Answer:
The answer is B - At dynamic equilibrium, the reactions continue but the amounts of reactants and products do not change.
Explanation:
got it right on Edge
Pressures simply add. If the partial pressure of gas is three and the partial pressure of water is five, the total pressure is eight. Find the partial pressure for water at the temperature of your experiment, subtract it from your <span>pressure reading.</span>

<span>2 NaOH(aq)+ 2 Al(s)+ 2 H</span>₂<span>O → 2 NaAlO</span>₂<span>(aq)+ 3 H</span>₂<span>(g)
</span> 2mol : 2mol : 3mol
2,14mol : 1,89mol : 2,835mol
remains completely consumed
2,14-1,89=0,25mol
A) Al
B)

C)
The "A)The water is part of the system" statement is true in the coffee-cup calorimeter used in the laboratory. The calorimeter is the device used for measuring the amount of the heat flow of a chemical reaction. The coffee-cup calorimeter is one of the most common types of calorimeter which has the water as a part of its system<span>.</span>
The answer is: [B]: 1.89 * 10²³ .
____________________________________________________
PV = nRT ; is the equation for "STP" conditions; that is, the "ideal gas equation" .
_______________________________________________
i.e. when Pressure, "P" = 1.00 atm;
Temperature, "T" = 273 K;
_____________________________________
R = the ideal gas constant = ((0.08206 L-atm/K-mol)
n = number of moles;
___________________________________
So, we plug in our known values:
______________________________________________
<span>(1.00atm) (7.02L) = ("n" mol) (0.08206 L-atm/K-mol) (273K); </span>
<span>_____________________________________________
</span>→ 7.02 L·atm = (? mol) (22.4 L·<span>atm/mol) .
</span>
(Note that t<span>he Molar Volume of a gas at STP is a constant using Avogadro's value of </span>22.4 L / mol. 1 mol of any ideal gas at STP occupies 22.4 L. An ideal gas takes the shape of its container)..
<span>_______________________________________________________
</span> → Divide EACH side of the equation by "(22.4 L·atm/mol)" ;
________________________________________________
→ 7.02 L·atm / = ("n" mol) (22.4 L·atm/mol) ;
_________________________________________________
→ 7.02 L·atm / (22.4 L·atm/mol) =
[("n" mol) (22.4 L·atm/mol)]/(22.4 L·atm/mol);
______________________________________________________
<span>to get:
_________________________________________________
</span>→ n = <span>0.313 mol ;
</span>_____________________________________________________
Note: 1 mole = 1 mol = 6.022 * 10²³ molecules.
____________________________________________
→ So, 0.313 mol Cl₂ (g) molecules * [(6.022 * 10²³ molecules) / (1 mol)] =
___________________________________________________________
→ [(0.313) * (6.022 *10²³) ] molecules of Cl₂ (g) ;
___________________________________________________________
→ = 1.88 * 10²³ molecules of Cl₂ g ;
___________________________________________________________
→ which most closely corresponds with answer choice:
___________________________________________________________
[B]: 1.89 * 10²³ .
___________________________________________________________