Answer:
7.98 × 10^3grams.
Explanation:
To find the mass of fluorine in the number of atoms provided, we first divide the number of atoms by Avagadros number (6.02 × 10^23atoms) to get the number of moles in the fluorine atom. That is;
number of moles (n) = number of atoms (nA) ÷ 6.02 × 10^23 atoms
n = 2.542 × 10^26 ÷ 6.02 × 10^23
n = 0.42 × 10^ (26-23)
n = 0.42 × 10^3
n = 4.2 × 10^2moles
Using mole = mass ÷ molar mass
Molar/atomic mass of fluorine (F) = 19g/mol
mass = molar mass × mole
Mass (g) = 19 × 4.2 × 10^2
Mass = 79.8 × 10^2
Mass = 7.98 × 10^3grams.
Answer:
a) heat it from 23.0 to 78.3
q = (50.0 g) (55.3 °C) (2.46 J/g·°C) =
b) boil it at 78.3
(39.3 kJ/mol) (50.0 g / 46.0684 g/mol) =
c) sum up the answers from the two calculations above. Be sure to change the J from the first calc into kJ
Explanation:
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
Learn more about pH of buffer:
brainly.com/question/21881762
C.) Radiant to Electrical
The grams of potassium chlorate that are required to produce 160 g of oxygen is 408.29 grams
<u><em>calculation</em></u>
2 KClO₃→ 2 KCl + 3O₂
Step 1: find the moles of O₂
moles = mass÷ molar mass
from periodic table the molar mass of O₂ = 16 x2 = 32 g/mol
moles = 160 g÷ 32 g/mol = 5 moles
Step2 : use the mole ratio to determine the moles of KClO₃
from equation given KClO₃ : O₂ is 2:3
therefore the v moles of KClO₃ = 5 moles x 2/3 = 3.333 moles
Step 3: find the mass of KClO₃
mass= moles x molar mass
from periodic table the molar mass of KClO₃
= 39 + 35.5 + (16 x3) =122.5 g/mol
mass = 3.333 moles x 122.5 g/mol =408.29 grams
