A1.1000 gram carbonate sample chosen from li2co3, k2co3, na2co3 and caco3 was reacted with h2so4 and was found to lose 0.3497 gram of co2. (1) show the calculation of the % co2 in the unknown carbonate sample. (2) show the calculation of the % co2 in each of the carbonate compounds and identify the unknown carbonate from the list. atomic weights: c = 12.01, o = 16.00. mws: li2co3 = 73.892, k2co3 = 138.21, na2co3 = 105.99 and caco3 = 100.09

For 1: The percentage composition of in unknown carbonate sample is 31.79 %

For 2: The unknown carbonate sample is potassium carbonate.

Explanation:

To calculate the percentage composition of carbon dioxide in sample, we use the equation:

    .....(1)

Mass of sample = 1.1000 g

Mass of carbon dioxide = 0.3497 g

Putting values in equation 1, we get:

Hence, the percentage composition of in unknown carbonate sample is 31.79 %

We are given:

Mass of carbon dioxide =

Mass of lithium carbonate= 73.892 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:

The percent composition of carbon dioxide in lithium carbonate is 59.55 %

Mass of potassium carbonate= 138.21 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:

The percent composition of carbon dioxide in potassium carbonate is 31.84 %

Mass of sodium carbonate= 105.99 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:

The percent composition of carbon dioxide in sodium carbonate is 41.52 %

Mass of calcium carbonate = 100.09 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:

The percent composition of carbon dioxide in calcium carbonate is 43.97 %

As, the percent composition of carbon dioxide in the unknown sample matches the percent composition of carbon dioxide in potassium carbonate.

Hence, the unknown carbonate sample is potassium carbonate.

i know insulin is primary but i dont know the others im sorry

explanation: