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viktelen [127]
3 years ago
8

How many silver atoms are in 1 mole of silver?

Chemistry
2 answers:
never [62]3 years ago
7 0

Answer:

6.022 x 10^23

Explanation:

The number of atoms in one mole of a  substance is equal to Avogadro's number, or 6.022 x 10^23.

So, calculating the number of silver atoms in one mole of silver is simple, it is just equal to Avogadro's number, 6.022 x 10^23.

Let me know if this helps!

jekas [21]3 years ago
5 0

Answer:

Explanation:

Silver has a molar mass of 107.87 g⋅mol−1 . What does this mean? It means that if I have such a mass of silver, there are Avogadro's number, NA=6.022×1023 , individual silver atoms

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Answer:

What Structures??

Explanation:

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2 years ago
Which of these phrases accurately describe disproportionation? Check all that apply.
Olin [163]

Answer:

A single compound is simultaneously oxidized and reduced.

Explanation:

In chemistry, disproportionation is a simultaneous oxidation and reduction of a single chemical specie.

What this means is that; in a disproportionation reaction, only one compound is both oxidized and reduced. This implies that two products are formed during disproportionation. One is the oxidized product while the other is the reduced product.

Consider the disproportionation of CuCl shown below;

2CuCl -----> CuCl2 + Cu

Here, CuCl2 is the oxidized product while Cu is the reduced product.

7 0
2 years ago
Read 2 more answers
Why is coal not classified as a mineral?
Fittoniya [83]

A mineral is a naturally occurring chemical compound, usually of crystalline form and abiogenic in origin. A mineral has one specific chemical composition. Coal is not a mineral because it is organic while a mineral is inorganic that have repeating crystalline structure.

4 0
3 years ago
If the actual yield of a reaction is 37.6 g while the theoretical yield is 112.8 g what is the percent yield
Zigmanuir [339]
<h2>Hello!</h2>

The answer is:

The percent yield of the reaction is 32.45%

<h2>Why?</h2>

To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.

We are given that:

ActualYield=37.6g\\TheoreticalYield=112.8g

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.

So, calculating we have:

PercentYield=\frac{ActualYield}{TheoreticalYield}*100\\\\PercentYield=\frac{37.6g}{112.8g}*100=0.3245*100=32.45(percent)

Hence, we have that the percent yield of the reaction is 32.45%.

Have a nice day!

8 0
3 years ago
Read 2 more answers
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
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