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Vanyuwa [196]
2 years ago
4

Chem question!! With reference to the "Chemistry at Work" box on explosives, use bond enthalpies to estimate the enthalpy change

for the explosion of 1.00 g of nitroglycerin.
I found that the answer is 7.85 KJ, but how?
Chemistry
1 answer:
Jlenok [28]2 years ago
3 0

The mass of nitroglycerin is 1 g,

Molar mass of nitroglycerin is 227 g/mol

The number of moles = moles/molar mass

= 1 g / 227 g/mol

= 0.004405 mol

The equation for the bond enthalpy of the reaction is shown below, thus, the total enthalpy change in the reaction is as follows:

ΔHreaction = Number of moles (∑ΔHbond break - ∑ΔH bond forms)

= 0.004405 mol × -3564.5 kJ/mol

= -7.85 kJ

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A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro
Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on  combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q

Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J

Heat released on  combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows  :

Heat released on  combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q'

Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J

Heat released on  combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0
2 years ago
Jill is doing an experiment on the movement of pill bugs. She will place the pill bugs on flat surfaces covered with diffirent m
Ostrovityanka [42]

Answer:D

Explanation:

6 0
2 years ago
A student has two samples of NaCl, each one from a different source. Assume that the only potential contaminant in each sample i
bija089 [108]

Answer:

The correct option is;

A. Which sample has the higher purity

Explanation:

The information given relate to the presence of two samples of NaCl, from different sources

The only potential contaminant in each of the sources = KCl

The content of the sample = NaCl

The molar mass of NaCl = 58.44 g/mol

The molar mass of KCl = 74.5513 g/mol

Let the number of moles of KCl in the sample = X

For a given mass of NaCl, KCl mixture, we have;

The molar mass of potassium = 39.0983 g/mol

The molar mass of chlorine = 35.453 g/mol

The molar mass of sodium ≈ 23 g/mol

Therefore;

Each mole of KCl, will yield 35.453 g/mol per 74.5513 g/mol of KCl

While each mole of NaCl will yield 35.453 g/mol per 58.44 g/mol of NaCl

Therefore, the pure sodium chloride sample will yield more chlorine per unit mass of sample.

As such if the two samples have the same mass, the sample with the contaminant of KCl will yield less mass of chlorine per unit mass of the sample, from which the student will be able to tell the purity of the solution.

The sample with the higher purity will yield  a higher mass chlorine per unit mass of the sample.

6 0
2 years ago
At the beginning of the school year, a chalk company receives an order for 2000 boxes which is the largest order ever placed. Ea
lys-0071 [83]

Answer:

  • <u>259,000 g of chalk.</u>

Explanation:

<u>1) Data:</u>

a) 2000 boxes

b) 175 g / box

c) % yield = 74%

<u>2) Formula: </u>

  • % yield = (theoretical yield / actual yield) × 100

<u>3) Solution:</u>

a) Calcualte the actual yield:

  • mass of product = 2000 box × 175 g/ box = 350,000 g

b) Solve for the theoretical yield from the % yield formula:

  • % yield = (theoretical yield / actual yield) × 100

        ⇒ theoretical yield = % yield × actual yield / 100

             theoretical yield = 74% × 350,000g / 100 = 259,000 g

6 0
2 years ago
A 12.0 g sample of a metal is heated to 90.0 ◦C. It is then dropped into 25.0 g of water. The temperature of the water rises fro
Liula [17]

Answer:

The specific heat of the metal is 0.335 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of the metal = 12.0 grams

Initial temperature of the metal = 90.0 °C

Mass of the water = 25.0 grams

Initial temperature of water = 22.5 °C

Final temperature of water (and metal) = 25.0 °C

The specific heat of water = 4.18 J/g°C

<u>Step 2:</u> Calculate the specific heat of the metal

Qgained  = -Qlost

Qwater = -Qmetal

Q= m*c*ΔT

m(metal) *c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒ mass of the metal = 12.0 grams

⇒ c(metal) = TO BE DETERMINED

⇒ ΔT(metal) = T2 - T1 = 25.0 - 90.0 °C = -65.0

⇒ mass of the water = 25.0 grams

⇒ c(water) = the specific heat of water = 4.18 J/g°C

⇒ ΔT(water) = T2 - T1 = 25.0 - 22.5 = 2.5°C

12.0 * c(metal) * -65.0 = -25.0 * 4.18 * 2.5

c(metal) = 0.335 J/g°C

The specific heat of the metal is 0.335 J/g°C

6 0
2 years ago
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