89.1% Au
The molar mass of Au2O3 is 2(196.97) + 3(16.00) = 441.94 g/mol
The mass percent of Au is therefore 2(196.97) / 441.94, which is 393.94/441.94 = 0.891 —> 89.1%
Answer:
The number of molecules is 1.4140*10^24 molecules
Explanation:
To know the number of molecules, we need to determine how many moles of water we have, water has molar mass of 18.015g/mol
This means that one mole of water molecules has a mass of 18.015g.
42.3g * 1 mole H2O/18.015g
= 2.3480 moles H2O
We are using avogadros number to find the number of molecules of water
2.3480 H2O * 6.022*10^ 23moles/ 1mole of H2O
That's 2.3480 multiplied by 6.022*10^23 divided by 1 mole of H2O
Number of molecules = 1.4140 *10^24 molecules
THE ANSWER
for solving this, we just use the formula of the equation of state
PV = n RT
where
P: pressure =10atm
V volume
n number of mole = 35.8 moles
R universal gas constant = 0.082
T: The temperature= 70°C= 343.15 Kelvin
therefore, for finding V
we have
V= (n RT) / P = 35.8 x 0.082 x 343.15 / 10 = 100.7 ≈ 101 L
the answer is V = 101L
The answer is: Dividing the number of molecules in the sample by Avogadro's number.
The Avogadro’s number is the number of atoms in 12 grams of the isotope carbon-12 (¹²C).
Na is Avogadro number or Avogadro constant (the number of particles, in this example carbon, that are contained in the amount of substance given by one mole).
The Avogadro number has value 6.022·10²³ 1/mol in the International System of Units; Na = 6.022·10²³ 1/mol.
For example:
N(Ba) = 2.62·10²³; number of atoms of barium.
n(Ba) = N(Ba) ÷ Na.
n(Ba) = 1.3·10²⁴ ÷ 6.022·10²³ 1/mol.
n(Ba) = 2.158 mol; amount of substance of barium.
CO + 2 H2 → CH3OH
<span> find # of mols in each reactants, </span>
<span>152500 g CO x 1 mol CO / 28.01g CO = 5444 mol CO </span>
<span>24500 g H2 x 1 mol H2 / 2.02 g H2 = 12129 mol H2 </span>
<span>mol ratio between CO and H2 is 1:2, which means each mol of production of CH3OH need 1 mol of CO and 2 mol of H. </span>
<span>H2 is enough to produce 6064 mols of CH3OH but there are only 5444mol of CO. </span>
<span>5444 mol CH3OH x molar mass of CH3OH / 1 mol CH3OH </span>
<span>= 174371 g = 174.4 kg</span>
