Answer:
You first need to construct a balanced chemical equation to describe the reaction:
KOH + HNO3 ---------> KNO3 + H2O
Work out the no. moles of HNO3 being neutralized:
Moles = Volume x Concentration = (25/1000) x 0.0150 = 0.000375 moles
From the balanced equation the molar ratio of KOH to HNO3 is 1:1 so you also need 0.000375 moles of KOH to neutralise the nitric acid
Now you can work out the volume of KOH required:
Volume = Moles/Concentration = (0.000375)/0.05 = 0.0075 dm^3 = 7.5 cm^3
Answer:
The correct answer will be "18.25 g".
Explanation:
The given values are:
Specific heat,
C = 0.45 J/g･°C
Heat involved,
q = 801 J
Temperature,
ΔT = 120.0°C-22.5°C
= 97.5°C
As we know,
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
Answer:
12.02 g
Explanation:
From the question given above, the following data were obtained:
Half life (t½) = 2 days
Original amount (N₀) = 96 g
Time (t) = 6 days
Amount remaining (N) =..?
Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:
Half life (t½) = 2 days
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 2
K = 0.3465 /day
Therefore, the rate of disintegration of the isotope is 0.3465 /day.
Finally, we shall determine the amount of the isotope remaining after 6 days as follow:
Original amount (N₀) = 96 g
Time (t) = 6 days
Decay constant (K) = 0.3465 /day.
Amount remaining (N) =.?
Log (N₀/N) = kt / 2.303
Log (96/N) = (0.3465 × 6) / 2.303
Log (96/N) = 2.079/2.303
Log (96/N) = 0.9027
Take the anti log of 0.9027
96/N = anti log (0.9027)
96/N = 7.99
Cross multiply
96 = N × 7.99
Divide both side by 7.99
N = 96 /7.99
N = 12.02 g
Therefore, the amount of the isotope remaining after 6 days is 12.02 g
Balance Chemical Equation for this reaction is,
2 CH₄ + O₂ → 2CH₃OH
According to this eq, 22.4 L (1 moles) of Oxygen requires 44.8 L (2 mole) CH₄ for complete reaction.
So, the volume of CH₄ required to consume 0.66 L of O₂ is calculated as,
22.4 L O₂ required to consume = 44.8 L CH₄
0.660 L O₂ will require = X L of CH₄
Solving for X,
X = (44.8 L × 0.660 L) ÷ 22.4 L
X = 1.320 L of CH₄
Result:
1.320 L of CH₄ <span>gas is needed to react completely with 0.660 L of O</span>₂<span> gas to form methanol (CH</span>₃OH<span>).</span>
Answer: The answer is the second one
