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san4es73 [151]
2 years ago
10

I need help on this may u help me?

Chemistry
1 answer:
sergij07 [2.7K]2 years ago
7 0

Answer:

just go to googel and it will tell you

Explanation:

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At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ / mol. When 1.697 g of compound
melisa1 [442]

Answer:

13.85 kJ/°C

-14.89 kJ/g

Explanation:

<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>

<em />

The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

1.697g.\frac{1mol}{101.67g} .\frac{(-3039.0kJ)}{mol} =-50.72kJ

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qcomb + Qcal = 0

Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ

The heat capacity (C) of the calorimeter can be calculated using the following expression.

Qcal = C . ΔT

where,

ΔT is the change in the temperature

Qcal = C . ΔT

50.72 kJ = C . 3.661 °C

C = 13.85 kJ/°C

<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>

Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ

The heat of combustion per gram of B is:

\frac{-56.09 kJ}{3.767g} =-14.89 kJ/g

4 0
3 years ago
15.50 g of NH4Cl reacts with an excess of AgNO3. In the reaction 35.50 g AgCl is produced. what is the theoretical yield of AgCl
Readme [11.4K]

Answer:

The answer to your question is 41.6 g of AgCl

Explanation:

Data

mass of NH₄Cl = 15.5 g

mass of AgNO₃ = excess

mass of AgCl = 35.5 g

theoretical yield = ?

Process

1.- Write the balanced chemical reaction.

              NH₄Cl  +  AgNO₃   ⇒   AgCl  +  NH₄NO₃

2.- Calculate the molar mass of NH₄Cl and AgCl

NH₄Cl = 14 + 4 + 35.5 = 53.5 g

AgCl = 108 + 35.5 = 143.5 g

3.- Calculate the theoretical yield

                 53.5 g of NH₄Cl -------------------- 143.5 g of AgCl

                  15.5 g of NH₄Cl  -------------------    x

                         x = (15.5 x 143.5) / 53.5

                         x = 2224.25 / 53.5

                         x = 41.6 g of AgCl

8 0
3 years ago
Which of the following is most likely to produce data that are not precise
IrinaK [193]
Precision is obtained by getting values that are very close together. If you mess around with the protocol, you'll end up with crazy values that probably are neither accurate or precise.
7 0
3 years ago
You need to prepare at 2.0 mL sample of a diluted drug for injection. The total amount of the drug to be injected in this 2.0 mL
pav-90 [236]

Answer:

a) The concentration of drug in the bottle is 9.8 mg/ml

b) 0.15 ml drug solution + 1.85 ml saline.

c) 4.9 × 10⁻⁵ mol/l

Explanation:

Hi there!

a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml

b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:

0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.

The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)

If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.

To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline

c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.

Let´s convert it to molarity:

0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l

3 0
3 years ago
Unscramble the word nlpituloo
nadezda [96]
Hello

The word is "pollution"

Hope this helps and have a great day!
3 0
3 years ago
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