Answer:
18 KJ
Explanation:
Data Given:
mass of Lead (m) = 21 g
Heat taken for vaporization (Q) = ?
Solution:
This problem is related to phase change and latent heat of vaporization.
Latent heat of vaporization is the amount of heat taken to convert one mole of substance at its boiling point to its vapor.
So, Latent heat of vaporization of lead has a constant value
Latent heat of vaporization of lead = 177.7 KJ/mol
Formula used
Q = m x Lv. . . . . (1)
where
Lv = specific latent heat of vaporization
here the value for latent heat of vaporization is for mole so instead of mass we will use moles in formula.
So,
Q = no. of mol x Lv. . . . . (2)
first find no. of moles for 21 g of lead
no. of moles = mass in grams / molar mass . . . . . . (3)
molar mass of lead (Pb) = 207 g/mol
put values in equation 3
no. of moles = 21g / 207 g/mol
no. of moles = 0.101 mol
so,
number of moles of lead (Pb) = 0.101 mol
Put values in the eq.2
Q = 0.101 mol x 177.7 KJ/mol
Q = 18 kJ
So, 18KJ of heat is taken to vaporize 21 g of lead (Pb)
Answer: A - Carbon
Explanation: Hope this helps!
Answer:
S.G = 0.79.
Explanation:
Hello there!
In this case, according to the given information in the problem, it turns out possible for us to calculate the specific gravity of this alcohol by simply dividing its density, 0.79 g/mL by that of the water, 1 g/mL just as a reference for us to work with:
Thus, we plug in the densities to obtain:
Which is dimensionless as g/mL is cancelled out due to its presence on both top and bottom of the previous formula.
Regards!
Answer:
See explanation
Explanation:
2HCl(aq) + CaCO3(aq) ------->CaCl2(aq) + CO2(g) + H2O(l)
Number of moles of acid present = 50/1000 * 0.15 = 0.0075 moles
Number of moles of calcium carbonate = 0.054g/100 g/mol = 0.00054 moles
2 moles of HCl reacts with 1 mole of calcium carbonate
x moles of HCl reacts with 0.00054 moles of calcium carbonate
x = 2 * 0.00054/1
x = 0.00108 moles of HCl
Amount of acid left = 0.0075 moles - 0.0075 moles = 0.00642 moles
Reaction of HCl and NaOH
HCl(aq) + NaOH(aq) ------> NaCl(aq) + H2O(l)
Since the reaction is in the mole ratio of 1:1
0.00642 moles of HCl is neutralized by 0.00642 moles of NaOH
