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3 years ago

If you have 1.4 grams of silver (Ag), how many moles of silver do you have?

Chemistry

1 answer:

creativ13 [48]3 years ago

3 0

Answer: There are 0.0129 moles of silver present in 1.4 grams of silver (Ag).

Explanation:

Given: Mass of silver = 1.4 g

Number of moles is the mass of substance divided by its molar mass.

As molar mass of silver is 107.86 g/mol. Therefore, moles of silver are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{1.4 g}{107.86 g/mol}\\= 0.0129 mol$

Thus, we can conclude that there are 0.0129 moles of silver present in 1.4 grams of silver (Ag).

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What happens inside a can of spoiled food as gas is produced?

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4 years ago

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The following shows the precipitation reaction of barium chloride (BaCl₂) and sodium hydroxide (NaOH):

Nadya [2.5K]

Answer:

The answer to your question is:

a) BaCl2

b) 0.8208 g

c) yield = 85.3 %

Explanation:

BaCl₂(aq) + NaOH(aq) ----> Ba(OH)₂(s) + 2NaCl(aq)

Data

a) 1 g of BaCl₂

1 g of NaOH

MW BaCl2 = 137 + (35.5x2) = 208 g

MW NaOH = 23 + 16 + 1 = 40 g

208 g of BaCl2 ------------- 1 mol

1 g of BaCl2 ------------- x

x = ( 1 x 1) / 208 = 0.0048 mol of BaCl2

40 g of NaOH ------------ 1 mol

1 g of NaOH ------------ x

x = (1 x 1) / 40

x = 0.025 mol of NaOH

The ratio BaCl2 to NaOH is 1:1 (in the equation)

But experimentally we have 0.0048 : 0.025, so the limiting reactant is BaCl2, because is in lower concentration.

1 mol of BaCl2 -------------- 1 mol of Ba(OH)2

0.0048 mol --------------- x

x = (0.0048 x 1) / 1

x = 0.0048 mol of Ba(OH)2

MW Ba(OH)2 = 137 + 32 + 2 = 171 g

171 g of Ba(OH)2 -------------------- 1 mol

x -------------------- 0.0048 mol

x = (0.0048 x 171) / 1

x = 0.8208 g

c)Data

Ba(OH)2 = 0.700 g

% yield = 0.700 / 0.8208 x 100

% yield = 85.3

Sorry, i don't understand this question

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3 years ago

For the first ionization energy for an N2 molecule, what molecular orbital is the electron removed from

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5. Find the number of moles equal to 1.90 x 1025 molecules of Mg(OH)2

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