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3 years ago

A volume is measured experimentally as 4.26mL. what is the percentage error, given that the correct value is 4.15mL

2 answers:

8 0

% error = | accepted value - experimental value | / accepted value × 100

Step one- | 4.15mL - 4.26mL | = 0.11

Step two- 0.11 / 4.15 = 0.027

Step three- -0.027 × 100 = 2.7%

kherson [118]3 years ago

5 0

Answer: The percentage error is 2.65 %.

Explanation:

To calculate the percentage error, we use the equation:

$\%\text{ error}=\frac{|\text{Experimental value - Accepted value}|}{\text{Accepted value}}\times 100$

We are given:

Experimental value of volume = 4.26 mL

Accepted value of volume = 4.15 mL

Putting values in above equation, we get:

$\%\text{ error}=\frac{|4.26-4.15|}{4.15}\times 100\\\\\%\text{ error}=2.65\%$

Hence, the percentage error is 2.65 %.

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Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f

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Answer: The mass of cryolite produced is 51.48 kg

Explanation:

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

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

For aluminium oxide:

Given mass of aluminium oxide = 12.5 kg = 12500 g (Conversion factor: 1 kg = 1000 g)

Molar mass of aluminium oxide = 101.96 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium oxide}=\frac{12500g}{101.96g/mol}=122.6mol$

For NaOH:

Given mass of NaOH = 55.4 kg = 55400 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{55400g}{40g/mol}=1389mol$

For HF:

Given mass of HF = 55.4 kg = 55400 g

Molar mass of HF = 20 g/mol

Putting values in equation 1, we get:

$\text{Moles of HF}=\frac{55400g}{20g/mol}=2770mol$

For the given chemical reaction:

$Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O(g)$

By Stoichiometry of the reaction:

1 mole of aluminium oxide reacts with 6 moles of sodium hydroxide and 12 moles of HF.

So, 122.6 moles of aluminium oxide will react with $(6\times 122.6)=735.6mol$ of sodium hydroxide and $(12\times 122.6)=1471.2mol$ of HF

As, given amount of NaOH and HF is more than the required amount. So, they are considered as an excess reagent.

Thus, aluminium oxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aluminium oxide produces 2 moles of cryolite

So, 122.6 moles of aluminium oxide will produce = $\frac{2}{1}\times 122.6=245.2mol$ of cryolite

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

Molar mass of cryolite = 209.94 g/mol

Moles of cryolite = 245.2 mol

Putting values in equation 1, we get:

$245.2mol=\frac{\text{Mass of cryolite}}{209.94g/mol}\\\\\text{Mass of cryolite}=(245.2mol\times 209.94g/mol)=51477.3g$

Converting this into kilograms, we use the conversion factor:

1 kg = 1000 g

So, $51477.3 g\times (\frac{1kg}{1000g})=51.48kg$

Hence, the mass of cryolite produced is 51.48 kg

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