Two samples of carbon tetrachloride were decomposed into their constituent elements. One sample produced 38.9 g of carbon and 463 g of chlorine, and the other sample produced 14.8 g of carbon and 144 g of chlorine. Part A Are these results consistent with the law of definite proportions?
Answers
Explanation:
Firstly, defining Law of definite proportion which states that in any chemical compound, the elements are the same and are in the same proportion by mass. It is also called Proust's law.
In sample A;
38.9 g of carbon
463 g of chlorine
In sample B;
14.8 g of carbon
144 g of chlorine.
Sample A, Ratio of mass of Carbon to mass of Chlorine
= 38.9 : 463
= 0.084
Sample B, Ratio of mass of Carbon to mass of Chlorine
= 14.8 : 144
= 0.103
These results above show that the ratio of the masses of Carbon to Chlorine in both samples A and B are not the same so therefore, the results are not consistent with the law of definite proportion.
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A mixture contains N a H C O 3 together with unreactive components. A 1.68 g sample of the mixture reacts with H A to produce 0.561 g of C O 2 . What is the percent by mass of N a H C O 3 in the original mixture
Answers
Answer:
C6H14 < C6H13Br < C6H13OH < C6H12(OH)2
Explanation:
Hello,
In this case, since the solubility in water is related with the presence of polar bonds in the given molecules we can see that C6H12(OH)2 has the presence two O-H bonds which promote the highest solubility via hydrogen bonds as well as the C6H13OH but in a lower degree as only on O-H bond is present. Next since the bond C-Br in is slightly close to the polar bond C6H13Br rather than the C-C bonds only had by C6H14 we can infer that C6H13Br is more soluble in water than C6H14, therefore the required order is:
C6H14 < C6H13Br < C6H13OH < C6H12(OH)2
Whereas C6H12(OH)2 is the most soluble and C6H14 the least soluble in water.
Best regards.
Answers
Answer:
Nhiệt dung riêng của đồng là
380
J
/
k
g
.
K
Explanation:
T4: Specific Heats and Molar Heat Capacities
Substance cp in J/g K cp in cal/g K or Btu/lb F
Aluminum 0.900 0.215
Bismuth 0.123 0.0294
Copper 0.386 0.0923
Brass 0.380 0.092
compound with a molecular mass of 46.07 amu causesthe temperature
in the calorimeter to rise from 25.000oC to 30.589
oC. The total heat capacity ofthe calorimeter and all
its contents is 3576 JoC-1. What is
the energy of combustion ofthe organic compound,
DU/ kJ
mol-1?
Answers
Answer:
1383.34 kJ/mol is the energy released on combustion of the organic compound.
Explanation:
Mass of an organic compound = 0.6654 g
Molar mass of organic compound = 46.07 g/mol
Moles of an organic compound =
Let heat evolved during burning of 0.6654 grams of an organic compound be -Q.
Heat absorbed by calorimeter = Q' = -Q
The total heat capacity of the calorimeter all its contents = C
C = 3576 J/°C
Change in temperature of the calorimeter =
ΔT = 30.589°C - 25.000°C = 5.589°C
Q' = 19.975 kJ
Q = -19.975 kJ (negative sign; energy released)
0.01444 moles of an organic compound gives 19.975 kilo Joule.
The 1 mole of an organic compound will give :
Answers
Answer:
9.09
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answers
Answer:
You could collect the mixture and pour it in water, stir it , ad filter out the sand. This uses the physical property of solubility.
Explanation:
The salt dissolved, the sand didn't.
Answers
Answer:
1.25 g
Explanation:
Now we have to use the formula;
N/No = (1/2)^t/t1/2
N= mass of cesium-137 left after a time t (the unknown)
No= mass of cesium-137 present at the beginning = 5.0 g
t= time taken for 5.0 g of cesium-137 to decay =60 years
t1/2= half life of cesium-137= 30 years
Substituting values;
N/5= (1/2)^60/30
N/5= (1/2)^2
N/5= 1/4
4N= 5
N= 5/4
N= 1.25 g
Therefore, 1.25 g of cesium-137 will remain after 60 years.