Answer:
6.17 g/cm³
Explanation:
Data given:
one side of cube = 0.53 cm
mass of the cube is 0.92 g
density of the cube = ?
Solution:
First we will calculate for volume the cube
As we know all the sides or edges of a cube are equal so volume equation will be
So,
V = length x width x height
V = e³
as on side = 0.53 cm
then
V = (0.53 cm)³
V = 0.149 cm³
Now we will calculate density of cube
To calculate density, formula will be used
d = m/v . . . . . (1)
where
d = density
m = mass
v = volume
put values in above formula 1
d = 0.92 g / 0.149 cm³
d = 6.17 g/cm³
so. the density of cube = 6.17 g/cm³
Answer:
= 25.05°C
Explanation:
Given:
the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.
mass = 0.905g of dimethylphthalate
molar mass = 194.18g dimethylphthalate
number of moles of dimethylphthalate = ???
= 21.5°C
= 6.15 kJ/°C
= ???
since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;
0.905g of dimethylphthalate × 
number of moles of dimethylphthalate = 0.000466 moles
Heat released = moles of dimethylphthalate × heat of combustion
= 0.000466 moles × 4685 kJ
= 21.84 kJ
∴ Heat absorbed by the calorimeter = 
21.84 kJ =6.15 kJ/°C 
21.84 KJ = 
21.84 KJ = 
- 132.225 kJ
21.84 KJ + 132.225 kJ =
154.065 kJ =
= 
=25.05°C
In the combustion process using excess oxygen, each mole of methane results to 1 mole of co2 while ethane produces 2 moles of Co2. Under same conditions, these can be translated to volume. Hence the total volume absorbed is 10 cm3 + 20 cm3 = 30 cm3.
Answer:
According to my research A molecule is two or more atoms held together by covalent bonds. An atom is the smallest part of an element. ... A sodium atom has one outer electron, and a carbon atom has four outer electrons.
Explanation:
Answer:
14.33 g
Explanation:
Solve this problem based on the stoichiometry of the reaction.
To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and finally calculate the mass of AgCl.
2 AgNO₃ + CaCl₂ ⇒ Ca(NO₃)₂ + 2 AgCl
mass, g 6.97 6.39 ?
MW ,g/mol 169.87 110.98 143.32
mol =m/MW 0.10 0.06 0.10
From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :
0.10 mol x 143.32 g/mol = 14.33 g
