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2 years ago

How much faster will ammonia vapor travel across a room then carbon dioxide

Chemistry

1 answer:

FromTheMoon [43]2 years ago

6 0

Answer:

D) 1.61 times faster

Explanation:

$vrms_{NH3}$ = √(3)RTM

R constant= 0.08206

T=constant, so in this problem we dont need a value for it

M=17.031 g/mol

√(3)(0.08206)(17.031)= 2.047

$vrms_{CO2}$ = √(3)RTM

R constant= 0.08206

T=constant, so in this problem we dont need a value for it

M= 44.01 g/mol

√(3)(0.08206)(44.01)= 3.29

Since we are trying to measure how much faster NH3 will be, we have to realize that mass and speed have an inverse relationship.

So instead of doing (2.047)/(3.29) = 6.22

we have to flip the values to get (3.29)/(2.047)= 1.61

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3 years ago

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How might a collision between Neptune and Halley's Comet affect Neptune's orbit?

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Answer:

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2 years ago

Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at

valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100 - x

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,

$121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042$

$120.9038x+122.9042\left(100-x\right)=12176.01$

Solving for x, we get that:

x = 57.2 %

Thus, percentage abundance of 121 Sb is = 57.2 %

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3 years ago

You are working in your laboratory and decide to do some cleaning. You find a test tube with some brown substance congealed at t

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Answer:

The mystery substance is C. Bromine (Br)

Explanation:

Argon (Ar) is a noble gas. Whose freezing point is -189 °C (very low), thus it cannot be the frozen substance. Also, it is not reactive, thus is would have not reacted with iron. Hence, argon is not the mystery substance.

Scandium (Sc) is a metal from group 3 of the periodic table, thus is will not react with iron. Thus, scandium is not the mystery substance.

Both bromine and iodine are halogens (group 17 of the periodic table).

The freezing point of bromine is −7.2 °C, and the freezing point of iodine is 113.7 °C. Thus, both could be solids (frozen) in the lab.

The reactivity of the halogens decrease from top to bottom inside the group. Bromine is above iodine. Then bromine is more reactive than iodine.

Bromine is reactive enough to react with iron. Iodine is not reactive enough to react with iron.

You can find in the internet that bromine vapour over hot iron reacts producing iron(III) bromide. Also, that bromine vapors are red-brown.

Therefore, the mystery substance is bromine (Br).

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3 years ago

An ordered list of chemical substances is shown. Chemical Substances 1 CuO 2 O2 3 CO2 4 NO2 5 Fe 6 H2O Which substances in the l

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Answer:

reactants: 2 O2

products: 3 CO2, 4 NO2, 6 H2O

Explanation:

In a combustion, a combustible material, which generally is composed of C, H, O, N, and S, is combusted, that is, react with oxygen after a spark was produced; obtaining fire, heat and subproducts, including ashes and gases.

Oxygen is always one of the reactants of a combustion.

If Nitrogen was present in the combustible, NO2 (or other nitrogen oxides) will be produced.

If Carbon was present in the combustible, CO2 will be produced (also CO can be produced).

If Hydrogen was present in the combustible, H2O will be produced.

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3 years ago