Answer:
Explanation:
A sound knowledge of specific heat capacity of the metals is required in this case.
The specific heat capacity of a metal is the quantity of heat required to the raise the temperature of a unit mass of it by 1°C.
It is related to quantity of heat using the expression below;
H = m c Δt
where m is the mass
c is the specific heat capacity
Δt is the temperature change
let us make the specific the subject of the expression;
c = 
we can see that there is an inverse relationship between specific heat and temperature change.
The specific heat capacity of a body is an intensive property that is unique to the metal.
The higher the specific heat capacity, the lower the amount of temperature change in it.
Let us find the specific heat capacity of the given metals;
Aluminium 0.897J/gK
Iron 0.412J/gK
Silver 0.24J/gK
After the heat is supplied,
Silver > Iron > Aluminium in terms of temperature change
Answer:
0.03
Explanation:
22.8 g Ba(OH)2 (1 mol Ba (OH)2/ 171.34 g) = 0.133 mol Ba (OH)2
77.2 g H2O (1 mol H2O/18 g) = 4.29 mol H2O
X= molar fraction= mol Ba(OH)2/ mol total
X= 0.133/ (0.133+4.29) = 0.03
Answer: If you think about it, B. would be the most reasonable answer with the given factors.
Answer:
The four coefficients in order, separated by commas are 1, 8, 5, 6
Explanation:
We count the atoms in order to balance this combustion reaction. In combustion reactions, the products are always water and carbon dioxide.
C₅H₁₂ + ?O₂→ ?CO₂ + ?H₂O
We have 12 hydrogen in right side and we can balance with 6 in the left side. But the number of oxygen is odd. We add 2 in the right side, so we have 24 H, and in the product side we add a 12.
As we add 2 in the C₅H₁₂, we have 10 C, so we must add 10 to the CO₂ in the product side.
Let's count the oxygens: 20 from the CO₂ + 12 from the water = 32.
We add 16 in the reactant side. Balanced equation is:
2C₅H₁₂ + 16O₂→ 10CO₂ + 12H₂O
We also can divide by /2 in order to have the lowest stoichiometry
C₅H₁₂ + 8O₂→ 5CO₂ + 6H₂O
Answer:
M = 20.5 g/mol
Explanation:
Given data:
Volume of gas = 1.20 L
Mass of gas = 1.10 g
Temperature and pressure = standard
Solution:
First of all we will calculate the density.
Formula:
d = mass/ volume
d = 1.10 g/ 1.20 L
d = 0.92 g/L
Now we will calculate the molar mass.
d = PM/RT
0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K
M = 0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K / 1 atm
M = 20.5 g/mol
