Chemical reactions can be broadly classified into two categories from the context of heat requirement. These include; endothermic and exothermic reactions.
Endothermic reactions proceed with the absorption of heat while exothermic reactions are accompanied by the evolution of heat.
1) Oxidation of iron involves reaction of elemental iron with oxygen to form iron oxide (rust). This is essentially a corrosion reaction in which the metallic bonds in Fe are broken and new bonds between Fe and O are formed, This is an exothermic reaction
4Fe + 3O2 → 2Fe2O3
Ans D)
2) The process of condensation i.e. phase change liquid to solid (or gas to liquid) involves release of heat to the surroundings, hence it will be an exothermic reaction.
Ans B)
3) Specific heat is the amount of heat required to raise the temperature of 1 g of a substance by a 1 C. Typically, liquids and gases are used as coolants. For a coolant to be efficient, it must have a high specific heat.
Ans B)
Answer:
The energy required is 3225 Joules.
Explanation:
Given,
mass of lead cube = 500 grams
T₁ = 25°C
T₂ = 75°C
specific heat of lead = 0.129 J/g°C
Energy required to heat the lead can be found by using the formula,
Q = (mass) (ΔT) (Cp)
Here, ΔT = T₂ - T₁ = 75 - 25 = 50
Substituting the values,
Q = (500)(50)(0.129)
Q = 3225 Joules.
Therefore, energy required is 3225 J.
Answer: ![Rate=k[L]^1[M]^2](https://tex.z-dn.net/?f=Rate%3Dk%5BL%5D%5E1%5BM%5D%5E2)
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
![Rate=k[L]^x[M]^y](https://tex.z-dn.net/?f=Rate%3Dk%5BL%5D%5Ex%5BM%5D%5Ey)
(1)
k= rate constant
x = order with respect to L
y = order with respect to M
n =( x+y)= Total order
a) If [L] is doubled, the reaction rate will increase by a factor of 2:
![2\times Rate=k[2L]^x[M]^y](https://tex.z-dn.net/?f=2%5Ctimes%20Rate%3Dk%5B2L%5D%5Ex%5BM%5D%5Ey)
(2)
b) If [M] is doubled, the reaction rate will increase by a factor of 4:
![4\times Rate=k[L]^x[2M]^y](https://tex.z-dn.net/?f=4%5Ctimes%20Rate%3Dk%5BL%5D%5Ex%5B2M%5D%5Ey)
(3)
Dividing 2 by 1:
![\frac{2\times Rate}{Rate}=\frac{k[2L]^x[M]^y}{k[L]^x[M]^y}](https://tex.z-dn.net/?f=%5Cfrac%7B2%5Ctimes%20Rate%7D%7BRate%7D%3D%5Cfrac%7Bk%5B2L%5D%5Ex%5BM%5D%5Ey%7D%7Bk%5BL%5D%5Ex%5BM%5D%5Ey%7D)
Dividing 3 by 1
![\frac{4\times Rate}{Rate}=\frac{k[L]^x[2M]^y}{k[L]^x[M]^y}](https://tex.z-dn.net/?f=%5Cfrac%7B4%5Ctimes%20Rate%7D%7BRate%7D%3D%5Cfrac%7Bk%5BL%5D%5Ex%5B2M%5D%5Ey%7D%7Bk%5BL%5D%5Ex%5BM%5D%5Ey%7D)
Thus rate law is: ![k[L]^1[M]^2](https://tex.z-dn.net/?f=k%5BL%5D%5E1%5BM%5D%5E2)
Answer:
(4) Water, stirring, and filtering
Explanation:
The added water will dissolve the sugar but not the sand. When you filter the mixture, the sand will be trapped in the filter paper and the dissolved sugar will pass through the pores of the paper.
(1) and (2) are wrong. You don't separate a mixture of sugar and sand by adding more sand. Furthermore, neither substance will boil at 100 °C.
(3) is wrong. You can dissolve the sugar in water but, if you boil the water away, the sugar and sand will still be together.
Answer:
A- In (NH4)3PO4 there are four ions in total as three ions of NH₄ and is one ion of PO₄
