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2 years ago

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Chemistry

2 answers:

stira [4]2 years ago

7 0

Answer:

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Anettt [7]2 years ago

5 0

Answer:

Thanks- :D

Explanation:

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What is the meaning of the word organism

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B) any complex thing with properties normally associated with living things

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3 years ago

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is carried out adiabatically in a constant-volume batch reactor. The rate law is Plot and analyze the conversion, temperature, a

Sphinxa [80]

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See explaination

Explanation:

The mole balance for a constant-volume batch reactor is given such as, For a first-order isothermal reaction, the time to reach a given conversion is the same for constant-pressure and constant-volume reactors. Also, the time is the same for a reaction of any order if there is no change in the number of moles.

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3 years ago

If 25.0 g NO are produced, how many grams of nitrogen gas are used?

ivanzaharov [21]

Based on the assumption that the reaction involves N and O to produce NO, if 25.0 g of NO are produced, the amount of N gas used would be 11.66 grams

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

N + O ---------> NO

Mole ratio of N to NO is 1:1

Mole of 25.0 g of NO = 25/30.01 = 0.833 moles

Equivalent mole of N = 0.833 moles

Mass of 0.833 moles N = 0.833 x 14 = 11.66 grams

More on stoichiometric calculations can be found here: brainly.com/question/8062886

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2 years ago

When the name of an anion that is part of an acid ends in -ite, the acid name includes the suffix _____.

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(B) suffixous

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3 years ago

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Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.

Elenna [48]

The given concentration of boric acid = 0.0500 M

Required volume of the solution = 2 L

Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.

Calculating the moles of 0.0500 M boric acid present in 2 L solution:

$2 L * \frac{0.0500 mol B(OH)_{3} }{1 L} = 0.100 mol B(OH)_{3}$

Converting moles of boric acid to mass:

$0.100 mol B(OH)_{3} * \frac{61.83 g}{mol B(OH)_{3}} = 6.183 g$

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.

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3 years ago