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True [87]
2 years ago
10

In which level of organization would the roots of a plant most likely be classified. Make sure to explain

Chemistry
1 answer:
Nookie1986 [14]2 years ago
6 0

Answer:

jsjsososjahshbsjakaokwwj

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The correct answer is the 3rd one  
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5. How much heat (in calories) is absorbed by a reaction when
Wewaii [24]

Answer:

1.2 × 10⁴ cal

Explanation:

Given data

  • Mass of water (m): 300 g
  • Initial temperature: 80 °C
  • Final temperature: 40°C

We can calculate the heat released by the water (Q_w) when it cools using the following expression.

Q_w = c \times m \times (T_f - T_i)

where

c is the specific heat capacity of water (1 cal/g.°C)

Q_w = \frac{1cal}{g.\°C}  \times 300g \times (40\°C - 80\°C) = -1.2 \times 10^{4} cal

According to the law of conservation of energy, the sum of the heat released by the water (Q_w) and the heat absorbed by the reaction (Q_r) is zero.

Q_w + Q_r = 0\\Q_r = -Q_w = 1.2 \times 10^{4} cal

7 0
3 years ago
What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a
Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})

Where:

Q: is the energy transferred = 5.0 MJ

k_{B}: is the Boltzmann constant = 1.38x10⁻²³ J/K  

T_{i}: is the initial temperature = 1000 K

T_{f}: is the final temperature = 500 K

Hence, the entropy change is:

\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}                                    

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!          

7 0
3 years ago
Describe the different types of storms.
Genrish500 [490]

Answer:

The different types of storms are hailstorms, ice storms, snowstorms, thunderstorms, wind storms, hurricanes, and tornadoes.

Explanation:

8 0
3 years ago
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