<span>Let's </span>assume that water vapor has ideal gas
behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>
</span><span>Where, P is the pressure of the gas (Pa), V
is the volume of the gas (m³), n is the number
of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 1 atm = 101325 Pa (standard pressure)
V = 13.97 L = 13.97 x 10</span>⁻³ m³<span>
n = ?
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 0 °C = 273 K (standard temperature)
<span>
By substitution,
</span>101325 Pa x 13.97x 10</span>⁻³
m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K<span>
n = 0.624 mol
<span>
Hence, the moles of water vapor at STP is 0.624 mol.
According to the </span></span>Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.
<span>
Hence, number of atoms in water vapor = 0.624 mol x </span>6.022 × 10²³ mol⁻¹
<span> = 3.758 x 10</span>²³<span>
</span>
Answer:
h2+O ---> H2O
reactants: H2 & O
products: H2O
Explanation:
The simple reaction that produces a water molecule from H2 and O would be the one written above, even though there are 2 hydrogen molecules, they will form an H2 molecule rather than 2 individual H molecules (almost never seen) the reactants would be your hydrogen and oxygen molecules individually before they bond to form a molecule of water (H2O) which is the product
<span>C6H12 = 6x12 + 6x1 = 78.
The equation indicates that 2x78 = 156g benzene will produce 6542kJ.
Using proportions you can then calculate that
x/6542kJ = 7.9g / 156g
x = 331.3kJ = 331300J.
heat = mass x ΔT x 4.18J/g°
ΔT = 331300J / (5691g x 4.18J/g°) = 13.9°
final temp = 21 + 14° = 35°C</span>
Answer:
9.6 mol AgCl2
Explanation:
You have to use Avogadro's number: 6.023 x 10^23
5.78 x 10^24 molecules (1 mol AgCl2/ 6.023 x 10^23 molecules) =9.6 mol AgCl2
Answer:
1 kg
Explanation:
prefix nano means 10^-9
kilo 10^3
mili 10^-3
centi 10^-2
the largest amount is 1 kg
