Answer:
Explanation:
a ) 15( ft)³ = 15 ( 12 inch )³ = 15 x 12 x 12 x 12 (inch)³ = 25920 (inch)³
=25920( 2.54 cm)³ = 25920 x 2.54 x 2.54 x 2.54 cm³ =424752.7 cc
424752.7/1000 litre = 424.75 liter
424.75 liters cost 1.30 dollars
1 liter costs 1.3/424.75 = 3.06 x 10⁻³ dollars.
So rate
= 3.06 x 10⁻³ dollars/liter
b ) one liter of water takes 0.304 ft³ of gas
2.1 liter will take 0.304 x 2.1 = 0.6384 ft³ of gas
cost = 1.3 x .6384 / 15 = 0.055328 dollar = 55.33 x 10⁻³ dollar.
Given ,
Mass of sample of cobalt = 27 g
density of sample of cobalt = 9g/cm^3
We know that ,
Density = mass of sample/volume of sample
From that relation ,
We can deduce the following as
Volume = mass of sample/density of sample
Hence , required volume of sample of cobalt = 27 g /9 g/cm^3 = 3 cm^3
The volume is
the balanced equation for the formation of water is as follows
2H₂ + O₂ ---> 2H₂O
molar ratio of oxygen to water is 1:2
mass of water produced is - 46 g
therefore the number of moles of water produced - 46 g / 18 g/mol = 2.56 mol
1 mol of oxygen reacts to give 2 mol of water assuming oxygen is the limiting reactant
for 2 mol of water to be produced 1 mol of oxygen has to react
therefore for 2.56 mol of water to be produced - 1/2 x 2.56 mol = 1.28 mol of oxygen should react
mass of oxygen reacted is - 1.28 mol x 32 g/mol = 41.0 g
mass of oxygen that should react is 41 g
Answer:NH
+
4
:
0.272
M
SO
2
−
4
:
0.136
M
Explanation:
We can solve this problem using some molarity calculations:
molarity
=
mol solute
L soln
We should convert the given mass of
(NH
4
)
2
SO
4
to moles using its molar mass (calculated to be
132.14
g/mol
):
10.8
g (NH
4
)
2
SO
4
⎛
⎝
1
l
mol (NH
4
)
2
SO
4
132.14
g (NH
4
)
2
SO
4
⎞
⎠
=
0.0817
mol (NH
4
)
2
SO
4
This is the quantity present in
100
mL soln
, so let's calculate the molarity of the solution (converting volume to liters):
molarity
=
0.0817
l
mol (NH
4
)
2
SO
4
0.100
l
L soln
=
0.817
M
10
mL
of this solution is added to
50
mL H
2
O
, which makes a
60
-
mL
total solution.
We can now use the dilution equation
M
1
V
1
=
M
2
V
2
to find the molality of the new,
60
-
mL
solution:
(
0.0817
M
)
(
10
l
mL
)
=
(
M
2
)
(
60
l
mL
)
M
2
=
(
0.817
M
)
(
10
mL
)
60
mL
=
0.136
M
This means that there are
0.136
moles of
(NH
4
)
2
SO
4
per liter of solution.
Let's recognize that
1
mol (NH
4
)
2
SO
4
contains
2
mol NH
+
4
1
mol SO
2
−
4
The concentrations of each ion is thus
(
2
)
(
0.136
M
)
=
0.272
M
NH
+
4
(
1
)
(
0.136
M
)
=
0.136
M
SO
2
−
4
Explanation:
Answer:
The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of C atoms in the chain.