Answer:
59.4 tons of iron can be isolated from this ore
Explanation:
mass of ore = 250 tons
mass of hematite = 34 % of 250
= 85 tons
mass of iron = 69.9 % of hematite
= 69.9 % of 85
= 59.4 tons
So, 59.4 tons of iron can be isolated from this ore .
Answer: Option (c) is the kinetic energy.
Explanation:
The kinetic energy present within the particles of a substance is known as thermal energy. Measure of average kinetic energy of particles of a substance is known as temperature.
Whereas heat defined as the transfer of thermal energy from a hot object to a cold object.
Enthalpy is defined as the heat gained or lost by a substance in a chemical reaction.
Thermodynamics is the study of relation between heat and work with chemical or physical changes.
Thus, we can conclude that a measure of kinetic energy of particle motion within a substance is temperature.
Answer:
The answer is
<h2>22 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
mass of liquid = 44 g
volume = 2 mL
The density of the liquid is
We have the final answer as
<h3>22 g/mL</h3>
Hope this helps you
Answer:
29 L
Explanation:
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
A) according to this reaction:
by using ICE table:
NH2OH(aq) + H2O(l) → HONH3+(aq) + OH-
initial 0.4 M 0 0
change -X +X +X
Equ (0.4-X) X X
when Kb = [OH-][HONH3+]/[NH2OH]
when we have Kb = 1.1x10^-8 so,
by substitution:
1.1x10^-8 = X^2/(0.4-X) by solving this equation for X
∴X = 6.6x10^-5 M
∴[OH] = 6.6x10^-5 M
when POH = - ㏒[OH]
∴POH = -㏒(6.6x10^-5)= 4.18
∴PH = 14 - POH = 14 - 4.18
= 9.82
when PH = -㏒[H+]
∴[H+] = 10^9.82 = 1.5x10^-10 M+0.02molHcl
= 0.02
∴ the new value of PH = -㏒(0.02)
∴PH = 1.7
B) according to this reaction:
by using ICE table:
HONH3+(aq) → H+(aq) + HONH2(aq)
intial 0.4 0 0
change -X +X +X
Equ (0.4-X) X X
when Ka HONH3Cl = 9.09x10^-7
and Ka = [H+][HONH2] / [HONH3+]
So by substitution and we can assume [HONH3+] = 0.4 as the value of Ka is so small so,
9.09x10^-7 = X^2 / 0.4 by solving for X
∴ X = 6 x 10 ^-4
∴[H+] = 6x10^-4
PH = -㏒[H+]
= -㏒ (6x10^-4) = 3.22
when [H+] = 6x10^-4 + 0.02 m HCl
∴new value of PH = -㏒(6x10^-4+0.02)
= 1.69
C) when we have pure H2O and PH of water = 7
So we can get [H+] when PH = -㏒[H+]
∴[H+] = 10^-7 + 0.02MHCl
= 0.02
∴new value of PH = -㏒0.02
PH = 1.7
d) when HONH2 & HONH3Cl have the same concentration and Hcl added to them so we can assume that PH=Pka
and when we have Ka for HONH3Cl = 9.09x10^-7
So we can get the Pka:
Pka = -㏒Ka
= -㏒9.09x10^-7
= 6.04
∴PH = 6.04
and because of the concentration of the buffer components, HONH2 & HONH3Cl have 0.4 M and the adding of HCl = 0.02 M So PH will remain very near to 6
