When a student mixes 5.00 g of NH4NO3 with 50.0 mL of water in a coffee-cup calorimeter, the temperature of the resultant solution decreases from 22.0 °C to 16.5 °C. Assume the density of water is 1.00 g/ml and the specific heat capacity of the resultant solution is 4.18 J/g·°C.
1) Calculate q for the reaction. You must show your work.
2) Calculate the number of moles of NH4NO3(s) which reacted. You must show your work.
3) Calculate ΔH for the reaction in kJ/mol. You must show your work.
Answer:
Explanation:
NH₄NO₃ = NH₄⁺ +NO₃⁻
heat released by water = msΔ T
m is mass , s is specific heat and ΔT is fall in temperature
= 50 x 4.18 x ( 22 - 16.5 ) ( mass of 50 mL is 50 g )
= 1149.5 J .
This heat will be absorbed by the reaction above .
q for the reaction = + 1149.5 J
2 )
molecular weight of NH₄NO₃ = 80
No of moles reacted = 5/80 = 1 / 16 moles.
3 )
5 g absorbs 1149.5 J
80 g absorbs 1149.5 x 16 J
= 18392 J
= 18.392 kJ.
= + 18.392 kJ
ΔH = 18.392 kJ / mol
Answer:
Explanation:
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In this case, considering the required product which is dinitrogen tetroxide whose molecular formula is N₂O₄, by considering its formation reaction which is starting by both gaseous nitrogen and oxygen as shown below:
Nevertheless, it should be balance since four oxygen atoms are present at the right side, thus, we obtain:
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Answer: The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:
where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.
We are given:
Mass of
Mass of
Mass of sample = 0.1153 g
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.3986 g of carbon dioxide, of carbon will be contained.
In 18g of water, 2 g of hydrogen is contained.
So, in 0.0578 g of water, of hydrogen will be contained.
To calculate the percentage composition of a substance in sample, we use the equation:
......(1)
Mass of sample = 0.1153 g
Mass of carbon = 0.1087 g
Putting values in equation 1, we get:
Mass of sample = 0.1153 g
Mass of hydrogen = 0.0066 g
Putting values in equation 1, we get:
Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.
Answer:
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B. A person who oversees the experiment to make sure it is following proper procedures.
C. The variable controlled by the scientist to affect the dependent variable.
D. The name for the set of independent and dependent variables that will be controlled by the scientist.
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The statement, that describes the ‘control’in an experiment is "the name for the set of independent and dependent variables that will be controlled by the scientist."
A control is an element in an experiment that remains intact or unaffected by other variables. An experiment or observation aiming to minimise the influence of variables other than the independent variable is referred to as a scientific control. It serves as a standard or point of reference against which other test findings are measured.
In a scientific experiment, an independent variable is the variable that is modified or manipulated in order to assess the effects on the dependent variable. In a scientific experiment, the dependent variable is the variable that is being tested and measured. The designation given to the set of independent and dependent variables that the scientist will regulate.
Hence the correct option is D.
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D. The name for the set of independent and dependent variables that will be controlled by the scientist.