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sladkih [1.3K]
3 years ago
11

A teaspoon of the carbohydrate sucrose (common sugar0contains 16 Calories (16 KCal). What is the mass of the teaspoon of the suc

rose if the average number of Calories for the carbohydrates is 4.1 Calories/g?
Chemistry
1 answer:
Tcecarenko [31]3 years ago
5 0

Answer:

3.902 grams is the mass of the teaspoon of the sucrose.

Explanation:

Calories in 1 teaspoon of carbohydrate sucrose = 16 calorie

The average number of calories for the carbohydrates is 4.1 Calories/g

Then number of grams in 1 calorie =\frac{1}{4.1} g

Mass of 16 calories:

16 caloire=16\times \frac{1}{4.1 }g=3.902 g

3.902 grams is the mass of the teaspoon of the sucrose.

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The decomposition of mercury (ii) oxide at high temperature, is it an endothermic or exothermic process? Write a chemical reacti
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Explanation:

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Read 2 more answers
If 0.500 mol neon at 1.00 atm and 273 K expands against a constant external pressure of 0.100 atm until the gas pressure reaches
trapecia [35]

Answer:

The work done on neon = -323 J

The internal energy change= -392.84 J

The heat absorbed by neon = -69.84 J

Explanation:

Step 1: Data given

Number of moles  = 0.500 moles

Pressure  = 1 atm

Temperature  = 273 Kelvin

The pressure will change from 1.00 atm to 0.200 atm. The temperature changes from 273 to 210 Kelvin.

a) calculate the work done on neon

W = -P(V2-V1)    

⇒ with P = the pressure = 0.1 atm

⇒ with V1 = the initial volume = nRTi /Pi

⇒ with V2 = the final volume = nRTf /Pf

W = -PnR((T2/P2) -(T1/P1))

⇒ with T2 = the final temperature = 210 K

 ⇒ with T1 = the initial temperature = 273 K

 ⇒ with P2 = the final pressure = 0.200 atm

 ⇒ with P1 = the initial pressure = 1.00 atm

W = -nR (210*(0.1/0.2) - 273*(0.1/1.00))

W = -nR*(105 - 27.3)

W= -(0.500)*(8.314)*(77.7)

W = -323 J

b) calculate the internal energy change

E = (3/2)*nRT

ΔE = Ef - Ei

ΔE =(3/2)*nR(T2-T1)

⇒ with n= number of moles = 0.500 moles

⇒ with T2 =the final temperature = 210 K

⇒ with T1 = the initial temperature = 273 K

ΔE = (3/2)*(0.5)*(8.314)(210-273)

ΔE = -392.84 J

c) Calculate the heat absorbed by neon

ΔE = q + W

q = ΔE -W

⇒ with ΔE = -392.84 J

⇒ with W = -323 J

q = -392.84 J -( -323 J)

q =-392.84 J + 323 J

q = -69.84 J

4 0
3 years ago
Are the properties of a specific element the same as the properties of a molecule made up of that element? Why or why not?
LuckyWell [14K]

Answer:

no molecule and properties or defent

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2 years ago
Nombre las dos ciudades involucradas en los peligros potenciales de la ciencia y tecnología que ocurrió en el año 1945.
gogolik [260]

Answer:

ver explicacion

Explanation:

En 1945, durante la Segunda Guerra Mundial, dos ciudades japonesas; Hiroshima y Nagasaki fueron bombardeadas con bombas atómicas durante el ataque ofensivo.

Hiroshima fue bombardeada el 6 de agosto de 1945 mientras que Nagasaki fue bombardeada el 9 de agosto de 1945.

Las bombas provocaron la muerte instantánea de aproximadamente 120.000 personas y obligaron al emperador japonés a rendirse incondicionalmente.

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3 years ago
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