Chemistry, 25.07.2019 11:30 thefandomarmy24

How many milliliters of 1.0m agno3 are needes to provide 169.9 g of agno3?

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The number of Ml of 1.0 AgNo3 needed to provide 169.9 0f AgNO3 is calculated as follows

find the moles of AgNo3

= mass/molar mass = 169.9/169.87 = 1 mole

volume in Ml = moles/ molarity x1000

= 1/1 x1000= 1000 ml

Answer from: avilaaa

The molar mass of silver nitrate is 169.87 g/mol
Therefore the number of moles in 169.9 g will be;
= 169.9/169.87
= 1 mole
Therefore; since the molarity is 1.0 M
then the volume= 1/1 =1 liter which is equivalent to 1000ml
Thus, the answer is 1000 ml

Answer from: Quest

може ли да зареждате повече информация, моля

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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