Answer:
148.1 mL.
Explanation:
We can use the general law of ideal gas: PV = nRT.where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
If n and P are constant, and have different values of V and T:(V₁T₂) = (V₂T₁)
Knowing that:
V₁ = 125 mL, T₁ = 25°C + 273 = 298 K,
V₂ = ??? mL, T₂ = 55°C + 273 = 353 K,
Applying in the above equation(V₁T₂) = (V₂T₁)
∴ V₂ = (V₁T₂)/(T₁) = (125 mL)(353 K)/(298 K) = 148.1 mL.
Imagine that you have a partially inflated
tire. If you add approximately two
pumps more air into it from your tire
pump, what do you expect to change
when you compare the starting and
ending state of the tire?
A. Volume
B. Temperature
C. Pressure
D. Moles of air molecules
Answer:
volume, pressure, moles of air molecules
Explanation:
The volume, Pressure and Moles of air molecules expect to change when you compare the starting and ending state of the tire.
So, option A , C and D is correct option.
What happen to pressure and volume when more air pumped in the tire?Since, according to Boyle's law, volume is inversely proportional to the pressure.As we pumped more air in the tire then gas molecule compressed and packed together.So, pressure inside the tire increases and volume decreases.As more air pumped in tire then number of moles of air molecule increases.Learn about pressure , volume and moles
https://brainly.com/question/12050285
#SPJ2
What is the limiting reactant in a reaction that produces sodium chloride from 8 g of sodium and 8 g of diatomic chlorine?
Answer:
Cl₂ is the limiting reactant.
Explanation:
Firstly, we need to write the balanced reaction:2Na + Cl₂ → 2NaCl,
It is clear that 2 mol of Na react with 1 mol of Cl₂ to produce 2 mol of NaCl.
Firstly, we need to calculate the no. of moles of 8.0 g Na and 8.0 g Cl₂:For Na:
n = mass/molar mass = (8.0 g)/(22.989 g/mol) = 0.348 mol.
For Cl₂:
n = mass/molar mass = (8.0 g)/(70.9 g/mol) = 0.113 mol.
From the stichiometry: 2 mol of Na react with 1 mol of Cl₂ with (2: 1) molar ratio.The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it.
So, 0.113 mol of Cl₂ react completely with 0.226 mol of Na which be in excess (0.384 - 0.226 = 0.158 mol).
So, the limiting reactant is Cl₂ which is the reactant has the lowest molar ratio.
Diatomic chlorine (Cl₂) is the limiting reactant for the reaction between 8 g of sodium and 8 g of diatomic chlorine.
The limiting reactant in a chemical reaction is simply defined as the reactant which is completely used up in the chemical reaction.
With the above information in mind, we can determine the limiting reactant for the reaction between 8 g of sodium and 8 g of diatomic chlorine (Cl₂). This can be obtained as follow:
2Na + Cl₂ —> 2NaCl
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of Cl₂ = 35.5 × 2 = 71 g/mol
Mass of Cl₂ from the balanced equation = 1 × 71 = 71 g
From the balanced equation above,
46 g of Na reacted with 71 g of Cl₂
Therefore,
8 g of Na will react with = [tex]\frac{8 * 71}{46}\\\\[/tex] = 12.3 g of Cl₂
From the calculation made above, we can see clearly that a higher mass of Cl₂ (i.e 12.3 g) than what was given (i.e 8 g) is needed to react completely with 8 g of Na.
Therefore, Cl₂ is the limiting reactant and Na is the excess reactant.
Learn more: https://brainly.com/question/14225536