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krek1111 [17]
3 years ago
7

You have 75.0 mL of a 2.50 M solution of Na2CrO4(aq). How many mL of a 1.74 M solution of AgNO3(aq) are needed to completely rea

ct both reactants?
Chemistry
1 answer:
Nonamiya [84]3 years ago
3 0

Answer:

215 mL of silver nitrate are needed.

Explanation:

Let's think the reaction:

Na₂CrO₄(aq) + 2AgNO₃(aq) → Ag₂CrO₄ (s) ↓ + 2NaNO₃(aq)

First of all, we determine the moles of chromate we have available.

Molarity = mol / volume(L)

We convert volume to L → 75mL . 1L / 1000mL = 0.075 L

Molarity . volume(L) = moles → 2.50 M . 0.075L = 0.1875 moles

Ratio is 1:2. For 1 mol of chromate I need the double of moles of nitrate.

So, for 0.1875 moles of chromate I would need (0.1875 . 2) = 0.375 moles of nitrate. Let's determine volume of solution

Molarity = mol/ volume(L) → volume (L) = mol / Molarity

volume (L) = 0.375mol / 1.74M → 0.215L

We convert the value to mL → 0.215L . 1000mL/ 1L = 215 mL

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Oxygen gas, generated by the reaction 2KClO3(s)---2KCl(s)+3O2(g), is collected over water at 27•C in 3.72L vassel at a total pre
Julli [10]

Answer:

moles = 0.093 moles

Explanation:

In this case, we know that this reaction is taking plave in a vessel that has a 730 torr of total pressure.

The total pressure is a value obtained by:

Pt = Pwater + PO2

We need to know the pressure of O2, because then, with stoichiometry, we can calculate the moles of KClO3

The pressure of oxygen is:

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Now, this pressure is in Torr, and we need to convert it to Atm, so:

704 Torr / 760 Torr = 0.9263 atm

Now, let's use the ideal gas equation:

PV = nRT

With this expression, we will calculate the moles of O2, and then, the moles of KClO3:

n = PV/RT

R = 0.082 L atm /K mol

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T = 27 + 273 = 300 K

Replacing the data:

n = 0.9263 * 3.72 / 300 * 0.082

n = 0.14 moles

Finally, by stoichiometry, we know that 2 moles of KClO3 produces 3 moles of O2, so:

moles of KClO3 = 0.14 * 2/3 = 0.093 moles of KClO3

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Answer:

A. 8

Explanation:

8 valence electrons means you have a full shell.

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