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1 year ago

Which of the following is not an example of temperature abuse ?

Chemistry

2 answers:

AnnyKZ [126]1 year ago

6 0

The following of what exactly?

docker41 [41]1 year ago

4 0

I’m not sure but I think food is not reheated enough to kill the pathogens

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A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c

Nata [24]

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

8 0

2 years ago

For some jobs, “good enough” is good enough. That’s sometimes true in chemistry. More often, though, careful planning, calculati

Lilit [14]

For the first task, lighting a campfire can be used. Quantities of firestarter and kindling do not need to be precise. For the second task, the process of titration can be used. In a titration, the precise amount of a substance may be determined by adding a precise amount of a reactive species

4 0

1 year ago

Read 2 more answers

Both black and white road surfaces radiate energy. at midnight on a starry night the warmer road surface is the

luda_lava [24]

The black road because during the day it absorbed more radiation than the with one

5 0

1 year ago

Use the PhET simulation to identify what happens to the concentrations of a dilute Drink mix mixture or a pure Drink mix solutio

elena-s [515]

Hello user, you did not add a picture of an attachment to enable me help you solve this problem. Please check the attachment I added i believe it is the question that needs to be solved.

Explanation:

1. In the first column

INCREASE CONCENTRATION: these are the items that needs to be placed in this bin

a. Add drink mix solid to a diluted mixture of drink mix in pure water

b. Add drink mix solution to a diluted mixture of drink mix in pure water

c. Add drink mix solid to pure drink mix solution

d Evaporate water from a diluted mixture of drink mix in pure water

e Evaporate water from pure drink mix solution

2. The items below are what should be placed in this bin,DECREASE CONCENTRATION:

a. Add water to a diluted mixture of drink mix in pure water

b. Add water to pure drink mix solution

3. These items here should be placed here. DOES NOT AFFECT CONCENTRATION:

a. Add pure drink mix solution to pure drink mix solution

b. Drain the pure drink mix solution

c. Drain the diluted mixture

7 0

2 years ago

A 251 g strip of glass wool is used to insulate a reaction flask. During the reaction the temperature of the glass wool increase

Ivahew [28]

Answer:

8.9 KJ

Explanation:

Given data:

Mass of strip = 251 g

Initial temperature = 22.8 °C

Final temperature = 75.9 °C

Specific heat capacity of granite = 0.67 j/g.°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 75.9 °C - 22.8 °C

ΔT = 53.1 °C

Q = 251 g × 0.67 j/g.°C × 53.1 °C

Q = 8929.8 J

Jolue to KJ.

8929.8J ×1 KJ / 1000 J

8.9 KJ

8 0

1 year ago