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scZoUnD [109]
2 years ago
6

Convert 22 ml in to liters

Chemistry
1 answer:
prisoha [69]2 years ago
8 0

Answer:

0.022

Explanation:

milliter (ml) = 1 cubic centimeter (cc)= 0.001 liters (l) = 0.000001 cubic meters (m3).

1 ml = 0.061024 cubic inches (in3) ; 1 in3 = 16.4 ml.

1 ml = 0.000035 cubic feet (ft3); 1 ft3 = 28,317 ml.

1 ml = 2.64 x 10-4 U.S. gallons (gal); 1 gal = 4.55 x 103 ml.

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How is chemical and physical properties of matter like in different
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When a person looks at a bright light, tiny muscles in the eye contract so less light can enter the eye.
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2 years ago
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A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure of 42.9 kPa. What is the partial pressure of
Lady bird [3.3K]
The total Pressure equals the sum of all pressures contained 

<span>Since total pressure and the pressure of nitrogen and oxygen is given, finding the pressure of carbon dioxide is given by: </span>

<span>Pressure of Carbon dioxide = 42.9- 6.6- 23.0 </span>
<span>=13.3kPa </span>
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3 years ago
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When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
2 years ago
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