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Radda [10]
3 years ago
11

How many atoms of phosphorus are represented in 2Ca3(PO4)2? 2 4 8 16

Chemistry
2 answers:
Angelina_Jolie [31]3 years ago
6 0
We will have *1 2of2 atoms =p
irinina [24]3 years ago
5 0
Hello !



<span>The number to the right of each element represents the number of the atoms of that element.
</span>

<span>PO4 is taken twice so you will multiply 1 and 4 by 2.
</span><span>
</span><span>Now we have 1*2=2 atoms of P
</span>


<span>To the left,there is 2,so you will multiply all the elements by 2.It means that you'll multiply 2 (atoms of P) by 2 :
</span><span>
</span><span>
</span><span>   2*2=4 atoms of phosphorus
</span>

ANSWER : 4
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3 years ago
How many grams of Br are in 335g of CaBr2
ASHA 777 [7]
The answer is 267.93 g

Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol

The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%

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x g in 79.98%
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7 0
3 years ago
How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them? (answ
Jet001 [13]

Answer:

7.12 mm

Explanation:

From coulomb's law,

F = kqq'/r².................... Equation 1

Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.

Make r the subject of the equation,

r = √(kqq'/F).......................... Equation 2

Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute into equation 2

r = √[ (75×10⁻⁹ )²9.0×10⁹/1]

r = 75×10⁻⁹.√(9.0×10⁹)

r = (75×10⁻⁹)(9.49×10⁴)

r = 711.75×10⁻⁵

r = 7.12×10⁻³ m

r = 7.12 mm

Hence the distance between the point charge = 7.12 mm

3 0
3 years ago
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