Answer:
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Explanation:
Gold does not tarnish, rust or corrode. Due to its wonderful qualities and its luster, gold is considered the most important metal in jewellery making. As pure gold is too soft for everyday wear, it is alloyed with a mixture of metals in order to make the gold harder, so it can be used for jewellery.
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Answer:
1. 0.02 M
2. 0.01 M
3. 4×10⁻⁶
Explanation:
We know that V₁S₁ = V₂S₂
1.
Concentration of HCl = 0.05 M
end point comes at = 10 ml
So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M
2.
2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)
[Ca²⁺] = 0.02 ÷ 2 = 0.01 M
3.
= [Ca²⁺(aq)] [OH⁻(aq)]²
Ca(OH)₂ (aq) ⇄ Ca²⁺ (aq) + 2OH⁻ (aq)
= [0.01 × (0.02)²] = 4×10⁻⁶
4.
If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.
5.
value describes the solubility of a particular ionic compound. The higher the value, the higher the Solubility will be.
6.
This may be due to uncommon ion effect. The process of other ions (K⁺ or Na⁺) may increase the solubility
Answer:
Mass = 9.58 g
Explanation:
Given data:
Mass of Zn = 2g
Theoretical yield of ZnI₂ = ?
Solution:
Chemical equation:
Zn + I₂ → ZnI₂
Number of moles of Zn:
Number of moles = mass/molar mass
Number of moles = 2g / 65.38 g/mol
Number of moles = 0.03 mol
Now we will compare the moles of Zn and ZnI₂.
Zn : ZnI₂
1 : 1
0.03 : 0.03
Mass of ZnI₂:
Mass = number of moles × molar mass
Mass = 0.03 mol × 319.22 g/mol
Mass = 9.58 g
The balanced equation is 2
AlI
3
(
a
q
)
+
3
Cl
2
(
g
)
→
2
AlCl
3
(
a
q
)
+
3
I
2
(
g
)
.
<u>Explanation:</u>
- Aluminum has a typical oxidation condition of 3+ , and that of iodine is 1- .
Along these lines, three iodides can bond with one aluminum. You get AlI3. For comparable reasons, aluminum chloride is AlCl3.
- Chlorine and iodine both exist normally as diatomic components, so they are Cl2( g ) also, I2( g ), individually. In spite of the fact that I would anticipate that iodine should be a strong.
Balancing the equation, we get:
2AlI
3( aq ) + 3Cl2
( g ) → 2AlCl3
( aq )
+ 3
I
2 ( g )
-
Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the AlCl 3 on the right.
-
Normally, presently we have two Al on the right, so I multiplied the AlI 3 on the left. Hence, I have 6 I on the left, and I needed to significantly increase I 2 on the right.
-
We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.
