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il63 [147K]
2 years ago
13

Label each of the following measurements by the quantity each represents. For instance, a measurement of 10.6kg/m3 represents de

nsity.
Measurement. Quantity the measurement represents




5.0g/mL

37s

47J

39.56g

25.3 cm3

325 m s

500m2

30.23 mL

2.7 mg

0.005L
Chemistry
1 answer:
postnew [5]2 years ago
3 0

The quantities represented by each of the measurements are

5.0g/mL - Density

37s - Time

47J - Energy or Work

39.56g - Mass

25.3 cm3 - Volume

325 m/s - Time

500m2 - Area

30.23 mL - Volume

2.7 mg - Mass

0.005L - Volume

To determine the quantity each of the measurements represents, we will observe the units.

  • 5.0g/mL

This represents density. Density is the ratio of mass to volume, and mass is measured in grams (g) while volume is measured in milliliters (mL)

  • 37s

This represents time because time is measured in seconds (s)

  • 47J

This represents Energy or Work because they are measured in Joules (J)

  • 39.56g

This represents mass because mass in measured in grams (g)

  • 25.3 cm3

This represents volume because volume is measured in cubic meters (cm³)

  • 325 m s

This represents time. Time can be measured in milliseconds (ms)

  • 500m2

This represents area because area is measured in square meters (m²)

  • 30.23 mL

This represents volume because volume is measured in milliliters (mL)

  • 2.7 mg

This represents mass because mass could be measured in milligrams (mg)

  • 0.005L

This represents volume because volume is measured in liters (L)

Hence, the quantities represented by each of the measurements are

5.0g/mL - Density

37s - Time

47J - Energy or Work

39.56g - Mass

25.3 cm3 - Volume

325 m s - Time

500m2 - Area

30.23 mL - Volume

2.7 mg - Mass

0.005L - Volume

Learn more here: brainly.com/question/1603371

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Suppose 0.701g of iron(II) chloride is dissolved in 50.mL of a 55.0mM aqueous solution of silver nitrate.
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Answer : The molarity of chloride anion in the solution is 0.05 mole/L

Explanation : Given,

Mass of FeCl_2 = 0.701 g

Volume of solution = 50 ml = 0.050 L

Molarity of AgCl = 55.0 mM = 0.055 M

Molar mass of FeCl_2 = 126.751 g/mole

First we have to calculate the moles of FeCl_2.

\text{Moles of }FeCl_2=\frac{\text{Mass of }FeCl_2}{\text{Molar mass of }FeCl_2}=\frac{0.701g}{126.751g/mole}=0.00553moles

Now we have to calculate the moles of AgNO_3.

\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume of solution}=0.055mole/L\times 0.050L=0.00275mole

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

FeCl_2(aq)+2AgNO_3(aq)\rightarrow 2AgCl(s)+Fe(NO_3)_2(aq)

From the balanced reaction we conclude that

As, 2 moles of AgNO_3 react with 1 mole of FeCl_2

So, 0.00275 moles of AgNO_3 react with \frac{0.00275}{2}=0.001375 moles of FeCl_2

From this we conclude that, FeCl_2 is an excess reagent because the given moles are greater than the required moles and AgNO_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgCl.

As, 2 moles of AgNO_3 react to give 2 moles of AgCl

So, 0.00275 moles of AgNO_3 react to give 0.00275 moles of AgCl

Now we have to calculate the molarity of AgCl.

\text{Molarity of }AgCl=\frac{\text{Moles of }AgCl}{\text{Volume of solution}}

\text{Molarity of }AgCl=\frac{0.00275mole}{0.055L}=0.05mole/L

As we know that, 1 mole of AgCl in solution gives 1 mole of silver ion and 1 mole of chloride ion.

So, the molarity of chloride ion = Molarity of AgCl = 0.05 mole/L

Therefore, the molarity of chloride anion in the solution is 0.05 mole/L

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If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4
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Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

<u>Step 1:</u> Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

<u>Step 2:</u> Calculate moles of Ca^2+

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Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

<u>Step 3: </u>Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

<u>Step 4: </u>Calculate total volume

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<u>Step 5: </u> Calculate Q

Q = [Ca2+] [SO42-]  

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Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

<u> Step 6:</u> Calculate molar solubility

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2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

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total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

<u>Step 8:</u> Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

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6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

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IrinaVladis [17]
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