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3 years ago

Friction, conduction, and induction are all methods through which _______ can be transferred.

Chemistry

2 answers:

Ugo [173]3 years ago

8 0

Friction, conduction, and induction are all methods through which energy can be transferred.

Hope that helps.

GREYUIT [131]3 years ago

4 0

Answer:

It is charge.

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Which of the following will result in a chemical change?

brilliants [131]

Answer:

i want to say It would be Abecause a chemical change is somthing that cant NOT be brought back to its areganal proproties.coal will never be able to be bavk the way it was!!

Explanation:

because a chemical change is somthing that cant not be brought back to its areganal proproties.coal will never be able to be back the way it was.

4 0

2 years ago

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Consider the assumption that the four molecules of interest have approximately the same molar mass and very similar polarizabili

Igoryamba

The supposition is substantial on the grounds that, for these whose enthalpies is more prominent than n-pentane, they have a positive ΔH(polar), which implies there are more polar than the n-pentane. for these whose enthalpies is littler than n-pentane, they have a negative ΔH(polar), which implies there are less polar than the n-pentane. This supposition has an impact when I consider the predominant IMFS and different IMFS in the outline segment.

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3 years ago

Liquid water at 325 K and 8000 kPa flows into a boiler at a rate of 10 kg⋅s− 1 and is vaporized, producing saturated vapor at 80

masha68 [24]

Answer:

0.4058

Explanation:

From the steam table

At 325 K, saturated liquid

Enthalpy, Hf = 217 kJ/kg

Entropy Sf = 0.7274 kJ/kg-K

Specific volume Vf = 1.013 cm3/gm

Saturated pressure Psat = 12.87 kPa

For compressed liquid

P1 = 8000 kPa

Temperature T = 325 K

Thermal expansion coefficient B = 460 x 10^{-6} K^{-1}

Enthalpy at initial conditions

H1 = Hf + Vf (1-BT)(P1 - P_sat)

= 217 + 1.013*10^-3 (1 - 460*10^{-6})(8000 - 12.87)

= 223.881 kJ/kg

Entropy at initial conditions

S1 = Sf - BVf (P1 - Psat)

= 0.7274 - 460×10^{-6}*1.013*10^-3 (8000 - 12.87)

= 0.724 kJ/kg-K

At 8000 kPa, saturated vapor

H2 = 2759.9 kJ/kg

S2 = 5.7471 kJ/kg-K

T = 300 K

Heat added Q = H_2 - H_1

= 2759.9 - 223.881

= 2536 kJ/kg

Maximum work

W = (H1 - H2) - T (S1 - S2)

= 223.881 - 2759.9 - 300(0.724 - 5.7471)

= - 1029 kJ/kg

Fraction of heat added = W/Q

= 1029/2536

= 0.4058

3 0

3 years ago

When the pressure that a gas exerts on a sealed container changes from 893 mm hg to 778 mm hg, the temperature changes from 49.3

solong [7]

Answer: The final temperature of the gas is 7.58 °C.

Explanation: We are given initial and final pressure of the system and we need to find the final temperature of the system.

To calculate it, we use the equation given by Gay-Lussac.

His law states that pressure is directly related to the temperature of the gas.

$P\propto T$

Or,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1$ = initial pressure = 893 mmHg = 1.175atm (Conversion factor: 1atm = 760mmHg)

$T_1$ = initial temperature = 49.3°C = [49.3 + 273.15]K = 322.45K

$P_2$ = Final pressure = 778mmHg = 1.023atm

$T_$ = Final temperature = ?°C

Putting values in above equation, we get:

$\frac{1.175atm}{322.45K}=\frac{1.023atm}{T_2}\\\\T_2=280.73K$

Converting Final temperature from kelvin to degree Celsius.

$T_2=280.73K=[280.73-273.15]^oC=7.58^oC$

Hence, the final temperature of the gas is 7.58 °C.

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3 years ago

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A group of students obtained the following data while trying to determine the effect of exercise on pulse rate.

ser-zykov [4K]

Answer:

The answer is A

Explanation: They can pick and chose what kind of exercise there is as well as change it.

7 0

3 years ago