The empirical formula of the alcohol is C₄H₄O
To solve the question given above, we'll begin by calculating the percentage of oxygen in the compound. This can be obtained as follow:
Carbon (C) = 70.57 %
Hydrogen (H) = 5.935 %
<h3>Oxygen (O) =? </h3>
O = 100 – (C + H)
O = 100 – (70.57 + 5.935)
O = 100 – 76.505
<h3>O = 23.495%</h3>
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Carbon (C) = 70.57 %
Hydrogen (H) = 5.935 %
Oxygen (O) = 23.495%
<h3>Empirical formula =?</h3>
Divide by their molar mass
C = 70.57 / 12 = 5.881
H = 5.935 /1 = 5.935
O = 23.495 / 16 = 1.468
Divide by the smallest
C = 5.881 / 1.468 = 4
H = 5.935 / 1.468 = 4
O = 1.468 / 1.468 = 1
Therefore, the empirical formula of the alcohol is C₄H₄O
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Answer:
A 0.001 molar solution of weak acid has a pH = 4.0,The pKa is :
B. 5
Explanation:
pH = It is negative logarithm of hydrogen ion concentration.
pH = -log([H+])
Ka = It is the dissociation constant.
pKa = -log(Ka)
<u>The acid dissociates according to the equation given below:</u>
Here , The concentration of A- and H+ will be same
So,
[H+] = [A-].....(1)
Now ,
pH = 4.0
pH = -log([H+])
4.0 = -log([H+])
- 4.0 = log([H+]) take antilog both side::"use scientific calculator"
From equation (1), [A-] =
[HA] = 0.001 M (given)
Insert the value of HA , H+ and A- in Ka
pKa = -log(Ka)
pKa = 5
The answer is <span>one liter.
</span>
The volume of the cube (V) is:
V = l · w · h (l - length, w - width, h - height)
It is known:
l = 10 cm
w = 10 cm
h = 10 cm
Therefore:
V = 10 · 10 · 10 = 1000 cm³
1 cm³ is equivalent to 0.001 liter, therefore 1000 cm³ is:
1 cm³ : 0.001 l = 1000 cm³ : x
x = 0.001 l · 1000 cm³ ÷ 1 cm³ = 1 l