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tatyana61 [14]
2 years ago
8

You have been asked to organize the following aqueous solutions: HCl, NaOH, H2SO4,Ca(OH)2, LiOH, and HCIO3.How would you group t

hem?On what are you basing your decision?
Chemistry
2 answers:
frutty [35]2 years ago
7 0

Answer & Explanation:

You could group them as acids and bases:

<u>Acids</u>

HCl

H₂SO₄

HClO₃

<u>Bases</u>

NaOH

Ca(OH)₂

LiOH

Oksanka [162]2 years ago
6 0

The easiest way to classify them is by separating into acids and bases.

\boxed{\begin{array}{c|c}\boxed{\bf{Acids}} &\boxed{\sf Bases}\\ &\\ \sf HCl &\sf NaOH\\ &\\ \sf H_2SO_4 &\sf Ca(OH)_2 \\&\\ \sf HClO_3 &\sf LiOH\end{array}}

Note it down

  • Acids give H+ ions and Bases give OH- ion while mixed up with water
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What is 9872002 in scientific notation?
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Answer:

= 9.872002 × 10^6

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Move the decimal point in your number until there is only one non-zero digit to the left of the decimal point. The resulting decimal number is a.

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If you moved the decimal to the left b is positive.

If you moved the decimal to the right b is negative.

If you did not need to move the decimal b = 0.

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Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of
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In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g)   (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

n = \frac{m}{M}

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m: is the mass

M: is the molar mass

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n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles

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n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles

We can see in reaction (1) that <u>3 moles of hydrogen</u> react with <u>1 mol of nitrogen</u>, so the number of hydrogen moles needed to react nitrogen is:

n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles

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We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that <u>3 moles of hydrogen</u> produces <u>2 moles of ammonia</u>, so:

n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles

Hence, hydrogen would produce <u>2.31 moles of ammonia</u>.

Therefore, hydrogen is the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

   

I hope it helps you!                        

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