H2SO4 + 2 NaOH ----> Na2SO4 + 2 H2O.
0.085 L * 0.176 mol/L = 0.01496 mol H2SO4
is neutralised by 0.01496 mol * 2
= 0.02992 mol NaOH.
1000 mL of 0.492 M NaOH
contains 0.492 moles NaPH.
0.02992 / 0.452 * 1000 mL
= 66.19 = 66 mL
Your answer is 3.80 moles
Answer:
3.4 × 10²³ molecules of CBr₄
Explanation:
Given data:
Mass of CBr₄ = 189 g
Number of molecules = ?
Solution:
First of all we will calculate the number of moles.
Number of moles = mass / molar mass
Number of moles = 189 g/ 331.63 g/mol
Number of moles = 0.6 mol
Now the given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
Foe 0.6 moles of CBr₄:
0.6 mol × 6.022 × 10²³ molecules of CBr₄ / 1 mol
3.4 × 10²³ molecules of CBr₄
Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:
Heat = mC(T2-T1)
Heat = 5(0.90)(32.1 - 22.1)
Heat = 45 J energy gained
Answer:
-1.78 V
Explanation:
There are several rules required to calculate the cell potential:
- given standard cell potential, we may reverse the equation: the products of a given reaction become our reactants, while reactants become our products in the reversed equation. For a reversed equation, we change the sign of the cell potential to the opposite sign;
- if we multiply the whole equation by some number, this doesn't influence the cell potential value. It only produces a different expression in the equilibrium constant.
That said, notice that the initial reaction with respect to the final reaction is:
- reversed: chromium(III) cation and chloride anion become our reactants as opposed to the products in the initial reaction, so we change the sign of the cell potential to a negative value of -1.78 V;
- each coefficient is multiplied by a fraction of
. It doesn't influence the value of the cell potential.
Thus, we have a cell of E = -1.78 V.
