Density is given by the equation D=m/V, were D is density, m is mass in grams, and V is volume in cubic centimeters.
In this problem, we have density and we have mass so we can plug into the equation and solve for V.
38.6=270.2/V
<em>*Multiply both sides by V*</em>
38.6V=270.2
<em>*Divide both sides by 38.6*</em>
V=7
The volume of the gold nugget is 7cm3.
Hope this helps!!
Divide 1.25/3.93 and then multiply by 100 to get the percent. This equals 31.81%
Answer:
19.28 g/cm^3 to the nearest hundredth.
Explanation:
The volume of water displaced = the volume of the metal.
density = mass / volume
0.0694 kg = 0.0694 * 1000
= 69.4 g.
Density = 69.4 / 3.6
= 19.28 g/cm^3.
Answer:
0.453 moles
Explanation:
The balanced equation for the reaction is:
2Fe(s) + 3O2(g) ==> 2Fe2O3
From the equation, mass of O2 involved = 16 x 2 x 3 = 96g
mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2
= 100g
Therefore 96g of O2 produced 100g of Fe2O3
32.2g of O2 Will produce 100x32.2/96
= 33.54g of Fe2O3
Converting it to mole using number of mole = mass/molar mass
but molar mass of Fe2O3 = 26 + (16 X 3)
= 74g/mole
Therefore number of mole of 33.54g of Fe2O3 = 33.54/74
= 0.453 moles
Answer:
Correct option is
B
5 liters of CH
4
(g)NO
2
at STP
No. of molecules=
22.4
5
mol=
22.4
5
×N
A
molecules
A) 5ℊ of H
2
(g)
No. of moles=
2
5
mol=
2
5
×N
A
molecules
B) 5l of CH
4
(g)
No. of moles of CH
4
=
22.4
5
mol=
22.4
5
N
A
molecules
C) 5 mol of O
2
=5N
A
O
2
molecules
D) 5×10
23
molecules of CO
2
(g)
Molecules of 5l NO
2
(g) at STP=5l of CH
4
(g) molecules at STP
Therefore, option B is correct.