Answer:
Option (3) 334J.
Explanation:
The following data were obtained from the question:
Mass (m) = 1g
Heat of fusion (ΔHf) = 334J/g
Heat (Q) =?
The heat released can be obtained as follow:
Q = m·ΔHf
Q = 1 x 334
Q = 334J.
Therefore, the heat released when 1g is converted to ice is 334J.
Answer:
b
d
a
c
Explanation:
More carbons in longest chain = higher BP
D has a BP of 0.8
A has a BP -0.5
C is the lowest because the more branched it is the lower BP
Tell the instructor and runs it out with cold water and call 911
question 1
by use of Avogadro law
that is,1 mole = 6.02 x10^23 atoms
what about 0.60 moles
by use of cross multipication
=(0.60 mole/ x 6.02 x10^23)/ 1mole = 3.612 x10^23 atoms of Zn
question 2
by use of Avogadro law constant
that is 1 mole =6.02 x10^23 molecules
what about 3.52 x10^24 molecules
by cross multiplication
=( 1 mole x3.52 x10^24 molecules/6.02 x10^23 molecules) = 5.847 moles of water
question 3
by use of Avogadro law constant
1 mole = 6.02 x10^23 atoms
what about 2 moles =? atoms
by use of cross multiplication
=( 2moles x 6.02 x10^23 )/1mole= 1.204 x10^24 atoms of Li
question 4
by use of Avogadro law constant
1 mole = 6.02 x10^23 atoms
what about 6.02 x10^23 atoms =? moles
cross multiplication
(1 mole x6.02 x10^23 atoms)/(6.02 x10^23 atoms)= 1 mole of carbon
question 5
by use Avogadro law constant
1 mole =6.02 x10^23 molecules
what about 4.9 x10^23 moles =? moles
by cross multipication
=( 1mole x 4.9 x10^23 molecules) /6.02 x10^23 molecules = 0.81 moles ZNCl2
Answer:
The answer to your question is 24.32 g
Explanation:
Data
Atomic weight = ?
HCl volume = 125 ml
Molarity = 0.2
mass of metal = 0.304 g
Balanced chemical equation
M + 2HCl ⇒ MCl₂ + H₂
Process
1.- Calculate the moles of HCl
Molarity = moles / volume (L)
- Solve for moles
moles = Molarity x volume
moles = 0.2 x 0.125
= 0.025
2.- Calculate the moles of the Metal
1 mol of M ----------------- 2 moles of HCl
x ----------------- 0.025 moles of HCl
x = (0.025 x 1) / 2
x = 0.0125 moles of HCl
3.- Calculate the atomic weight of the metal
atomic weight ---------------- 1 mol
0.304 g ---------------0.0125 moles
Atomic weight = (1 x 0.304) / 0.0125
Atomic weight = 24.32 g
