Answer:
a) 0.714g of bicarbonate of soda are required.
b) 0.221g of Al(OH)₃ are required
Explanation:
The reactions of HCl with bicarbonate of soda and aluminium hydroxide are:
HCl + NaHCO₃ → H₂O + NaCl + CO₂
3 HCl + Al(OH)₃ → 3H₂O + AlCl₃
The moles of HCl that we need neutralize are:
50mL = 0.050L * (0.17mol / L) = 0.0085 moles HCl
To solve these problem we need to find the moles of the antacid using the chemical reaction and its mass using its molar mass;
<em>a) </em><em>Moles NaHCO₃ = Moles HCl = 0.0085 moles </em>
The mass is -Molar mass NaHCO₃: -84g/mol-
0.0085 moles * (84g / mol) = 0.714g of bicarbonate of soda are required
b) 0.0085 moles HCl * (1mol Al(OH)₃ / 3mol HCl) = 2.83x10⁻³ moles Al(OH)₃
The mass is -Molar mass: 78g/mol-:
2.83x10⁻³ moles Al(OH)₃ * (78g/mol) =
<h3>0.221g of Al(OH)₃ are required</h3>
Answer:
Geothermal energy has a faster rate because a vast amount of heat energy is always present inside Earth's core.
Geothermal energy has a slower rate because the water present below Earth's surface cools down periodically
Explanation:
Answer:
Sunlight
Explanation:
Photosynthesis is the process where by plants manufacture their own food through conversion of carbon(iv)oxide and water in presence to sunlight to produce glucose and oxygen as by product.
The reaction is photo-catalyzed and would only take place in the presence of sunlight.
6CO₂ + 6H₂O + sunlight → C₆H₁₂O₆ + 6O₂
White phosphorus melts and then vaporizes at high temperatures. The gas effuses at a rate that is 0.404 times that of neon in the same apparatus under the same conditions-There are 4 atoms of P in the molecule
Explanation:
Ar=30,97g/mol
/==0,404
0,404=
=20,18/30,97*x
X=20,18/30,97*0,163
X=4
There are 4 atoms of P in the molecule
White phosphorus melts and then vaporizes at high temperatures. The gas effuses at a rate that is 0.404 times that of neon in the same apparatus under the same conditions-There are 4 atoms of P in the molecule
Answer:
0.46 V
Explanation:
The emf for the cell is given by:
Eº cell = Eº oxidation + Eº reduction
From the given balanced chemical equation, we can deduce that Fe²⁺ has been oxidized to Fe³⁺, and O reduced from 0 to negative 2, according to the half cell reactions:
4Fe²⁺ ⇒ Fe³⁺ + 4e⁻ oxidation
O₂ + 4H⁺ + 4 e⁻ ⇒ 2 H₂O reduction
From reference tables for the standard reduction potential, we get
Eº red Fe³⁺ / Fe²⁺ Eºred = 0.77 V
Eº red O₂ / H₂O Eºred = 1.23 V
Now all we need to do is change the sign of Eº reduction for the species being oxidized ( Fe²⁺ ) and add it to Eº reduction O₂:
Eº cell = Eº oxidation + Eº reduction = - (0.77 V ) + 1.23 V = 0.46 V