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Katena32 [7]
2 years ago
10

True or false; some organisms are not affected by any abiotic factors

Chemistry
2 answers:
777dan777 [17]2 years ago
8 0

Answer:

True

Explanation:

makkiz [27]2 years ago
3 0

Answer:

False

Explanation:

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Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 l of carbon dioxide at 20.0 °c an
kvv77 [185]

Answer:

Percentage by mass of calcium carbonate in the sample is 93.58%.

Explanation:

Assumptions:

  • calcium carbonate was dissolved completely and the amount of carbon dioxide released was proportional to the amount of calcium carbonate.
  • the sample did not contain any other compound that released or reacted with carbon dioxide.

PV = nRT

n = \frac{PV}{RT}

P = 791 mmHg = \frac{791 mmHg}{760 mmHg} x 1 atm = 1.040789 atm.

R = 8.314 \frac{J}{K*mol} = 0.082 \frac{L*atm}{K*mol}

T = 20^{0}C + 273^{0} C = 293 K

n = \frac{1.040789 atm * 1.14 L}{0.082 \frac{L * atm}{K * mol} * 293 K }

n = 0.0493839 mol.

given the equation of the reaction:

CaCO_{3}  + 2HCL =  CaCl_{2}  + CO_{2} + H_{2}O

from assumption:

amount carbon dioxide = amount of calcium carbonate

n(CaCO_{3} ) = n(CO_{2} )

reacting mass ( m )  = Molar Mass ( M ) * Amount ( n )

m(CaCO_{3} )   =   n(CaCO_{3} )  *   M(CaCO_{3} )

m = 0.04938397 mol * 100 \frac{g}{mol} =  4.93839 g

percentage by mass of   CaCO_{3} = \frac{mass of pure  }{mass of impure}*100  = \frac{4.93839g}{5.28g}*100  = 93.5802%.

8 0
2 years ago
If you have 10.0 grams of citric acid with enough baking soda (nahco3 how many moles of carbon dioxide can you produce?
kap26 [50]
Easy stoichiometry conversion :)

So, for stoichiometry, we always start with our "given". In this case, it would be the 10.0 grams of NaHCO3. This unit always goes over 1.

So, our first step would look like this:

10.0
------
  1

Next, we need to cancel out grams to get to moles. To do this, we will do grams of citric acid on the BOTTOM of the next step, so it cancels out. This unit in grams will be the mass of NaHCO3, which is 84.007. Then, we will do our unit of moles on top. Since this is unknown, it will be 1.

So, our 2nd step would look like this:

1 mole CO2
-----------------
84.007g NaHCO3

When we put it together: our complete stoichiometry problem would look like this:

10.0g NaHCO3     1mol CO2
---------------------- x -------------------------
            1                  84.007g NaHCO3

Now to find our answer, all we need to do is:
Multiply the two top numbers together (which is 10.0)
Multiply the two bottom numbers together (Which is 84.007)

And then....

Divide the top answer by the bottom answer.

10.0/84.007 is 0.119

So, from 10.0 grams of citric acid, we have 0.119 moles of CO2.

Hope I could help!
6 0
3 years ago
Read 2 more answers
CO2 + H2O - H2CO3<br> The reaction above is classified as
Elena L [17]

Answer:

Nonmetal oxides react with water to form oxyacids. Ex. CO2 + H2O → H2CO3 Page 3 Decomposition - compound (reactant) breaks down into 2 or more simpler substances.

3 0
2 years ago
Read 2 more answers
How to separate a mixture of Sodium chloride (salt) and sugar​
Zigmanuir [339]
By using chromatography
8 0
2 years ago
Which one of the following substances is a solution?
Nostrana [21]
Hdhdhdjdndjsjshshshhd
7 0
2 years ago
Read 2 more answers
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