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2 years ago

What change(s) would increase the number of deaths in the weebug population?

Chemistry

1 answer:

tangare [24]2 years ago

3 0

The change that would cause an increase in the no. of deaths in the little population of wee bug would be:

b). Decrease Greenleaf

c). Increase Furbil

<u>Wee bug Population</u>

As per the question, the decrease in Greenleaf would result in reduced eatables for the wee bugs.

While the increase in Furbil would cause a threat to the population of the wee bug.

As a result, due to decreased food and increased energy storage by Furbils would increase the death of wee bug population.

Thus, <u>options b and c </u>are the correct answers.

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3 years ago

A rock falls off a cliff and hits the ground after three seconds. What is its velocity right before it hits the ground?

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4 years ago

Read 2 more answers

Calculate the total energy, in kilojoules, that is needed to turn a 46 g block

neonofarm [45]

Answer: The total energy, in kilojoules, that is needed to turn a 46 g block of ice at -25 degrees C into water vapor at 100 degrees C is 11.787 kJ.

Explanation:

Given: Mass = 46 g

Initial temperature = $-25^{o}C$

Final temperature = $100^{o}C$

Specific heat capacity of ice = 2.05 $J/g^{o}C$

Formula used to calculate the energy is as follows.

$q = m \times C \times (T_{2} - T_{1})$

where,

q = heat energy

m = mass

C = specific heat capacity

$T_{1}$ = initial temperature

$T_{2}$ = final temperature

Substitute the values into above formula as follows.

$q = m \times C \times (T_{2} - T_{1})\\= 46 g \times 2.05 J/g^{o}C \times (100 - (-25))^{o}C\\= 11787.5 J (1 J = 0.001 kJ)\\= 11.787 kJ$

Thus, we can conclude that the total energy, in kilojoules, that is needed to turn a 46 g block of ice at -25 degrees C into water vapor at 100 degrees C is 11.787 kJ.

7 0

3 years ago

How much heat is required to raise 36 g ice at – 10.0oC to steam at 110oC? (get your answers from question #1)

Gnom [1K]

Answer:

The total heat required to raise ice at -10°C to steam at 110°C = 91606.8 J

Explanation:

The heat involved in this process involves the following:

1. Heat to change ice at -10°C to ice at 0°C;

2. Heat to change ice at 0°C to water at 0°C

3. Heat to change water at 0°C to water at 100°C

4. Heat to change water at 100°C to steam at 100°C

5. Heat to change steam at 100°C to steam at 110°C

Specific heat capacity of ice, c = 2040 J/K/kg, Latent heat of fusion of ice, L = 3.35 × 10⁵ J/kg, specific heat capacity of water, c = 4182 J/K/kg, latent heat of vaporization of water, l = 2.26 × 10⁶ J/kg, specific heat capacity of steam, c = 1996 J/K/kg

Step 1: H = mcθ; where m = 30.0 g = 0.03 g, c = 2040 J/K/kg, θ = (0 - -10) = 10 K

H = 0.03 * 2040 * 10 = 612 J

Step 2: H = mL, where L = 3.35 × 10⁵ J/kg

H = 0.03 * 3.35 × 10⁵ = 10050 J

Step 3: H = mcθ, where c = 4182 J/K/kg, θ = (100 - 0) = 100 K

H = 0.03 * 4182 * 100 = 12546 J

Step 4: H = ml, where l = 2.26 × 10⁶

H = 0.03 * 2.26 × 10⁶ = 67800 J

Step 5: H = mcθ, where c = 1996 J/K/kg, θ = (110 - 100) = 10 K

H = 0.03 * 1996 * 10 = 598.8 J

Total heat required to raise ice at -10°C to steam at 110°C = (612 + 10050 + 12546 + 67800 + 598.8) J = 91606.8 J

Therefore, the total heat required to raise ice at -10°C to steam at 110°C = 91606.8 J

8 0

3 years ago