Answer: 70.0°C
Explanation:
Quantity of heat = Mass * Specific heat * Change in temperature
Quantity of heat = 104.6 KJ
Mass = 500.0 g
Specific heat of water is 4.18 J/g°C
Change in temperature assuming final temperature is x = x - 20
Units should be in grams and joules:
104,600 = 500 * 4.18 * (x - 20)
104,600 = 2,090 * (x - 20)
x - 20 = 104,600/2,090
x = 104,600/2,090 + 20
x = 69.8
= 70.0°C
Answer:
Option E. Zirconium
Explanation:
From the question given above, the following data were obtained:
Length of side (L) of cube = 0.2 cm
Mass (m) of cube = 52 mg
Name of the unknown metal =?
Next, we shall determine the volume of the cube. This can be obtained as follow:
Length of side (L) of cube = 0.2 cm
Volume (V) of the cube =?
V = L³
V = 0.2³
V = 0.008 cm³
Next, we shall convert 52 mg to g. This can be obtained as follow:
1000 mg = 1 g
Therefore,
52 mg = 52 mg × 1 g / 1000 mg
52 mg = 0.052 g
Thus, 52 mg is equivalent to 0.052 g.
Next, we shall determine the density of the unknown metal. This can be obtained as follow:
Mass = 0.052 g.
Volume = 0.008 cm³
Density =?
Density = mass / volume
Density = 0.052 / 0.008
Density of the unknown metal = 6.5 g/cm³
Comparing the density of the unknown metal i.e 6.5 g/cm³ with those given in table in the above, we can conclude that the unknown metal is zirconium
Answer is: sulfuric acid is the limiting reactant.
Chemical reaction: 3H₂SO₄ + 2Al(OH)₃ → Al₂(SO₄)₃ + 6H₂O.
m(H₂SO₄) = 34 g.
n(H₂SO₄) = m(H₂SO₄) ÷ M(H₂SO₄).
n(H₂SO₄) = 34 g ÷ 98 g/mol.
n(H₂SO₄) = 0,346 mol.
m(Al(OH)₃) = 33 g.
n(Al(OH)₃) = 33 g ÷ 78 g/mol.
n(Al(OH)₃) = 0,423 mol.
From chemical reaction: n(H₂SO₄) : n(Al(OH)₃) = 3 : 2.
Answer:
The number of neutron in the Aluminium Isotope is :
B. 14
Explanation:
Isotopes : These are the atoms which have same atomic number but have different mass number.
<u>This image shows the average atomic mass of Al element because it is in decimals</u>.
Atomic mass = 26.98154
(Note : mass number of single isotope can never be in decimals)
It is the average of mass of different isotopes of Al
Major Isotopes of 
are :
......atomic mass = 26
.......atomic mass = 27
mass of Al given in image(26.98) is nearly equal to mass of 2nd isotope(27)
mass of 
Now calculate the neutron in
Number of neutron = mass number - atomic number
= 27 - 13
Number of neutron = 14
(Atomic mass is same as mass number)
The types of intermolecular forces that occur in a substance will affect its physical properties, such as its phase, melting point and boiling point.
