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3 years ago

In your OWN words explain how season are formed

Chemistry

2 answers:

MaRussiya [10]3 years ago

6 0

I am assuming that you mean seasons.
Seasons are formed through the Earth orbiting the sun and the Earth rotating.

astra-53 [7]3 years ago

3 0

Answer: Weather currents change wind tempature and there for change the weather of the seasons.

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Margarita [4]

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3 years ago

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noname [10]

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Lines of latitude run from east to west.

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3 years ago

Decide which element probably has a boiling point most and least similar to the boiling point of cesium.

natka813 [3]

Answer:

Take a look at the attachment below

Explanation:

Take a look at the periodic table. As you can see, Rubidium is the closest element to Cesium, and happens to have the closest boiling point to Cesium, with only a difference of about 30 degrees.

Respectively, you would think that fluorine should have the least similarity to Cesium with respect to it's boiling point, considering it is the farthest away from the element out of the 4 given. This is not an actual rule, there are no fixed trends of boiling points in the periodic table, there are some but overall the trends vary. However in this case fluorine does have the least similarity to Cesium with respect to it's boiling point, a difference of about 1,546.6 degrees.

<em>Hope that helps!</em>

5 0

3 years ago

If 0.55 g of a gas dissolved in 1.0 l of water at 20.0 kpa of pressure, how much will dissolve at 110.0 kPa of pressure?

GaryK [48]

Answer:sup

Explanation:

4 0

4 years ago

Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reac

stich3 [128]

Answer:

Value of $Q_{p}$ for the given redox reaction is $1.0\times 10^{-8}$

Explanation:

Redox reaction with states of species:

$14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)$

Reaction quotient for this redox reaction:

$Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}$

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of $H_{2}O$ is taken as 1 due to the fact that $H_{2}O$ is a pure liquid.

$pH=-log[H^{+}]$

So, $[H^{+}]=10^{-pH}$

Plug in all the given values in the equation of $Q_{p}$ :

$Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}$

7 0

3 years ago