Answer:
234.35 °C
Explanation:
Given data:
Volume of balloon = 125000 mL
Moles of oxygen = 3 mol
Pressure = 1 atm
Temperature = ?
Solution:
Formula:
PV = nRT
P = Pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Volume of balloon = 125000 mL × 1 L /1000 mL
Volume of balloon = 125 L
Now we will put the values:
Ideal gas constant = R = 0.0821 atm.L/mol.K
PV = nRT
T = PV/nR
T = 1 atm × 125 L/ 0.0821 atm.L/mol.K × 3 mol
T= 125 /0.2463 /K
T = 507.5 K
K to °C
507.5 K - 273.15 = 234.35 °C
Answer:
pH = 10.75
Explanation:
To solve this problem, we must find the molarity of [OH⁻]. With the molarity we can find the pOH = -log[OH⁻]
Using the equation:
pH = 14 - pOH
We can find the pH of the solution.
The molarity of Ca(OH)₂ is 2.8x10⁻⁴M, as there are 2 moles of OH⁻ in 1 mole of Ca(OH)₂, the molarity of [OH⁻] is 2*2.8x10⁻⁴M = 5.6x10⁻⁴M
pOH is
pOH = -log 5.6x10⁻⁴M
pOH = 3.25
pH = 14-pOH
<h3>pH = 10.75</h3>
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Answer:
35Cl = 75.9 %
37Cl = 24.1 %
Explanation:
Step 1: Data given
The relative atomic mass of Chlorine = 35.45 amu
Mass of the isotopes:
35Cl = 34.96885269 amu
37Cl = 36.96590258 amu
Step 2: Calculate percentage abundance
35.45 = x*34.96885269 + y*36.96590258
x+y = 1 x = 1-y
35.45 = (1-y)*34.96885269 + y*36.96590258
35.45 = 34.96885269 - 34.96885269y +36.96590258y
0.48114731 = 1,99704989y
y = 0.241 = 24.1 %
35Cl = 34.96885269 amu = 75.9 %
37Cl = 36.96590258 amu = 24.1 %
