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1 year ago

What is the relative rate of diffusion of hydrogen and nitrogen

Chemistry

1 answer:

vredina [299]1 year ago

3 0

Answer:

2.645

Explanation:

Rate of diffusion formula:

Sqrt(mass2/mass1)

>>sqrt(14/2)

(Note:Hydrogen must exist in dwiatomic, [H2])

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Are mass vs. density the same

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Mass is the amount of matter in an object. Density is the degree of compactness of a substance

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2 years ago

In an experiment Teresa's measures 15.5 mL of water she must have used a

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transfer pipet that had markings every 0.1 mL.

Explanation:

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3 years ago

Formula semidesarrollada de los hidrocarburos aromáticos de 1 a 10 átomos de carbono

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Answer:

El principal componente del gas natural es también el hidrocarburo más simple: el metano. Este compuesto está formado por un átomo de carbono y cuatro átomos de hidrógeno y se representa de dos formas:

El hidrocarburo que le sigue en simplicidad es aquel que está constituido por dos átomos de carbono. Su fórmula condensada es C2H6 y se le conoce como etano.

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3 0

2 years ago

Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%, respectively)

Reika [66]

Answer:

Average atomic mass = 79.9034 amu

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:

For first isotope:

% = 50.69 %

Mass = 78.9183 amu

For second isotope:

% = 49.31 %

Mass = 80.9163 amu

Thus,

$Average\ atomic\ mass=\frac{50.69}{100}\times {78.9183}+\frac{49.31}{100}\times {80.9163}\ amu$

$Average\ atomic\ mass=40.0036+39.8998\ amu$

Average atomic mass = 79.9034 amu

4 0

3 years ago

An unknown amount of Al203 decomposed producing 215 g of solid aluminum. 2Al2O3=4Al+3O2 How many grams of oxygen gas should be p

natulia [17]

Answer:

191.11 grams of oxygen gas should be produced.

Explanation:

The balanced reaction is:

2 Al₂O₃ → 4 Al + 3 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al₂O₃: 2 moles
Al: 4 moles
O₂: 3 moles

Being the molar mass of each compound:

Al₂O₃: 102 g/mole
Al: 27 g/mole
O₂: 32 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al₂O₃: 2 moles* 102 g/mole= 204 grams
Al: 4 moles* 27 g/mole= 108 grams
O₂: 3 moles* 32 g/mole= 96 grams

Then you can apply the following rule of three: if by stoichiometry 108 grams of aluminum are produced along with 96 grams of oxygen, 215 grams of aluminum are produced along with how much mass of oxygen?

$mass of oxygen=\frac{215 grams of aluminum*96 grams of oxygen}{108grams of aluminum}$

mass of oxygen= 191.11 grams

191.11 grams of oxygen gas should be produced.

6 0

2 years ago

What is the relative rate of diffusion of hydrogen and nitrogen ​

What is the relative rate of diffusion of hydrogen and nitrogen