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Anna [14]
3 years ago
6

students used a balance and a graduated cylinder to collect the data shown in table 7. calculate the density of the sample. if t

he accepted density of this sample is 6.95 g/mL, calculate the percent error.

Chemistry
1 answer:
Licemer1 [7]3 years ago
3 0

Answer:

              Percentage error  =  1.88 %

Solution:

Data Given:

                 Mass of Sample  =  20.46 g

                 Volume of Sample  =  43.0 mL - 40.0 mL  =  3.0 mL

Formula Used:

                 Density  =  Mass / Volume

Putting values,

                 Density  =  20.46 g /  3.0 mL

                 Density  =  6.82 g.mL⁻¹

Percentage Error:

                 Experimental Value  =  6.82 g.mL⁻¹

                 Accepted Value  =  6.95 g.mL⁻¹

                 = 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100  =  98.12 %

                 Percentage Error  =  100 % - 98.12 %

                Percentage error  =  1.88 %

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Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o
sattari [20]

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is  4.50  ×  10 ⁻⁴.</em>

<em />

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50  ×  10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4}  } = 0.0106 M

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%

4 0
3 years ago
Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel ro
melamori03 [73]

<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For nickel:</u>

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol

For the given chemical reaction:

3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)

  • <u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from \frac{3}{3}\times 0.252=0.252mol of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g

  • <u>For aluminium:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from \frac{2}{3}\times 0.252=0.168mol of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

4 0
3 years ago
The theoretical yield for a reaction is 55.9 g LiCl. The actual yield is 24.6 g LiCl. What is the percent yield of the reaction?
xenn [34]
In chemical reactions, the actual yield is not the same as the expected yield . Actual yield is lower than the theoretical yield . Then we have to find the yield percentage. To see what percentage of the theoretical yield is the actual yield.
Percent yield = actual yield / theoretical yield x 100%
Percent yield = 24.6/55.9 x100%
Percent yield = 44%
7 0
3 years ago
Read 2 more answers
Calculate the volume of 0.10 g of titanium (4.51 g/cm³).
Juli2301 [7.4K]

The volume of titanium with mass of 0. 10g and density of 4. 51 g/cm³ is 0. 02 cm³

<h3>What is volume?</h3>

Volume is known to be equal to the mass divided by the density.

It is written thus:

Volume = Mass / density

<h3>How to calculate the volume</h3>

The volume is calculated using the formula:

Volume = mass ÷ density

Given the mass = 0. 10g

Density = 4.51 g/cm³

Substitute the values into the formula

Volume of titanium = 0. 10 ÷ 4.51 = 0. 02 cm³

Thus, the volume of titanium with mass of 0. 10g and density of 4. 51 g/cm³ is 0. 02 cm³

Learn more about volume here:

brainly.com/question/1762479

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4 0
2 years ago
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shusha [124]

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Explanation:

8 0
2 years ago
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