Answer:
25.157 cm³
Explanation:
Data Given:
Mass of Sugar (m) = 40g
Density of sugar given in literature = 1.59 g/cm³
Volume of Sugar = ?
The formula will be used is
d = m/v ........................................... (1)
where
D is density
m is the mass
v is the volume
So
Rearrange the Equation (1)
d x v = m
v = m/ d ................................................ (2)
put the given values in Equation (2)
v = 40g / 1.59 g/cm³
v = 25.157 cm³
volume of 40 g of sugar = 25.157 cm³
Answer:
The Ka is 9.11 *10^-8
Explanation:
<u>Step 1: </u>Data given
Moles of HX = 0.365
Volume of the solution = 835.0 mL = 0.835 L
pH of the solution = 3.70
<u>Step 2:</u> Calculate molarity of HX
Molarity HX = moles HX / volume solution
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
<u />
<u>Step 3:</u> ICE-chart
[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4
Initial concentration of HX = 0.437 M
Initial concentration of X- and H3O+ = 0M
Since the mole ratio is 1:1; there will react x M
The concentration at the equilibrium is:
[HX] = (0.437 - x)M
[X-] = x M
[H3O+] = 1.995*10^-4 M
Since 0+x = 1.995*10^-4 ⇒ x=1.995*10^-4
[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M
[X-] = x = 1.995*10^-4 M
<u>Step 4: </u>Calculate Ka
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
The Ka is 9.11 *10^-8
Answer:(a) 6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)
(b) 5.55*10^37photons
Explanation:
(a) Here we have to write a balanced thermochemical equation for formation of 1.00 mol of glucose.
In this question it has been given that Chlorophyll absorbs light in the 600 to 700 nm region,
1st we will write a chemical equation for biochemical process of photosynthesis is that CO2 and H2O form glucose (C6H12O6) and O2.
6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)
The heat change off the reaction can be calculated as,
={(1 mol)(6 mol) }- {(6 mol) [H2O]}
=[1 1273.3 kJ + 6(0)] - [6 (-39.5 kJ) + 6 (-285.840 kJ)]
= 2802.74 or 2802.7 kJ
Thus the balanced equation can be written as,
6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g) = 2802.7 kJ for 1.00 mol of glucose.
Answer:
1.36
Explanation:
HClO2 ⇄ H+ + ClO2-
[HClO2] [H+} [ClO2-]
initial: 0.12 0 0
change: -x +x +x
equil: 0.12-x x x
[H+][ClO2-] / [HClO2] = Ka = 1.2 * 10^-2
x* x / 0.12 - x = 1.2 * 10^-2
x^2 = (2.4 * 10^-3) - (1.2 * 10^-2)x
x^2 + (1.2 * 10^-2)x – (2.4 * 10^-3) = 0
Upon solving for x,
x = 0.04335
[H+] = 0.04335
pH = -log[H+]
pH = -log(0.04335) = - (-1.36) = 1.36
