Answer : (C) Hafnium is the most likely identity of the given substance.
Solution : Given,
Mass of given substance (m) = 46.9 g
Volume of given substance (V) = 3.5 
First, find the Density of given substance.
Formula used :
Now,put all the values in this formula, we get
= 13.4 g/
So, we conclude that the density of given substance (13.4 g/) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ respectively).
According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.
Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.
So, Hafnium is the most likely element which is the identity of the given substance.
Answer:
[C₆H₅COO⁻][H₃O⁺]/[C₆H₅COOH] = Ka
Explanation:
The reaction of dissociation of the benzoic acid in water is given by the following equation:
C₆H₅-COOH + H₂O ⇄ C₆H₅-COO⁻ + H₃O⁺ (1)
The dissociation constant of an acid is the measure of the strength of an acid:
HA ⇄ A⁻ + H⁺ (2)
![K_{a} = \frac{[A^{-}][H^{+}]}{[HA]}](https://tex.z-dn.net/?f=%20K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%5BH%5E%7B%2B%7D%5D%7D%7B%5BHA%5D%7D%20)
(3)
<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA]. </em>
So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:
![K_{a} = \frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}](https://tex.z-dn.net/?f=%20K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BC_%7B6%7DH_%7B5%7DCOO%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BC_%7B6%7DH_%7B5%7DCOOH%5D%7D%20)
I hope it helps you!
Answer:
The answer to your question is 0.5 moles
Explanation:
Data
moles of Glucose = ?
moles of carbon dioxide = 3
Balanced chemical reaction
6CO₂ + 6H₂O ⇒ C₆H₁₂O₆ + 6O₂
Process
To solve this problem, use proportions, and cross multiplication.
Use the coefficients of the balanced equation.
6 moles of CO₂ ----------------- 1 mol of C₆H₁₂O₆
3 moles of CO₂ ---------------- x
x = (3 x 1) / 6
-Simplification
x = 3/6
-Result
x = 0.5 moles of Glucose
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