1. covalent bonds are strong
2. hydrogen bonds are weak
3. Ionic bonds are weak
Leftover: approximately 11.73 g of sulfuric acid.
<h3>Explanation</h3>
Which reactant is <em>in excess</em>?
The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,
- Al(OH)₃ is the limiting reactant.
- H₂SO₄ is in excess.
How many <em>moles</em> of H₂SO₄ is consumed?
Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
How many <em>grams</em> of H₂SO₄ is consumed?
The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.
How many <em>grams</em> of H₂SO₄ is in excess?
40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.
Question:
1. (NH)2CrO
a) Number of moles of H:
b) Number of moles of N:
Answer:
a) Number of moles of H: 2
b) Number of moles of N: 2
Explanation:
The 
is ammonium Chromate which is monoclinic and yellow Crystal that is formed due to the reaction of ammonium Hydroxide and ammonium di-chromate. It is used as catalyst, corrosion inhibitor as well as analytical inhibitors.
Question:
2. Ag.SO.
a) Molar Mass:
b) Percent Composition of Ag:
c) Percent Composition of S:
d) Percent Composition of O:
Answer:
a) Molar Mass: 155.93 Kg
b) Percent Composition of Ag: 69%
c) Percent Composition of S: 20.5%
d) Percent Composition of O: 10.2%
Explanation:
Molar mass = molar mass of Ag + molar mass of S + molar mass of O
=>107.87+32.06+16
=> 155.93 Kg
Percent Composition of Ag
= 
= 
= 0.69 \times 100
= 69%
Percent Composition of S:
= 
=
= 0.205 \times 100
= 20.5%
Percent Composition of O:
= 
= 
= 0.102 \times 100
= 10.2%
Answer:
Option C. 10m/s
Explanation:
Data obtained from the question include:
Wavelength = 5m
Frequency = 2Hz
Velocity =..?
Velocity of a wave is related to the wavelength and frequency by the following equation:
Velocity = wavelength x frequency
With the above equation, we can easily calculate the velocity of the ocean wave as follow:
Velocity = 5 x 2
Velocity = 10m/s
From the calculation made above, we can see that the velocity of the ocean wave is 10m/s.
Explanation:
(w/w) % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.
1) 100 g of 0.500% (w/w) NaI
Mass of solution = 100 g
Mass of solute = x
Required w/w % of solution = 0.500%
0.500 grams of solute needed to make 100 g of 0.500% (w/w) NaI.
2) 250 g of 0.500% (w/w) NaBr
Mass of solution = 250 g
Mass of solute = x
Required w/w % of solution = 0.500%
1.25 grams of solute needed to make 250 g of 0.500% (w/w) NaBr
3) 500 g of 1.25% (w/w) glucose
Mass of solution = 500 g
Mass of solute = x
Required w/w % of solution = 1.25%
6.25 grams of solute needed to make 500 g of 1.25% (w/w) (glucose)
4) 750 g of 2.00% (w/w) sulfuric acid.
Mass of solution = 750 g
Mass of solute = x
Required w/w % of solution = 2.00%
15.0 grams of solute needed to make 750 g of 2.00% (w/w) sulfuric acid.
