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2 years ago

Match the followings.

Chemistry

1 answer:

insens350 [35]2 years ago

5 0

Answer:

Explanation:

Here, we are to copy and complete the paragraph with the correct options as given below in the question.

Before that, let us have an understanding of what a conversion factor is all about.

A conversion factor refers to a formula that is being used to change over an estimation in a given arrangement of units to its comparable estimation in another arrangement of units. Subsequently, the formula comprises both a number and its unit.

So;

$\text{You can check that you have written the correct conversion factors for an }$

$\text{equality by looking at the numbers and units in each; the information on }$

$\text{one side of the equity appears in the of a conversion factor while the }$

$\text{information on the other side of the equity appears in the denominator.}$

$\text{To generate a second conversion factor, we "turn over" the first }$

$\text{conversion factor so the numbers and units in the numerator and }$

$\text{denominator switch places.}$

You might be interested in

Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation: 3NO2(g)+H2O(l)→2HNO3(l)+

Vera_Pavlovna [14]

Answer:

5 moles of NO₂ will remain after the reaction is complete

Explanation:

We state the reaction:

3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g)

3 moles of nitric oxide can react with 1 mol of water. Ratio is 3:1, so we make this rule of three:

If 3 moles of nitric oxide need 1 mol of water to react

Then, 26 moles of NO₂ may need (26 .1) / 3 = 8.67 moles of H₂O

We have 7 moles of water but we need 8.67 moles, so water is the limiting reactant because we do not have enough. In conclusion, the oxide is the reagent in excess. We can verify:

1 mol of water needs 3 moles of oxide to react

Therefore, 7 moles of water will need (7 .3)/1 = 21 moles of oxide

We have 26 moles of NO₂ and we need 21, so we still have oxide after the reaction is complete. We will have (26-21) = 5 moles of oxide that remains

5 0

2 years ago

Rank the solutions in order of decreasing [H3O ]. Rank solutions from largest to smallest hydronium ion concentration. To rank i

Usimov [2.4K]

Answer:

0.10M HCN < 0.10 M HClO < 0.10 M HNO₂ < 0.10 M HNO₃

Explanation:

We are comparing acids with the same concentration. So what we have to do first is to determine if we have any strong acid and for the rest ( weak acids ) compare them by their Ka´s ( look for them in reference tables ) since we know the larger the Ka, the more Hydronium concentration will be in these solutions at the same concentration.

HNO₃ is a strong acid and will have the largest hydronium concentration.

HCN Ka = 6.2 x 10⁻¹⁰

HNO₂ Ka = 4.0 x 10⁻⁴

HClO Ka = 3.0 x 10⁻⁸

The ranking from smallest to largest hydronium concentration will then be:

0.10M HCN < 0.10 M HClO < 0.10 M HNO₂ < 0.10 M HNO₃

5 0

3 years ago

PLEASE HELP! what are the steps that occur during the combustion of hydrogen?

Softa [21]

Answer:

Explanation:

Combustion releases energy in a single step in the form of light and heat. Whereas in respiration, energy is released in steps and is stored in the form of ATP.

5 0

3 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

Calculate the approximate number of atoms in a bacterium, assuming the average mass of an atom is ten times the mass of a hydrog

mezya [45]

Answer: -

The approximate number of atoms in a bacterium is 10¹¹

Explanation: -

We are given the mass of a bacterium is 10⁻¹⁵ kg.

We are told that the mass of a hydrogen atom is 10⁻²⁷ kg.

Finally we learn that the average mass of an atom of the bacterium is ten times the mass of a hydrogen atom.

Mass of an atom of bacterium = 10 x mass of hydrogen atom

= 10 x 10⁻²⁷ kg.

= 10⁻²⁶ kg.

Thus the number of atoms in a bacterium = $\frac{Total mass}{mass of 1 bacterium atom}$

= $\frac{10-15 kg}{10-26 kg}$

= 10¹¹

8 0

3 years ago