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Aliun [14]
3 years ago
13

Calculate the percentage of limestone that dissolved from each solution. Start by subtracting the final mass from the initial ma

ss. Divide that number by the initial mass. Then multiply the result by 100 to make it a percent. Use this formula:
% dissolved= initial mass- final mass/ initial mass x 100

ORIGINAL: 0% Acid 33% Acid 66% Acid 100% Acid

Initial Mass (g) 10.0 10.1 10.0 9.9

Final Mass (g) 10.0 10.0 9.8 9.5


Record the percentage of limestone dissolved in each acid concentration.
Look at attched...

Chemistry
1 answer:
Leno4ka [110]3 years ago
7 0

9514 1404 393

Answer:

  see attached

Explanation:

It works nicely to let a spreadsheet do the calculations described in your problem statement.

You don't actually have to multiply by 100 to make the number a percent. Spreadsheets have the capability to show a value as a percentage just by changing the format.

You might be interested in
Which choice can be classified as a pure substance?
ozzi

Answer:

d. compound

Explanation:

compound and elements are pure substances

8 0
3 years ago
(I)how many atoms are present in 7g of lithium?
ICE Princess25 [194]

Answer :

(i) The number of atoms present in 7 g of lithium are, 6.07\times 10^{23}

(ii) The number of atoms present in 7 g of lithium are, 1.204\times 10^{24}

(iii) The number of moles of F_2 is, 1 mole

The number of moles of CO_2 is, 0.5 mole

The number of moles of OH^- is, 1 mole

Explanation :

<u>Part (i) :</u>

First we have to calculate the moles of lithium.

\text{Moles of }Li=\frac{\text{Mass of }Li}{\text{Molar mass of }Li}

Molar mass of Li = 6.94 g/mole

\text{Moles of }Li=\frac{7g}{6.94g/mol}=1.008mole

Now we have to calculate the number of atoms present.

As, 1 mole of lithium contains 6.022\times 10^{23} number of atoms

So, 1.008 mole of lithium contains 1.008\times 6.022\times 10^{23}=6.07\times 10^{23} number of atoms

Thus, the number of atoms present in 7 g of lithium are, 6.07\times 10^{23}

<u>Part (ii) :</u>

First we have to calculate the moles of carbon.

\text{Moles of }C=\frac{\text{Mass of }C}{\text{Molar mass of }C}

Molar mass of C = 12 g/mole

\text{Moles of }C=\frac{24g}{12g/mol}=2mole

Now we have to calculate the number of atoms present.

As, 1 mole of carbon contains 6.022\times 10^{23} number of atoms

So, 2 mole of carbon contains 2\times 6.022\times 10^{23}=1.204\times 10^{24} number of atoms

Thus, the number of atoms present in 7 g of lithium are, 1.204\times 10^{24}

<u>Part (iii) :</u>

<u>To calculate the moles of </u>F_2<u> :</u>

\text{Moles of }F_2=\frac{\text{Mass of }F_2}{\text{Molar mass of }F_2}

Molar mass of F_2 = 38 g/mole

\text{Moles of }F_2=\frac{19g}{19g/mol}=1mole

Thus, the number of moles of F_2 is, 1 mole

<u>To calculate the moles of </u>CO_2<u> :</u>

\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}

Molar mass of CO_2 = 44 g/mole

\text{Moles of }CO_2=\frac{22g}{44g/mol}=0.5mole

Thus, the number of moles of CO_2 is, 0.5 mole

<u>To calculate the moles of </u>OH^-<u> ions :</u>

\text{Moles of }OH^-=\frac{\text{Mass of }OH^-}{\text{Molar mass of }OH^-}

Molar mass of OH^- = 17 g/mole

\text{Moles of }OH^-=\frac{17g}{17g/mol}=1mole

Thus, the number of moles of OH^- is, 1 mole

4 0
3 years ago
What is the correct answer?
Ivahew [28]
C - bc as the water boils the cold water particles melt and become the hot water particles!
3 0
3 years ago
Read 2 more answers
Help ASAP need help and thank you:)
Akimi4 [234]

Answer:

50 N

Explanation:

Because It not in motion means it standing still. So the answer is 50 Newton

4 0
3 years ago
Given: C + O2 → CO2 Bond Bond Energy (kJ/mol) C=O 799 O=O 494 Calculate the enthalpy change for the chemical reaction. The chang
aliya0001 [1]

Answer:

-1104 kJ/mol

Explanation:

The change in the enthalpy of a reaction is equal to the difference between: the sum of the enthalpy changes of the bonds broken and the sum of the enthalpy changes of the bonds formed.

The bonds broken correspond to the cleavage of bonds of the reactants, the bonds formed correspond to the bonds of the products:

  • we only break oxygen O=O bond, since carbon is not bonded to anything;
  • we form two C=O bonds in carbon dioxide.

Therefore, the enthalpy change is calculated by:

\Delta H^o = 1 mol\cdot 494 kJ/mol - 2 mol\cdot 799 kJ/mol = -1104 kJ/mol

3 0
3 years ago
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