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Ira Lisetskai [31]
2 years ago
14

D) Evaluate your answer and explain why your answer makes sense​

Chemistry
1 answer:
Alik [6]2 years ago
4 0

Answer:

?

Explanation:

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A mixture of 454 kg of applesauce at 10 degrees Celsius is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet
stira [4]

Explanation:

The given data is as follows.

           Mass of apple sauce mixture = 454 kg

           Heat added (Q) = 121300 kJ

 Heat capacity (C_{p}) of apple sauce at 32.8^{o}C = 4.0177 kJ/kg^{o}C

So, Heat given by heat exchanger = heat taken by apple sauce

                            Q = mC_{p} \Delta T

or,                    Q = mC_{p} (T_{f} - T_{i})  

Putting the given values into the above formula as follows.

                     Q = mC_{p} (T_{f} - T_{i})  

              121300 kJ = 454 kg \times 4.0177 kJ/kg^{o}C \times (T_{f} - 10)

                      T_{f} = 76.5^{o}C

Thus, we can conclude that outlet temperature of the apple sauce is 76.5^{o}C.

3 0
3 years ago
How do magnets move objects?​
shutvik [7]

Answer: Magnets have a force which is known as no contact force. This force tries to attract the metallic objects near it, which makes it move.

Explanation:

7 0
3 years ago
After 11.5 days, 12.5% of a sample of radon-222 that originally weighed 42g remains. What is the half-life of this isotope?
Bond [772]
Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value.  The equation to describe the decay is
Nt=N0(1/2) ^{t/t(1/2)}  where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time.  So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days
5 0
3 years ago
A student has the following data: V1 = 822 mL; T1 = 75.0 °C; T2 = 25.0 °C. She uses this data to calculate V1 and gets -274 mL.
luda_lava [24]
Answer to this question is C. Regarding the volume.
4 0
3 years ago
2Al + 6HCl --> 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L
dem82 [27]

Answer:

8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

<em>Moles H2:</em>

PV = nRT; PV/RT = n

<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>

Replacing:

1atm*11L/0.082atmL/molK*273.15K = n

n = 0.491 moles of H2 must be produced

<em />

<em>Moles Al:</em>

0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required

<em />

<em>Mass Al -Molar mass: 26.98g/mol-:</em>

0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries

7 0
3 years ago
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