The answer to this question would be C. I hope this helps!
I believe the atomic level increases hope this helps.
<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:
Explanation:
We are given two reactions which are the two steps of a mechanism:
......(1)
.......(2)
To determine the net chemical equation, we will multiply equation 2
.......(3)
Adding (1) and (3)
Thus, the net balanced chemical equation is:
Answer:
The chemistry will need 2*10⁶ moles of antimony trifluoride.
Explanation:
The balanced reaction is:
3 CCl₄ (g) + 2 SbF₃ (s) → 3 CCl₂F₂(g) + 2 SbCl₃ (s)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- CCl₄: 3 moles
- SbF₃: 2 moles
- CCl₂F₂: 3 moles
- SbCl₃: 2 moles
You can apply the following rule of three: if by reaction stoichiometry 3 moles of freon are produced by 2 moles of antimony trifluoride, 3*10⁶ moles of Freon are produced from how many moles of antimony trifluoride?

moles of antimony trifluoride= 2*10⁶
<u><em>The chemistry will need 2*10⁶ moles of antimony trifluoride.</em></u>