Answer:
12.02 g
Explanation:
From the question given above, the following data were obtained:
Half life (t½) = 2 days
Original amount (N₀) = 96 g
Time (t) = 6 days
Amount remaining (N) =..?
Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:
Half life (t½) = 2 days
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 2
K = 0.3465 /day
Therefore, the rate of disintegration of the isotope is 0.3465 /day.
Finally, we shall determine the amount of the isotope remaining after 6 days as follow:
Original amount (N₀) = 96 g
Time (t) = 6 days
Decay constant (K) = 0.3465 /day.
Amount remaining (N) =.?
Log (N₀/N) = kt / 2.303
Log (96/N) = (0.3465 × 6) / 2.303
Log (96/N) = 2.079/2.303
Log (96/N) = 0.9027
Take the anti log of 0.9027
96/N = anti log (0.9027)
96/N = 7.99
Cross multiply
96 = N × 7.99
Divide both side by 7.99
N = 96 /7.99
N = 12.02 g
Therefore, the amount of the isotope remaining after 6 days is 12.02 g
The change in pH is calculated by:
pOH = Protein kinase B + log [NH4+]/ [NH3]
Protein kinase B of ammonia = 4.74
initial potential of oxygen hydroxide= 4.74 + log 0.100/0.100 = 4.74
pH = 14 - 4.74=9.26
moles NH4+ = moles NH3 = 0.100 L x 0.100 M = 0.0100
moles H+ added = 3.00 x 10^-3 L x 0.100 M=0.000300
NH3 + H+ = NH4+
moles NH3 = 0.0100 - 0.000300=0.00970
moles NH4+ = 0.0100 + 0.000300=0.0103
pOH = 4.74 + log 0.0103/ 0.00970= 4.77
oH = 14 - 4.77 = 9.23
the change is = 9.26 - 9.23 =0.03
A big difference is that for the Greeks philosophy was almost a fresh start. For us, doing philosophy cannot avoid taking into consideration what the great thinkers of the past have thought (or how they have thought).
The hydronium and hydroxide concentrations of a solution that is 5.0 x 10-3 M H2SO4 is 2.7.
pH= -log[H+] - (i)
10^-3=H2So4
H+= 2×10-3
here ,
h2so4 ——— 2[H+] + so4^2-
thus [H+]= 2*10^(-3) because hydrogen ion has two moles
pH= -log[H+]
pH= -log(2×10^-3)
pH= 3-log2
pH= 3-log2pH= 2.7
The pH is 2.7
<h3>What is pH?</h3>
PH is the degree of alkalinity and acidicity in a solution.
Therefore, The hydronium and hydroxide concentrations of a solution that is 5.0 x 10-3 M H2SO4 is 2.7
Learn more about pH from the link below.
https://brainly.in/question/9937410
Answer:
1.) 0.1 M
2.) 0.2 M
3.) 1 M
4.) Solution #3 is the most concentrated because it has the highest molarity. This solution has the largest solute to solvent ratio. The more solvent there is, the lower the concentration and molarity.
Explanation:
To find the molarity, you need to (1) convert grams NaOH to moles (via molar mass from periodic table) and then (2) calculate the molarity (via the molarity equation). All of the answers should have 1 sig fig to match the given values.
Molar Mass (NaOH): 22.99 g/mol + 16.00 g/mol + 1.008 g/mol
Molar Mass (NaOH): 39.998 g/mol
4 grams NaOH 1 mole
---------------------- x ------------------ = 0.1 moles NaOH
39.998 g
1.)
Molarity = moles / volume (L)
Molarity = (0.1 moles) / (1 L)
Molarity = 0.1 M
2.)
Molarity = moles / volume (L)
Molarity = (0.1 moles) / (0.5 L)
Molarity = 0.2 M
3.)
Molarity = moles / volume (L)
Molarity = (0.1 moles) / (0.1 L)
Molarity = 1 M
