Assuming that the initial pH is pHi.
The initial concentration of [OH-] is
[OH]- = 10^(14 - pHi)
Since half of the Na2HPO4 is neutralized, the concentration of [OH-] will be reduced to half. The new pH will be
[OH-]/2 = 10^(14 - pHi)/2 = 10^(14 - pH)
14 - pHi - log 2 = 14 - pH
In terms of pHi,
pH = pHi + log 2
Group I and Group II have the same number of outermost electrons as you go down each group but the shells increase therefore as you go downward it becomes much easier to remove electrons because of its wide radius however group 7 and 6 have seven and six electrons in their outermost shell respectively. Therefore down the group it is much easier to attract electrons and across the period it is much harder because the number of electrons on the outermost shell increase
Answer:
3.94 L
Explanation:
From the question given above, the following data were obtained:
Mass of O₂ = 5.62 g
Volume of O₂ =?
Next, we shall determine the number of mole present in 5.62 g of O₂. This can be obtained as follow:
Mass of O₂ = 5.62 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mole of O₂ =?
Mole = mass / molar mass
Mole of O₂ = 5.62 / 32
Mole of O₂ = 0.176 mole
Finally, we shall determine the volume of 5.62 g (i.e 0.176 mole) of O₂ at STP. This can be obtained as follow:
1 mole of O₂ occupied 22.4 L at STP.
Therefore, 0.176 mole of O₂ will occupy = 0.176 × 22.4 = 3.94 L at STP.
Thus 5.62 g (i.e 0.176 mole) of O₂ occupied 3.94 L at STP
Answer:
12 moles of water will be produced
Explanation:
Given data:
Number of moles of NH₃ = 8.00 mol
Number of moles of O₂ = 14.0 mol
Number of moles of H₂O produced = ?
Solution:
Chemical equation:
4NH₃ + 7O₂ → 4NO₂ +6H₂O
Now we will compare the moles of reactant with product.
NH₃ : H₂O
4 : 6
8 : 6/4×8 = 12
O₂ : H₂O
7 : 6
14 : 6/7×14 = 12
12 moles of water will be produced.
Answer
false
Explanation:
sorry I tbought i put false I hope I helped ....