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2 years ago

1. What is the empirical formula for a compound containing 63.8% N and 36.2% O?

Chemistry

1 answer:

Fittoniya [83]2 years ago

3 0

$M\ N_2=28g/mol\\
63,8\%--28g\\
100\%--x\\
x=43,89g$

43,89g≈44g
44g == 2x 14g (N) + 1x 16x(O)
M N₂O = 44g

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When a solution of 0.1 M Mg(NO3)2 was mixed with a limited amount of aqueous ammonia, a light white, wispy solid was observed, i

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Answer: The net ionic equation is written below.

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Net ionic equation of any reaction does not include any spectator ions.

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3 years ago

What does potassium hydroxide and sulphuric acid make?

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Base + Acid = Water + Salt

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Balanced rxn is:

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2 years ago

Read 2 more answers

A solution of H2SO4(aq) with a molal concentration of 4.80 m has a density of 1.249 g/mL. What is the molar concentration of thi

Naddika [18.5K]

Imagine we have mass of solvent 1kg (1000g)
According to that: $molality =n(solute) / m(solvent) =\ \textgreater \ 4.8m = 4.8mole / 1kg$
= 4.8 mole * 98 g/mole = 470g
$m(H2SO4) = n(H2SO4)*Mr(H2SO4) =\ \textgreater \$ $=\ \textgreater \ Molarity = 4.8 mole / 1.2 L = 4 M$ m(H2SO4) which is =470g

m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =>
$Molarity = n(solute) / V(solution) =\ \textgreater \$

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