Answer:
81.97 g of NaAl(OH)₄
Solution:
The reaction for the preparation of Sodium Aluminate from Aluminium Metal and NaOH is as follow,
2 NaOH + 2 Al + 6 H₂O → 2 NaAl(OH)₄ + 3 H₂
According to this equation,
2 Moles of NaOH produces = 163.94 g (2 mole) of NaAl(OH)₄
So,
1 Mole of NaOH will produce = X g of NaAl(OH)₄
Solving for X,
X = (1 mol × 163.94 g) ÷ 2 mol
X = 81.97 g of NaAl(OH)₄
1.Magnesium (Mg) has two electrons in its outer shell. Mg will <span>try to give two electrons to another atom (because the Mg is metal).
</span>2. When an atom loses electrons, it becomes <span>positively charged because its protons outnumber its electrons. ( In an atom, number of protons = number of electrons, when electrons are lost, the number of protons >the number of electrons.)</span>
Answer:
V₂ = 520.42 mL
Explanation:
Given data:
Initial volume = 350.0 mL
Initial pressure = 840 mmHg
Initial temperature = 33°C (33 +273 = 306 K)
Final temperature = 52°C (52+273 = 325 K)
Final volume = ?
Final pressure = 600 mmHg
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 840 mmHg × 350.0 mL × 325 K / 306 K × 600 mmHg
V₂ = 95550000 mmHg.mL.K /183600 K.mmHg
V₂ = 520.42 mL
The corrosion process is a result of the undesired reaction of metals with agents in the environment. The corrosion reaction involves the metal:
becoming reduced
becoming oxidized
<u>becoming the oxidizing agent </u>
becoming more metallic
