume 0.46 L final 273 K

Torr is 1/760 atm 332 torr = 0.437 atm 488 Torr =0.642 atm

PV = nRT so n=RT/PV

INITIAL n= 0.082 X 223/(0.437)(0.46) = 91 moles

final n= 0.082 X 273 / (.437)(488) = 105 moles

2AsH3----------> 2As + 3H2

x moles of Arsine decomposed to make 1.5 moles of H2

the final number of moles was

(91 -X)+ 1.5 X = 105 moles

91 + 0.5 X = 105

0.5 X = 14

X =28

CHECK

if 28 moles of Arsine , then the container would have

91 --28 + 1.5(28) = 91 +14 =105 check

so 28/95 = 29.,5 % of Arsine decomposed

Your answer

(quit)

polyalchemVirtuoso

Answer:

Explanation:

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4 0

2 years ago