From Melting! Hope this helped
Answer: 3m/s2
Explanation:
Force =24 N
Mass = 8kg
Acceleration =?
Force = Mass x Acceleration
Acceleration = Force /Mass = 24/8 = 3m/s2
Answer:
New temperature T2 = 707.5 K (Approx.)
Explanation:
Given:
Old pressure P1 = 2 atm
Old temperature T1 = 283 K
New Pressure P2 = 5 atm
Find:
New temperature T2
Computation:
Using Gay-Lussac law;
P1 / T1 = P2 / T2
So,
2 / 283 = 5 / T2
New temperature T2 = 707.5 K (Approx.)
Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)
Answer:
-3670.33 J/K
Explanation:
Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.
Mathematically, change of Entropy can be expressed as,
ΔS = ΔH/T ....................................... Equation 1
Where ΔS = Change of entropy, ΔH = heat change, T = temperature.
ΔH = -(Lf×m).................................... Equation 2
Note: ΔH is negative because heat is lost.
Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg
Substitute into equation
ΔH = -(3.34×10⁵×3.0)
ΔH = - 1002000 J.
But T = 0 °C = (0+273) K = 273 K.
Substitute into equation 1
ΔS = -1002000/273
ΔS = -3670.33 J/K
Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C
Answer:
Explanation:
The result will be affected.
The mass of KHP weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).
Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.
During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.
Thus, NaOH will appear at a higher concentration than it actually is.
