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kobusy [5.1K]
3 years ago
9

I NEED HELP PLEASE, THANKS! :)

Chemistry
1 answer:
marin [14]3 years ago
6 0

Answer:

\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}

Explanation:

1. Calculate the moles of copper(II) hydroxide

\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}

2. Calculate the molecules of copper(II) hydroxide

\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}

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How many grams of water react to form 7.21 moles of Ca(OH)2
oksano4ka [1.4K]

You have the stoichiometric equation. This tells you unequivocally that an  

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If you don't from where I am getting these numbers, you should know, and someone will be willing to elaborate.

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6.21

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×

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How many dm3 of hydrogen are released when 3 g of potassium is reacted with hydroiodic acid?
nlexa [21]

Answer:

V = 0.0859dm³

Explanation:

Hydroiodic acid, HI, reacts with potassium, K, to produce potassium iodide, KI, and hydrogen, as follows:

2HI + 2K → 2KI + H₂(g)

To solve this question we have to find the moles of hydrogen produced knowing that 2 moles of K produce 1 mole of H₂. With the moles of hydrogen we can find the volume of hydrogen assuming there are STP conditions:

<em>Moles K -Molar mass: 39.0983g/mol-</em>

3g * (1mol / 39.0983g) = 0.0767 moles of K

<em>Moles H₂:</em>

0.0767 moles of K * (1mol H₂ / 2mol K) = 0.03836 moles H₂

Using: PV = nRT; V = nRT / P

<em>Where V is volume in dm³,</em>

<em>n are moles of gas: 0.03836 moles,</em>

<em>R is gas constant = 0.082atm*dm³/molK</em>

<em>T is absolute temperature = 273.15K at STP</em>

<em>and P is pressure = 1atm</em>

The volume of the gas is:

V = 0.03836mol*0.082atm*dm³/molK*273.15K / 1atm

<h3>V = 0.0859dm³</h3>

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