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How many joules of heat are removed from a 21.0 g sample of water if it is cooled from 34.0°C

yaroslaw [1]

Answer:

527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.

In other words, the sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a change in temperature without there being a change in physical state (solid, liquid or gaseous). The equation that allows to calculate this heat exchange is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT=Tfinal-Tinitial is the change in temperature.

In this case:

c= 4.184 $\frac{J}{g*C}$
m=21 g
ΔT=Tfinal-Tinitial=28 °C - 34 °C=-6 °C

Replacing:

Q= 4.184 $\frac{J}{g*C}$ * 21 g* (-6 C)

Q= - 527.184 J

To lower the temperature, heat has to be given, for that the final temperature must be lower than the initial temperature; and it receives the name of transferred heat and has a negative value, as in this case.

527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.

4 0

3 years ago

An ice cream company decides to make a new lower calorie flavor,but testers reported that the flavor wasnt as good as the origin

IgorLugansk [536]

Answer:

Yes, the given instance is an example of a trade-off.

Explanation:

The trade-off seems to be a strategy to minimize or surrender one or more favorable results in consideration for increasing either receiving other favorable results to optimize cumulative benefit or efficacy within such conditions.
The above is indeed a trade-off since the participants are experimenting in exchange again for ice cream individuals have to enjoy. The ice cream manufacturer always gets a review of the technician in exchange for free ice cream.

3 0

3 years ago

When of a certain molecular compound X are dissolved in of dibenzyl ether , the freezing point of the solution is measured to be

kondaur [170]

This question is incomplete, the complete question is;

When 4.28 g of a certain molecular compound X are dissolved in 60.0 g of dibenzyl ether [(C₆H₅CH₂)₂0] , the freezing point of the solution is measured to be -3.2°C . Calculate the molar mass of X.

If you need any additional information on dibenzyl ether, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to significant digit.

Answer: molar mass of solute (X) is 88.03 g/mol

Explanation:

Given that;

mass of solute = 4.28 g

mass of solvent = 60.0 g = 0.060 kg (Dibenzyl ether)

depression constant kf = 6.17 °CKg/mol

Freezing Point of solvent T₀ = 1.80°C (Dibenzyl ether)

freezing point of solution Tsol = -3.20°C

Now we know that

Depression in freezing point ΔTf = depression constant kf × molaity m

and (ΔTf = T₀-Tsol)

so T₀ - Tsol = kf × m

we substitute

1.80 - (-3.20) = 6.17 × m

5 = 6.17 × m

m = 5 / 6.17

m = 0.8103 kg/mol

so molaity m = 0.8103 kg/mol

we know that

Molaity of solute m = (mass of solute / M.wt of solute) × ( 1 / mass of solvent in Kg)

solve for molar mass of solute

molar mass of solute = (mass of solute / molaity) × ( 1 / mass of solvent in Kg)

now we substitute

molar mass = (4.28g / 0.8103 kg/mol) × (1 / 0.060kg)

molar mass = ( 5.2839 × 16.66 ) g/mol

molar mass = 88.0297 g/mol ≈ 88.03 g/mol

Therefore molar mass of solute (X) is 88.03 g/mol

3 0

3 years ago

HELP WITH CHEMISTRY PLEASE!

maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

In our problem:

P1 = ??? (is needed to be calculated) and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.

2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

In our problem:

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? (is needed to be calculated)

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.

3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

In our problem:

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? (is needed to be calculated) and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.

4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; 101.325 kPa = 1.0 atm, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.

5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

In our problem:

P1 = 2200.0 mmHg and T1 = ??? (is needed to be calculated) .

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

6 0

3 years ago

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This is actually fun. First comment is brainliest.

Diano4ka-milaya [45]

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6 0

3 years ago

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