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3 years ago

3.0 cm x 4.0 cm x 1.0 cm[?]cm^3

Chemistry

1 answer:

CaHeK987 [17]3 years ago

4 0

Explanation:

Hi there!!

you asked to multiply these all right,

you can simply multiply it ,

=3cm × 4 cm × 1cm

= 12cm^2×1cm (4×3=12)

= 12cm^3 (12×1=12)

Therefore, theanswer is 12 cm^3.

Hope it helps..

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The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa

Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of $MgCl_2$ (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of $MgCl_2$ (solute) = 3.37 g

First we have to calculate the mass of solute.

$\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2$

$\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g$

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg$

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

$\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%$

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

$\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%$

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

8 0

3 years ago

How many moles of MgCl2 are there in 343 g of the compound?

mash [69]

The answer is 3.592?

6 0

3 years ago

A student is asked to determine the molarity of a strong base by titrating it with 0.250 M solution of H2SO4. The students is in

tamaranim1 [39]

Answer:

The molarity of the strong base is 0.625 M

Which procedural error will result in a strong base molarity that is too high?

⇒ Using a buret with a tip filled with air rather than the H2SO4 solution

Explanation:

Step 1: Data given

Molarity of H2SO4 = 0.250 M

The initial buret reading is 5.00 mL

The final reading is 30.00 mL

Step 2: Calculate volume of H2SO4 used

30.00 mL - 5.00 mL = 25.00 mL

Step 3: Calculate moles of H2SO4

0.250 M = 0.250 mol/L

Since there are 2 H+ ions per H2SO4

0.250 mol/L * 2 = 0.500 mol/L

The number of moles H2SO4 = 0.500 mol/L * 0.025 L

Number of moles H2SO4 = 0.0125 mol

Step 4: Calculate moles of OH-

For 1 mol H2SO4, we need 1 mol of OH-

For 0.0125 mol of H2SO4, we have 0.0125 mol of OH-

Step 5: Calculate the molarity of the strong base

Molarity = moles / volume

Molarity OH- = 0.0125 mol / 0.02 L

Molarity OH - = 0.625 M

Which procedural error will result in a strong base molarity that is too high?

⇒ Using a buret with a tip filled with air rather than the H2SO4 solution

5 0

3 years ago

1. Take the reaction: NH3 + O2 + NO + H2O. In an experiment, 3.25g of NH3 are allowed

Rudiy27

Answer:

5.74g of NO

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

4NH3 + 5O2 —> 4NO + 6H2O

Step 2:

Determination of the masses of NH3 and O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

Molar Mass of NO = 14 + 16 = 30g/mol

Mass of NO from the balanced equation = 4 x 30 = 120g

From the balanced equation above,

68g of NH3 reacted with 160g of O2 to produce 120g of NO.

Step 3:

Determination of the limiting reactant.

We need to determine the limiting because it will be used to calculate the maximum yield of the reaction. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Therefore, 3.25g of NH3 will react with = (3.25 x 160)/68 = 7.65g of O2.

From the simple illustration above, we can see that lesser mass of O2 is needed to react with 3.25g of NH3. Therefore, NH3 is the limiting reactant while O2 is the excess reactant.

Step 4:

Determination of the mass of NO produced from the reaction.

In this case the limiting reactant will be used because all of it were used in the reaction.

The limiting reactant is NH3.

From the balanced equation above,

68g of NH3 reacted to produce 120g of NO.

Therefore, 3.25g of NH3 will react to produce = (3.25 x 120)/68 = 5.74g of NO.

From the calculations made above, 5.74g of NO is produced.

4 0

3 years ago

Radiometric dating is used to determine

PolarNik [594]

Answer:

used to date rocks and other objects based on the known decay rate of radioactive isotopes.

Explanation:

Different methods of radiometric dating can be used to estimate the age of a variety of natural and even man-made materials.

4 0

3 years ago

3.0 cm x 4.0 cm x 1.0 cm[?]cm^3​

3.0 cm x 4.0 cm x 1.0 cm[?]cm^3