Answer:
for the balanced equation
<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:
Option d. 2615.0g
Explanation:
Let M1, M2, and M3 represent the masses of the three different samples
M1 = 0.1568934 kg = 156.8934g
M2 = 1.215mg = 1.215x10^-3 = 0.001215g
M3 = 2458.1g
Total Mass = M1 + M2 + M3
Total Mass = 156.8934 + 0.001215 + 2458.1
Total Mass = 2614.994651g
Total Mass = 2615.0g
