Answer: 1.414x10^24 molecules in 94.4g MgO
Explanation: molar mass MgO 40.204
molecules in 40.204 g MgO = avogadro number
molecules in 94.4 g MgO = (94.4/40.204)*avogadro number
(94.4/40.204)*6.02214076*10^23 = 14.14x10^23
Answer:
Approximately 1.9 kilograms of this rock.
Explanation:
Relative atomic mass data from a modern periodic table:
To answer this question, start by finding the mass of Pb in each kilogram of this rock.
89% of the rock is
. There will be 890 grams of
in one kilogram of this rock.
Formula mass of
:
.
How many moles of
formula units in that 890 grams of
?
.
There's one mole of
in each mole of
. There are thus
of
in one kilogram of this rock.
What will be the mass of that
of
?
.
In other words, the
in 1 kilogram of this rock contains
of lead
.
How many kilograms of the rock will contain enough
to provide 1.5 kilogram of
?
.
In oil and gas industry:
When crude oil get extracted from well, salt water and some other stuff needs to be removed before oil can be sued in the car
Answer:
V2 = 35.967cm^3
Explanation:
Given data:
P1 = 0.2atm
P2 = 1.4atm
V1 = 250cm^3
V2 = ?
T1 = 10°C + 273 = 283K
T2 = 12°C + 273 = 285K
Apply combined law:
P1xV1/T1 = P2xV2/T2 ...eq1
Substituting values:
0.2 x 250/283 = 1.4 x V2/285
Solve for V2:
V2 = 14250/396.2
V2 = 35.967cm^3