Answer:
It has a different value at different temperatures.
Explanation:
For any system in equilibrium, the molar concentration of all the species on the right hand side are related to the molar concentrations of those at the left side by a constant known as the equilibrium constant.
The equilibrium is a constant at a given temperature as it is temperature dependent.
A change in temperature of an equilibrium system shifts the system to a new equilibrium point. A rise in temperature actually shifts equilibrium position to the direction that absorbs heat and vice versa.
The shift in equilibrium as a result of temperature change is actually a change in the value of the equilibrium constant. Equilibrium constant is represented as [tex]K_{eq}[/tex]
The derivation of the equilibrium constant is based on the Law of Mass Action which states: the rate of a chemical reaction is proportional to the product of the concentrations of the reacting substances.
Answer:
It has a different value at different temperatures.
Explanation:
I took the test, and this answer was correct
f a hypothesis can stand the test of repeated examination, it can become a _____.
Answer:
theory
If it has been tested enough, then it can be considered a scientific theory.
An expandable container of oxygen gas has a volume of 125 mL and a temperature of 25.0ºC. What volume will the gas occupy at 55°C?
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
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need help with 14 and 16 pls asap!! this is my friends test and im taking it tomorrow!!
Answer:
Q14: 17,140 g = 17.14 kg.
Q16: 504 J.
Explanation:
Q14:
To solve this problem, we can use the relation:Q = m.c.ΔT,
where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).
m is the mass of the ice (m = ??? g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).
∵ Q = m.c.ΔT
∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)
∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.
Q16:
To solve this problem, we can use the relation:Q = m.c.ΔT,
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 12.0 g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).
∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.
How many moles are in 9.8 grams of calcium
Answer:
0.24 moles.
Explanation:
number of moles = mass of sample /mass of 1 mole
9.8g /40.08g = 0.2445 moles <3
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.
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Please can someone help me with this. I am lost and I don't understand how can I do this.
What volume, in L, of 0.219 M NiCl2 solution is required to produce 68.4 g of precipitate.
3NiCl2(aq) + 2Na3PO4(aq) → Ni3(PO4)2(s) + +6NaCl(aq)
Answer:
2.56 L NiCl2
Explanation:
This is a stoichiometry problem. You need to find a way to go from g of Ni3(PO4)2 to L of NiCl2. To do this you travel
mass Ni3(PO4)2 > moles Ni3(PO4)2 > moles NiCl2 > L NiCl2
so you need to know the key that lets you travel from one to the next.
Mass > moles, use molar mass
moles > moles, use the mole ratio
moles > L, use the molarity (moles / liter) in this case. (With gases, you use the STP gas volume of 22.4L / mole).
The attached picture shows what this problem looks like.
The problem requires application of stoichiometry using a balanced chemical equation. About 2.51 L of 0.219 M NiCl2 solution is needed to produce 68.4 g of Ni3(PO4)2 precipitate.
Explanation:The subject of the problem concerns stoichiometry in chemistry. Specifically, the use of a balanced chemical equation to calculate volumes of a solution needed, in this case - 0.219M of NiCl2 solution required to produce 68.4g of Ni3(PO4)2 precipitate.
First, we need to convert the given mass of precipitate (Ni3(PO4)2) to moles. We use the molar mass of Ni3(PO4)2 for this which is roughly 374.09 g/mol. Thus, 68.4g of Ni3(PO4)2 is approximately 0.183 moles.From the balanced chemical equation, we know that three moles of NiCl2 are required to produce one mole of Ni3(PO4)2. So, we need about 0.183 x 3 = 0.549 moles of NiCl2.Finally, we need to determine the volume of a 0.219 M NiCl2 solution that contains 0.549 moles of NiCl2. Using the molarity formula (moles/volume), we find that the volume is approximately 2.51 L.Learn more about Stoichiometry here:https://brainly.com/question/30215297
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