Chemistry, 05.05.2020 00:20 emeraldchest99

How many grams of sodium nitrate (NaNO3, sorry subscripts don't work) would you use to prepare 5.00 L of a 2.500 M aqueous solution (dissolved in water)?
A. 1062.5 g
B. 170.0 g
C. 425.0 g
D. 6.8g

Answers: 2

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Answers

Answer from: Quest

$\frac{239}{94}pu+\frac{1}{0}n ====> \frac{y}{z}x+\frac{134}{54}xe+3\frac{1}{0}$

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137 subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54 subtract 54 from both sides.

40 = z

answer

$\frac{103}{40}x=\frac{103}{40}zr$

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

Answer from: Quest

answer: mole of the gas

answer: mole of the gasexplanation:

answer: mole of the gasexplanation: this is because 1 mole of a gas is 22.4dm³, which is use in ideal gas

Answer from: Quest

Ch4 and ph3 are both covalent since they are composed both of non-metals. ch4 is non- polar because of london forces and has a geometry of tetrahedral ph3 , on the other hand, is umbrella shaped polar and has dipole-dipole intermolecular force (imf) . since ph3 is polar, it has a stronger imf

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How many grams of sodium nitrate (NaNO3, sorry subscripts don't work) would you use to prepare 5.00...

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