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NemiM [27]
3 years ago
10

How many moles are in 6.80 x 10^23 atoms of gold, Au?

Chemistry
1 answer:
frosja888 [35]3 years ago
8 0

Answer:

1.13 moles Au

Explanation:

Moles Au = 6.80x10²³atoms / 6.023x10²³atoms/mole = 1.13 moles Au

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Nhóm nào gồm tất cả các kim loại tan trong axit sunfuric đặc nóng nhưng không tan trong axit sunfuric loãng là
MakcuM [25]

Answer:Nhóm gồm tất cả các kim loại tan trong axit sunfuric đặc nóng nhưng không tan trong axit sunfuric lo

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Explanation:

4 0
3 years ago
What is Type of compound of ICI
valentina_108 [34]

Answer:

ICI 204448 hydrochloride | C23H27Cl3N2O4 | CID 129407 - structure, chemical names, physical and chemical properties, classification, patents, literature, etc...

hope this helps!! have an amazing day <3

8 0
3 years ago
Read 2 more answers
What is a boundary between geologic units?
Airida [17]
The answer to this question is Contact
depending on the geological event that put into place, this will have a specific semantic value.
A contact usually represented by different kind of lines on the geological map and usually use to determine <span>where rocks come into contact across fault zones</span>
3 0
3 years ago
Read 2 more answers
Into a 0.25 M solution of Ba3(PO4)2(aq), excess Na2SO4(aq) was added to form BaSO4(s). Ba3(PO4)2(aq) + 3Na2SO4(aq) → 3BaSO4(s) +
kogti [31]

Answer:

Answer is in the explanation.

Explanation:

For the reaction:

Ba₃(PO₄)₂(aq) + 3Na₂SO₄(aq) → 3BaSO₄(s) + 2Na₃PO₄(aq)

As Na₂SO₄(aq) is in excess, limiting reactant is Ba₃(PO₄)₂(aq). As the molarity of the solution is 0,25M and you knew the volume of the solution, you can obtain the moles of Ba₃(PO₄)₂ doing 0,25M×volume.

As 1 mol of Ba₃(PO₄)₂(aq) react with 3 moles of BaSO₄ the moles of BaSO₄ are three times moles of Ba₃(PO₄)₂.

As BaSO₄ molar mass is 233,38g/mol. The mass of BaSO₄ is given by moles of BaSO₄ × 233,38g/mol

I hope it helps!

6 0
3 years ago
what is the theoretical mass of xenon tetrafluoride that should form when 130.g of xenon is reacted with 100.g of F2? what is th
serg [7]

Answer:

The answer to your question is: 70.7 %

Explanation:

Equation

                  Xe   +   2F₂    ⇒    XeF₄

limiting reactant = Xe

Xe is the limiting reactant because the ratio is:

theoretical = 131/ 76 = 1.72 g

experimental ratio = 130/100 = 1.3  the amount of F increased.

                     131.3 g of Xe ------------------ 207 g of XeF₄

                     130 g of Xe  -------------------  x

                         x = (130 x 207) / 131.3

                        x = 205 g of XeF₄

% yield = 145 / 205 x 100

% yield = 70.7

5 0
3 years ago
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