3 years ago

Imagine that hotter magma is lying beneath an area of cooler magma deep in the mantle. What do you predict will happen?

Chemistry

1 answer:

allsm [11]3 years ago

3 0

Answer:

it will explod because off all of the prussere but on to it.

Explanation:

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Assume that 8.5 L of iodine gas (I2) are produced at STP according to the following balanced equation:2KI(aq) + Cl2(

Aliun [14]

Answer: 0.3794 moles of Iodine gas are produced.

Explanation:

$2KI(aq)+Cl_2(
g)\rightarrow 2KCl(aq)+I_2$

Volume of iodine gas produced at STP =8.5 L

At STP, the 1 mol of gas occupies volume = 22.4 L

So, 8.5 L of volume will be occupied by: $\frac{1}{22.4 L}\times 8.5=0.3794 moles$

0.3794 moles of Iodine gas are produced.

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Why are red dwarfs not able to fuse helium to heavier elements?

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Fusing helium into heavier elements requires extreme pressure and temperature. However red dwarfs' gravity is simply not enough to bring helium atom close enough to each other for fusion to occur. Therefore, red dwarfs not able to fuse helium to heavier elements.☺

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When two identical nonmetal atoms are bonded together the result is a?

Irina-Kira [14]

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Sodium thiosulfate, Na 2S 2O 3, is used as a "fixer" in black and white photography. Identify the reducing agent in the reaction

Tatiana [17]

Answer: Reducing agent in the given reaction is $S_{2}O^{2-}_{3}$ .

Explanation:

A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.

In the given reaction, oxidation state of sulfur in $S_{2}O^{2-}_{3}$ is +2 and $I_{2}(aq)$ has 0 oxidation state.

In $S_{4}O^{2-}_{6}(aq)$ oxidation state of S is 2.5 and in $2I^{-}(aq)$ oxidation state of I is -1.

Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.

Thus, we can conclude that reducing agent in the given reaction is $S_{2}O^{2-}_{3}$ .

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3 years ago