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
<u>a) Answer: </u>
<em>Number of molecules in 1 mole</em>
<u>Explanation:</u>
a) Whether we take any of the substance among all three of the given substances they will have the same number of molecules in 1 mole of the substance is considered and the value for this will be 
<u>b) Answer: </u>
<em>In the given question </em><em>mass of the substance</em><em> which is </em><em>greatest</em><em> is asked for </em><em>one mole</em><em> and we also know that </em><em>mass of one mole is given by molar mass. </em>
<u>Explanation:</u>
b) It is known that 
is the molar mass for oxygen which is greater than that of hydrogen while fluorine has a molar mass of 
which on comparison shows that, it is the highest amongst all three.
Answer: Objects with like charge repel each other.
Answer:
No, there is no evidence that the manufacturer has a problem with underfilled or overfilled bottles, due that according our results we cannot reject the null hypothesis.
Explanation:
according to this exercise we have the following:
σ^2 =< 0.01 (null hypothesis)
σ^2 > 0.01 (alternative hypothesis)
To solve we can use the chi-square statistical test. To reject or not the hypothesis, we have that the rejection region X^2 > 30.14
Thus:
X^2 = ((n-1) * s^2)/σ^2 = ((20-1)*0.0153)/0.01 = 29.1
Since 29.1 < 30.14, we cannot reject the null hypothesis.
Answer:
The answer is "2%"
Explanation:
Equation:
Formula:
![Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BNO_2%5E%7B-%7D%5D%7D%7B%5BHNO_2%5D%7D)
Let
![[H^{+}] = [NO_2^{-}] = x](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5BNO_2%5E%7B-%7D%5D%20%3D%20x)
at equilibrium
therefore,
![[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%202.0%5Ctimes%2010%5E%7B-2%7D%20%5C%20M%20%3D%200.02%20%5C%20M)
Calculating the % ionization:
![= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%28%5BH%5E%7B%2B%7D%5D%7D%7B%5BHNO_2%5D%29%7D%20%5Ctimes%20100%20%5C%5C%5C%5C%3D%20%5Cfrac%7B0.02%7D%7B1%7D%5Ctimes%20100%20%5C%5C%5C%5C%3D%202%5C%25%5C%5C%5C%5C)
