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3 years ago

One mole of ANY element contains the:

Chemistry

2 answers:

Oksi-84 [34.3K]3 years ago

6 0

Answer:all thing will be moving around

VMariaS [17]3 years ago

3 0

Answer:

1)The molar mass of an atom is simply the mass of one mole of identical atoms. However, most of the chemical elements are found on earth not as one isotope but as a mixture of isotopes, so the atoms of one element do not all have the same mass.

2)Equally important is the fact that one mole of a substance has a mass in grams numerically equal to the formula weight of that substance. Thus, one mole of an element has a mass in grams equal to the atomic weight of that element and contains 6.02 X 1023 atoms of the element.

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The volume of a fixed amount of gas is double and the absolute temperature os doubled according to the ideas gas law how has the

olya-2409 [2.1K]

Given that P=nRT/V, for an ideal gas, the pressure would be unchanged.

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3 years ago

What happens to the rate constant as activation energy increases?

Rina8888 [55]

Answer:

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Explanation:

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3 years ago

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15. Ammonium nitrate, NH4NO3, and aluminum powder react explosively producing nitrogen gas, water vapor and aluminum oxide. Writ

Lunna [17]

Answer:

$2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)$

$Entalpy=-2861.9~KJ$

Explanation:

In this case, we have to start with the reagents:

$Al~+~NH_4NO_3$

The compounds given by the problem are:

-) Nitrogen gas = $N_2$

-) Water vapor = $H_2O$

-) Aluminum oxide = $Al_2O_3$

Now, we can put the products in the reaction:

$Al_(_S_) ~+~NH_4NO_3_(_aq_) ->N_2_(_g_) ~+~H_2O_(_g_) ~+~Al_2O_3_(_S_)$

When we balance the reaction we will obtain:

$2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)$

Now, for the enthalpy change, we have to find the standard enthalpy values:

$Al_(_S_)=0~KJ/mol$

$NH_4NO_3_(_a_q_)=-132.0~KJ/mol$

$N_2_(_g_)=0~KJ/mol$

$H_2O_(_g_)=~-~241.8~KJ/mol$

$Al_2O_3_(_S_)=~-~1675.7~KJ/mol$

With this in mind, if we multiply the number of moles (in the balanced reaction) by the standard enthalpy value, we can calculate the energy of the reagents:

$(0*2)~+~(-132*3)=~-396~KJ$

And the products:

$(0*3)~+~(-241.8*6)~+~(-1675.7*1)=-3125.9~KJ$

Finally, for the total enthalpy we have to subtract products by reagents :

$(-3125.9~KJ)-(-396~KJ)=-2729.9~KJ$

I hope it helps!

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3 years ago

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Hydrogen gas

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3 years ago

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