Answer:
14.8 × 10²³ molecules
Explanation:
Given data:
Mass of sulfuric acid = 240 g
Number of molecules = ?
Solution:
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
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
Number of moles of sulfuric acid
<em>Number of moles = mass/ molar mass</em>
Number of moles = 240 g/ 98 g/mol
Number of moles = 2.45 mol
Number of molecules:
1 mole = 6.022 × 10²³ molecules
2.45 × 6.022 × 10²³ molecules
14.8 × 10²³ molecules
Solids maintain their shape, whereas fluids do not because <span>the molecules in solids maintain a regular pattern and only vibrate, or move very slowly. The correct option among all the options that are given in the question is the last option or option "d". I hope the answer has come to your help.</span>
Answer:
Explanation:
After the glorious appearance of Full Moon, the lunar shape starts to wane, meaning it gets smaller. It's visible later at night and into the early morning, and we see a steadily shrinking shape of the lunar surface that's being lit up.
Answer:
1.02mole
Explanation:
The reaction equation is given as:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
Given:
Mass of H₂SO₄ = 50g
Unknown:
Number of moles of NaOH = ?
Solution:
To solve this problem, we first find the number of moles of the acid given;
Number of moles = 
Molar mass of H₂SO₄ = 2(1) + 32 + 4(16) = 98g/mol
Now;
Number of moles = 
= 0.51mole
From the balanced reaction equation:
1 mole of H₂SO₄ will be neutralized by 2 mole of NaOH
0.51 mole of H₂SO₄ will be neutralized by 2 x 0.51 = 1.02mole of NaOH
The pH of the solution : 12
<h3>Further explanation</h3>
Reaction
HCOOH + NaOH ⇒ HCOONa + H₂O
mol HCOOH =
mol NaOH =
Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]
ICE method :
HCOOH + NaOH ⇒ HCOONa + H₂O
4 5
4 4 4 4
0 1 1 1
Concentration of [OH⁻] from NaOH :
pOH=-log[OH⁻]
pOH=-log 10⁻²=2
pH+pOH=14
pH=14-2=12
