Answer: 0.4M
Explanation:
Given that,
Amount of moles of NaOH (n) = ?
Mass of NaOH in grams = 40.0g
For molar mass of NaOH, use the atomic masses: Na = 23g; O = 16g; H = 1g
NaOH = (23g + 16g + 1g)
= 40g/mol
Since, n = mass in grams / molar mass
n = 40.0g / 40.0g/mol
n = 1 mole
Volume of NaOH solution (v) = 2.5 L
Concentration of NaOH solution (c) = ?
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
c = 1 mole / 2.5 L
c = 0.4 mol/L (Concentration in mol/L is the same as Molarity, M)
Thus, the concentration of a solution of a 40.0 g of NaOH in 2.5 L of solution is 0.4 mol/L or 0.4M
Answer:
The answer to your question is: C2 = 0.0004 M
Explanation:
Data
pH = 2.5; V = 1.0 L
V2 = 8.0 L C2 = ?
Formula
C1V1 = C2V2
C2 = C1V1 / V2
pH = -log[H⁺]
Process
[H⁺] = antilog -pH
[H⁺] = antilog (-2.5)
[H⁺] = 0.003 M = C1
Finally
(0.003)(1 l) = C2(8)
C2 = 0.003 / 8
C2 = 0.0004 M
Answer:
The answer is below
Explanation:
Sulfuric acid is used for various purposes, some of which include the following:
1. It is used in the production of various manufactured goods.
2. It is used in the manufacturing of chemicals
3. It is also used in the making of fertilizer
4. It is used in the refining process of petroleum products
5. It is used in the processing of metals
Answer : The enthalpy of the reaction is, -2552 kJ/mole
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The given enthalpy of reaction is,

The intermediate balanced chemical reactions are:
(1)

(2)

(3)

(4)

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :
(when we are reversing the reaction then the sign of the enthalpy change will be change.)
The expression for enthalpy of the reaction will be,



Therefore, the enthalpy of the reaction is, -2552 kJ/mole