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Alinara [238K]
3 years ago
6

Why is calcium carbonate used as a building material

Chemistry
2 answers:
Yuliya22 [10]3 years ago
7 0
Calcium Carbonate is very cheap to produce in bulk and has some properties that can be favorable for building a house.  That being said, im guessing you got this question from the acid rain post with calcium carbonate.  Many things will break down in contact with an acidic substance but it will take a very long time in the case of acid rain (because it is dilute)
VMariaS [17]3 years ago
3 0
Because it is used for building - making concrete and cement - and the manufacture of glass, steel and iron. 
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The “fatal flaw” in Rutherford’s model of the atom was the idea that if the electrons were ________.
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Which of the following compounds will produce H2O as one of the products when it decomposes?
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2 years ago
Read 2 more answers
Explain how the information given below could be used to determine the concentration of an unknown sample.
Vika [28.1K]

Answer:

See below

Explanation:Plot the known concentrations and adsorbance data.  Draw a best fit line through thwe points.  When the absorbance of a solution of unknown concentration (but same substance) is determined, find the concentration from the line at that absorption value.  See attached graph.

E.g., an sample of the same substance had an absorbance of 0.35.  Find that on the x scale and then determine the concentration that would be required to produce that level of absorbance.  0.483M in this case.

8 0
1 year ago
An unknown compound contains only C, H, and O. Combustion of 8.50 g of this compound produced 20.0 g CO2 and 5.46 g H2O. What is
Galina-37 [17]

Answer:

The answer to your question is: C₃H₄O

Explanation:

Data

CxHyOz = 8.5 g

CO2 = 20 g

H2O = 5.46 g

Reaction

             CxHyOz + O2     ⇒    CO2   +    H2O

CO2

    MW = 44g

                       44g CO2 ----------------- 12g of C

                        20g CO2 ----------------   x

                       x = (20 x 12) / 44

                       x = 5.45 g of C

                     # of moles = n = 5.45 / 12 = 0.454 mol of C

H2O

     MW = 18 g

                       18 g H2O ------------------- 2g of H

                       5.46 g    --------------------   x

                       x = (5.46 x 2) / 18 = 0.61 g of H

                       n = 0.61 / 1 = 0.61 moles of H2

Mass of O2

              mass CxHyOz = mass CO2 + mass H2 + mass O2

              mass O2 = 8.5 - 5.45 - 0.61

              mass O2 = 2.44g

              n = 2.44 / 16 = 0.153 mol of O2

Now, divide by the lowest number of moles

0.454 mol of C/ 0.153 = 2.97 ≈ 3

0.61 moles of H2/ 0.153 = 3.99 ≈ 4

0.153 mol of O2/ 0.153 =  1

Then, the empirical formula is: C₃H₄O

7 0
3 years ago
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