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luda_lava [24]
2 years ago
9

10. A solution contains 130 grams of KNO3 dissolved in 100 grams of water When 3 more grams of KNO3 is added, none of it dissolv

es, nor do any additional crystals appear. The temperature of the solution is closest to
A.65 Celsius

B. 68 Celsius

C. 70 Celsius

D. 72 Celsius
Chemistry
2 answers:
krek1111 [17]2 years ago
6 0

Answer:

B. 68 Celsius

Explanation:

Murljashka [212]2 years ago
3 0

Answer:

Option B

Explanation:

We will check the solubility graph for potassium nitrate,  KNO 3. Based on the graph it can be said that the temperature of solution when 130 grams of KNO3 dissolves in 100 grams of water is near to 65 degree Celsius. Now if three grams of solute is increased then the temperature of the solution will increase by a degree or so and hence the most probable temperature would be 68 degree Celsius.

Hence, option B is correct

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The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ <span>joule = 1 eV

Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules

From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
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4 0
2 years ago
Read 2 more answers
Write the reduction reaction of glucose to form sorbitol. List and explain the side effects caused by too much sorbitol consumpt
Lemur [1.5K]

Answer:

The product of reduction of glucose is sorbitol

The side effects caused by too much sorbitol consumption include: Diarrhea, Nausea, stomach discomfort

Explanation:

Please find attached the reaction of glucose with NADPH to produce sorbitol

5 0
2 years ago
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Q1) A vapor-compression refrigeration system operates on the cycle of Fig. 9.1. The refrigerant is 1,1,1,2-Tetrafluoroethane. Gi
hoa [83]

Answer:

i)   0.5071 (kg/s)

ii)  -1407.1 kj/kg

iii)  204.05 Kw

iv)  5.881

v)    9.238

Explanation:

Given Data:

evaporation temperature ( T ) = 4°c = 277.15 K

Condensation Temperature ( T ) = 34°c = 307.15 K

<em>n</em> ( compressor efficiency ) = 0.76

refrigeration rate = 1200 kJ.s^-1

i) determine the circulation rate of the refrigerant

m = \frac{Q}{H2 - H1}  = \frac{Q}{H2 - H4\\}  ------- 1

Q = 1200 Kj.s^-1

H2 = entropy at step 2 = 2508.9 (kJ / kg ) ( gotten from Table F )

H4 = entropy at step 4 = 142.4 ( kJ/ kg )

back to equation 1

m ( circulation rate of refrigerant ) = 0.5071 (kg/s)

ii) heat transfer rate in the condenser

Q = m ( H4 - H3 )

    = 0.5071 ( 142.4 - 2911.27 )

    = -1407.1 kj/kg

where H3 = H2 + ΔH23 = 2911.27 (kj/kg) ( as calculated )

iii) power requirement

w = m * ΔH23

   = 0.5071 (kg/s) * 402.37 (kj/kg) =  204.05 Kw

where: ΔH23 = \frac{H'3 - H2 }{0.76} = \frac{2814.7-2508.9}{0.76} = 402.37 (kj/kg)

iv) coefficient of performance of a cycle

W = Qc / w

  = 1200 Kj.s^-1/ 204.05 kw

  = 5.881

v) coefficient of performance of a Carnot refrigeration cycle

w_{carnot} = \frac{T2}{T4 - T2}

            =  277.15 / ( 307.15 - 277.15 )

            = 9.238

4 0
2 years ago
Recording station x has determined that the epicenter of a recent earthquake was 250 km away. no information is available yet fr
KatRina [158]
The epicenter was located somewhere on a circle centered at Recording station X, with a radius of 250 km.<span>
</span>
7 0
2 years ago
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The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm? Use Ideal Gas Law (PV =
Scorpion4ik [409]

Answer:

26.3 mL

Explanation:

Step 1:

Obtaining an appropriate gas law from the ideal gas equation.

This is illustrated below:

From the ideal gas equation:

PV = nRT

Divide both side by T

PV/T = nR

At this stage, we'll assume the number of mole (n) to be constant.

Note: R is the gas constant.

PV/T = constant.

We can thus, write the above equation as:

P1V1/T1 = P2V2/T2

The above equation is called the general gas equation.

Step 2:

Data obtained from the question. This includes the following:

Initial volume (V1) = 27.5 mL

Initial temperature (T1) = 22.0°C = 22.0°C + 273 = 295K

Initial pressure (P1) = 0.974 atm.

Final temperature (T2) = 15.0°C = 15.0°C + 273 = 288K

Final pressure (P2) = 0.993 atm

Final volume (V2) =..?

Step 3:

Determination of the final volume of the gas using the general gas equation obtained. This is illustrated below:

P1V1 /T1 = P2V2/T2

0.974 x 27.5/295 = 0.993 x V2/288

Cross multiply to express in linear.

295x0.993xV2 = 0.974x27.5x288

Divide both side by 295 x 0.993

V2 = (0.974x27.5x288)/(295x0.993)

V2 = 26.3 mL

Therefore, the new volume of the gas is 26.3 mL

4 0
2 years ago
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