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Kobotan [32]
3 years ago
7

What are the main causes for seasons on earth?

Chemistry
2 answers:
galben [10]3 years ago
7 0
The tilt of the earth
yuradex [85]3 years ago
4 0
As the earth turns one of the sides moves closer to the sun, creating summer in that area, at the same time the part of the earth on the other side moves farther away, hence making it colder and the season becomes winter
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1 Balance each equation. Paragraph Styles Voice Sensity 2) Indicate the type of reaction by classifying each reaction as single
nika2105 [10]

1. Double replacement (DR)

2. Decomposition (D)

<h3>Further explanation</h3>

1. Al2(SO4)3 + Ca3(PO4)2 -> 2AIPO4 + 3CaSO4

Double replacement (DR) : there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product

General form :

AB + CD -> AD + CB

2.  2NaCIO3 → 2NaCl + 3O2

Decomposition (D) : Reactant breakdown into simpler ones(reverse of combination)

General form :

AB ---> A + B

6 0
3 years ago
How can the rate constant be determined from the rate law?
Vinil7 [7]

Answer:

B. The rate constant is the reaction rate divided by the concentration

terms.

Explanation:

The rate constant can be determined from the rate law because it is the reaction rate divided by the concentration terms. I hope I could help! :)

7 0
2 years ago
A student mixes a white solid and a clear liquid together. The mixture turning orange is evidence of a chemical reaction.
yulyashka [42]

Answer:

true

Explanation:

7 0
2 years ago
Read 2 more answers
Why do farmers spray water over their crops before a frost?
Nezavi [6.7K]

Answer:

When water freezes and turns into ice, it releases latent heat. Then, the ice that builds up on the plant will insulate it from the colder surrounding air temperatures. Because of this, some growers choose to spray their crop with water before the freeze occurs.

Explanation:

5 0
3 years ago
A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
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