144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.
<h3>What is Ideal Gas Law ? </h3>
The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.
PV = nRT
where,
P = Presure
V = Volume in liters
n = number of moles of gas
R = Ideal gas constant
T = temperature in Kelvin
Here,
P = 1 atm [At STP]
R = 0.0821 atm.L/mol.K
T = 273 K [At STP]
Now first find the number of moles
F₂ + CaBr₂ → CaF₂ + Br₂
Here 1 mole of F₂ reacts with 1 mole of CaBr₂.
So, 199.89 g CaBr₂ reacts with = 1 mole of F₂
1.28 g of CaBr₂ will react with = n mole of F₂
n = 0.0064 mole
Now put the value in above equation we get
PV = nRT
1 atm × V = 0.0064 × 0.0821 atm.L/mol.K × 273 K
V = 0.1434 L
V ≈ 144 mL
Thus from the above conclusion we can say that 144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.
Learn more about the Ideal Gas here: brainly.com/question/20348074
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Answer:
a. CH3NH2(aq) + H⁺ → CH3NH3⁺
Explanation:
The mixture of a weak base as CH3NH2 with its conjugate acid CH3NH3Cl produce a buffer. As the weak acid is in equilibrium with water, the mixture of the weak base and its conjugate base produce that the acid or base released react avoiding the change in pH.
For example, when a strong acid as HNO3 reacts, the weak base will react producing the conjugate base, that is:
CH3NH2(aq) + H⁺ → CH3NH3⁺
Right answer is:
<h3>a. CH3NH2(aq) + H⁺ → CH3NH3⁺</h3>
Answer:
The calorimeter constant is = 447 J/°C
Explanation:
The heat absorbed or released (Q) by water can be calculated with the following expression:
Q = c × m × ΔT
where,
c is the specific heat
m is the mass
ΔT is the change in temperature
The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.
The heat absorbed by the calorimeter (Q) can be calculated with the following expression:
Q = C × ΔT
where,
C is the calorimeter constant
The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).
Qabs + Qrel = 0
Qabs = - Qrel
Qcal + Qw₁ = - Qw₂
Qcal = - (Qw₂ + Qw₁)
Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)
Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]
Ccal = 447 J/°C
C is the answer hope this helps
Answer A: Connect a wire coil to an ammeter. Move a bar magnet into and out of the wire coil as you observe the ammeter.
