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neonofarm [45]
3 years ago
13

Which factors cause earth to experience seasons? Check all that apply.

Chemistry
2 answers:
Nataly [62]3 years ago
4 0

Answer:

the speed of the earth's rotation, the tilt of the earth's axis and the directness of the sun

Ainat [17]3 years ago
3 0

Answer:

B, C

Explanation:

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__C8H18(I) + __ O2(g) —> __Co2(g) + __H2O(g) Balance out the equation
Ymorist [56]

Answer:

2C8H18(l) + O2(g)--->CO2(g)+H2O

3 0
2 years ago
An equinox occurs when the:
xxTIMURxx [149]

Answer: sun is directly over the equator

Explanation:

There are only two times of the year when the Earth's axis is tilted neither toward nor away from the sun, resulting in a "nearly" equal amount of daylight and darkness at all latitudes.

3 0
3 years ago
Can you guess this place? In Arizona.
Temka [501]

Answer:

a high school classroom?

Explanation:

a high school classroom prob chemistry

8 0
2 years ago
Read 2 more answers
If 2g of zinc granules was reacted with excess dilute HCl to evolve hydrogen gas which came to completion after 5 minutes. Calcu
nordsb [41]

Answer:

25 g/hr

Explanation:

Remember that the rate of reaction refers to the rate at which reactants are used up or or the rate at which products appear.

Hence;

Rate of reaction = mass of reactant used up/time taken

Mass of reactant used up= 2g

Time taken = 5 minutes or 0.08 hours

Rate of reaction = 2g/0.08 hours = 25 g/hr

4 0
2 years ago
What is the electric force on a proton 2.5 fmfm from the surface of the nucleus? Hint: Treat the spherical nucleus as a point ch
sammy [17]

Explanation:

It is known that charge on xenon nucleus is q_{1} equal to +54e. And, charge on the proton is q_{2} equal to +e. So, radius of the nucleus is as follows.

            r = \frac{6.0}{2}

              = 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

              d = r + 2.5

                 = (3.0 + 2.5) fm

                 = 5.5 fm

                 = 5.5 \times 10^{-15} m     (as 1 fm = 10^{-15})

Therefore, electrostatic repulsive force on proton is calculated as follows.

              F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

Putting the given values into the above formula as follows.

           F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

              = (9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}

              = (9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}

              = 411.2 N

or,           = 4.1 \times 10^{2} N

Thus, we ca conclude that 4.1 \times 10^{2} N is the electric force on a proton 2.5 fm from the surface of the nucleus.

8 0
3 years ago
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