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Vesnalui [34]
2 years ago
12

I need help with 4, 5, 8, 9, and 6. Quickly I need it before class starts. Worth points!!!!!! HelP

Chemistry
1 answer:
GalinKa [24]2 years ago
3 0

Answer:

4. 264.6J

5. 37.5J

6. 96J

7. 55Watts

8. 77.14m

9. 6s

10. 750Watts

Explanation:

4). Mechanical energy (potential energy) = mass (m) × acceleration due to gravity (g) × height (h)

m = 3kg, h = 9m, g = 9.8m/s²

P.E = 3 × 9 × 9.8

= 264.6J

5). Kinetic energy (K.E) = 1/2 × m × v²

Where;

m = mass (kg) = 3kg

v = velocity (m/s) = 5m/s

K.E = 1/2 × 3 × 5²

K.E = 1/2 × 3 × 25

K.E = 1/2 × 75

K.E = 37.5J

6). Work done (J) = Force (N) × distance (m)

Force = 12N, distance = 8m

Work done = 12 × 8

= 96J

7). Power = work done (J) ÷ time (s)

Work done = 550J, time = 10s

Power = 550/10

= 55Watts.

8). Work done = force (F) × distance (m)

Work done = 540J, force = 7N, distance = ?

540 = 7 × d

540 = 7d

d = 540/7

d = 77.14m

9). Power = work done (J) ÷ time (s)

Work done = 300J, time = ?, Power = 50Watts.

50 = 300/t

50t = 300

t = 300/50

t = 6seconds.

10). Power = work done (J) ÷ time (s)

This means that;

Power = force × distance / time

Force = 300N, distance = 5m, time = 2s

Power = 300 × 5 ÷ 2

Power = 1500 ÷ 2

Power = 750Watts

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Salt is often added to water to raise the boiling point to heat food more quickly. if you add 30.0g of salt to 3.75kg of water,
sammy [17]

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

<h3>What is the boiling-point elevation?</h3>

Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.

  • Step 1: Calculate the molality of the solution.

We will use the definition of molality.

b = mass solute / molar mass solute × kg solvent

b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m

  • Step 2: Calculate the boiling-point elevation.

We will use the following expression.

ΔT = Kb × m × i

ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C

where

  • ΔT is the boiling-point elevation
  • Kb is the ebullioscopic constant.
  • b is the molality.
  • i is the Van't Hoff factor (i = 2 for NaCl).

The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:

100 °C + 0.140 °C = 100.14 °C

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

Learn more about boiling-point elevation here: brainly.com/question/4206205

7 0
2 years ago
When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

m=1.00 g/mL\times 100.0 mL = 100 g

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

q=m\times c\times (T_{final}-T_{initial})

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = 4.18 J/^oC

T_{final} = final temperature = 23.1^oC

T_{initial} = initial temperature = 28.9^oC

Now put all the given values in the above formula, we get:

q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0
3 years ago
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