Answer:
C3H6O2
Explanation:
To find the empirical formula of the compound, we divide the amount in moles of each of the elements by the amount in mole of the element with the smallest number of mole. In this question, the element with the smallest number of moles is oxygen with 1.36 mole. Hence, we divide the number of moles of each element by this.
H = 4.10/1.36 = 3
O = 1.36/1.36 = 1
C = 2.05/1.36 = 1.5
We then multiply through by 2 to yield the compound with the empirical formula C3H6O2
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
When an element losses its electron its called a cation. When an element accepted that electron it called anion. This is called an ionic bond.
the equation is p1 x v1 divided by T1 = p1 x v2 = T2 but since the pressure is kept constant you do not even need it so the equation would now be v1 divided by t1 = v2 divided by t2
2135 cm3 divided by 127 degrees celcius = x divided by 206
answer: 3460 cm3
Answer:
Explanation:
<u>1) Data:</u>
a) Hypochlorous acid = HClO
b) [HClO} = 0.015
c) pH = 4.64
d) pKa = ?
<u>2) Strategy:</u>
With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.
<u>3) Solution:</u>
a) pH
b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)
c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]
d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M
e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M
f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46
