Answer:
Number of moles of sodium reacted = 0.707 moles
Explanation:
P(H₂) = P(T) – P(H₂O)
P(H₂) = 754 – 17.5 = 736.5 mm Hg
Use the ideal gas equation which
PV= nRT, where P is the pressure V is the volume, n is the number of moles R is the Gas Constant and T is temperature
<u>Re- arrange to calculate the number of moles and using the data provided</u>
n = P x V/R x T
n =736.5 x 8.77/62.36367 x (mmHg/mol K) x (20 + 273)
n = 0.35348668
n = 0.353 moles H₂
<u>from the equation we know that</u>
0.353 mole H₂ x 2mole Na/1mole H₂, So
0.353 x 2 = 0.707 mole Na
The number of moles of Sodium metal reacted were 0.707 moles.
Answer:
50 ml (5x TBE) + 540 ml (water)
Explanation:
To prepare 0.5x TBE solution from 5x TBE solution we need to use the following dilution formula:
C1 x V1 = C2 x V2, where:
- C1, V1 = Concentration/amount (start), and Volume (start)
- C2, V2 = Concentration/amount (final), and Volume (final)
* So when we applied this formula it will be:
5 x V1 = 0.5 x 500
V1= 50ml
- To prepare 0.5x we will take 50ml from 5x and completed with 450ml water and the final volume will going to be 500ml.
Covalent and ionic bonds are two different types of chemical bonding. Covalent bonds involve the sharing of electrons between 2 atoms while ionic bonds involve the complete transferring of electrons from one atom to another. Covalent bonds usually form between two nonmetals while ionic bonds usually form between a metal and a nonmetal.
I hope this helps. Let me know if anything is unclear.
Answer:
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.
Explanation:
Moles of NaOH = 
Molarity of the nitric acid solution = 0.250 M
Volume of the nitric solution = 0.150 L
Moles of nitric acid = n
According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :
of NaOH
Moles of NaOH left unreacted in the solution =
= 0.375 mol - 0.0375 mol = 0.3375 mol
1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.
Then 0.3375 moles of NaOH will give :
of hydroxide ion
The molarity of hydroxide ion in solution ;
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.
40×19.32/100=7.7=8×2=16Ca
35.5×34.30/100=12.1=12×2=24Cl
16×46.38/100=7.4=7×2=14O
