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2 years ago

6

How many moles of sodium carbonate in 18.06x10 to the power 22

2 answers:

igor_vitrenko [27]2 years ago

4 0

Answer:

$moles = \frac{ number \: of \: particles}{6.02 \times {10}^{23} }$

=1.806×10^22/6.02×10^23

=0.03 moles

hope this helps :)

aksik [14]2 years ago

4 0

Answer 1.354
Explanation none

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3 years ago

How many formula units are in<br> 35.0 g KNO3?<br> (KNO3, 101.11 g/mol)<br><br> [?]×10¹²] fun KNO3

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1 year ago

The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo

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<u>Answer:</u> The solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

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Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

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The chemical equation for the ionization of silver (I) phosphate follows:

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3s s

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=(3s)^3\times s$

We are given:

$s=2.44\times 10^{-3}M$

Putting values in above expression, we get:

$K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}$

Hence, the solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

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2 years ago