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Artist 52 [7]
3 years ago
8

a compound contains two elements, sulfur and oxygen. in a sample of the compound, the mass of sulfur is 6.41 grams. the mass of

the oxygen is 9.60 grams. what is the percentage composition of the compound
Chemistry
1 answer:
Scrat [10]3 years ago
4 0
Percent composition is mass of element over mass of compound times 100. 6.41 + 9.60 = 16.01 g of compound, so % composition of sulfur is 6.41/16.01 x100 or 40.0%. % oxygen is 9.60/16.01 x 100 or 60.0%.
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GenaCL600 [577]

Answer:

remaining still during the night.

Explanation:

3 0
3 years ago
When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin
loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point

K_f = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9

8.2^0C=1\times K_f\times 1.9

K_f=4.32^0C/m

Now Depression in freezing point for sodium chloride is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=20.0^0C = Depression in freezing point

K_f = freezing point constant  

m= molality = \frac{136g\times 1000}{950g\times 58.5g/mol}=2.45

20.0^0C=i\times 4.32^0C\times 2.45

i=1.9

Thus vant hoff factor for sodium chloride in X is 1.9

3 0
3 years ago
How many mL of a stock 50% (w/v) KNO3 solution are needed to prepare 250 mL of a 20% (w/v) KNO3 solution?
Andre45 [30]

Answer:

100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.

Explanation:

In the given question it is mentioned that

     S1=50%

      V2=250ml

      S2= 20%

We all know that

                     V1S1=V2S2

                     ∴V1=  V2×S2÷S1

                     ∴V1=  V2S2×1/S1

                      ∴V1= 250×20÷50

                       ∴V1= 100ml

 

6 0
4 years ago
How many grams of oxygen are there in 45.7 grams of Ba(NO2),?
Flauer [41]
I’m assuming you mean barium nitrite, Ba(NO2)2.

First convert grams of Ba(NO2)2 to moles using the molar mass of Ba(NO2)2. Then use the mole ratio of 4 moles of oxygen per 1 mole of Ba(NO2)2 to convert to moles of oxygen. Then use the molar mass of oxygen to convert to grams of oxygen.

45.7 g Ba(NO2)2 • 1 mol Ba(NO2)2 / 229.35 g Ba(NO2)2 • 4 mol O / 1 mol Ba(NO2)2 • 16.0 g O / 1 mol O = 12.8 g oxygen
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3 years ago
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Answer:

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Explanation:

7 0
3 years ago
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