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iren2701 [21]
3 years ago
12

How many atoms are in 1 mole of gold

Chemistry
1 answer:
kirza4 [7]3 years ago
8 0

Answer:

1 mole of gold will have exactly 6.022.1023 atoms of gold.

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When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio
8090 [49]

Answer:

Volume of stock solution needed = 6.0299 mL

Explanation:

<u> </u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.

This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.

<u>Data:</u>

M1 = 6.01 M stock solution concentration

M2 = 0.3624 M diluted solution concentration

V2 =100 mL diluted solution volume

V1 = ? stock solution volume

M1 * V1 = M2 * V2

V1=\frac{M2*V2}{M1} =\frac{0.3624M*100mL}{6.01M} =6.0299 mL

4 0
3 years ago
What equipment might you use to find the SHC of a material? Need to know the answer ASAP PLZ
algol [13]
Theses can include the power supply circuit a joule meter to measure the energy transferred which makes the calculations a lot easier.
4 0
3 years ago
Determine the work done by an ideal gas while expanding by a volume of 0.25 L against an external pressure of 1.50 atm (assume a
Sliva [168]

Answer : The value of work done by an ideal gas is, 37.9 J

Explanation :

Formula used :

Expansion work = External pressure of gas × Volume  of gas

Expansion work = 1.50 atm × 0.25 L

Expansion work = 0.375 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 0.375 × 101.3 = 37.9 J

Therefore, the value of work done by an ideal gas is, 37.9 J

6 0
3 years ago
What is the Law of Conservation of energy?
ladessa [460]

Answer:

<u>~</u><u>Law of Conservation of </u><u>energy~</u>

The law of conservation of energy states that energy can neither be created nor destroyed, only energy can be converted from one form to another.

3 0
2 years ago
As a part of a clinical study, a pharmacist is asked to prepare a modification of a standard 22g package of a 2% mupirocin ointm
Vesna [10]

Answer:

226.8 mg of mupirocin powder are required

Explanation:

Given that;

weight of standard pack = 22 g

mupirocin by weight = 2%

so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g

so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment

mg of mupirocin powder are required = ?, lets rep this with x

Total weight of ointment = 22 + x g

Amount of mupirocin  = 0.44 + x g

percentage of mupirocin  in ointment is 3?

so

3/100 = 0.44 + x g / 22 + x g

3( 22 + x g ) = 100( 0.44 + x g )

66 + 3x g = 44 + 100x g

66 - 44 = 100x g - 3x g

97 x g = 22

x g = 22 / 97

x g = 0.2268 g

we know that; 1 gram = 1000 Milligram

so 0.2268 g = x mg

x mg = 0.2268 × 1000

x mg  = 226.8 mg

Therefore, 226.8 mg of mupirocin powder are required

8 0
3 years ago
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