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Answer:
55.75g
Explanation:
From
m/M = CV
Where
m= required mass of solute
M= molar mass of solute
C= concentration of solution
V= volume of solution=675ml
Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1
Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles
Since 1 mole of Na3PO4 contains 3 moles of Na+
It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles
mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g
Answer:
Moles of NaCl formed is 6.0 moles
Explanation:
We are given the equation;
2 Na(s) + Cl₂(g) → 2 NaCl(s)
- Moles of Na is 6.0 moles
- Moles of Cl₂ is 4.0 moles
From the reaction;
2 moles of sodium reacts with 1 mole of chlorine gas to form 2 moles of NaCl
In this case;
6 moles of Na would require 3 moles of Cl₂, this means that chlorine gas is in excess.
Thus, the rate limiting reagent is sodium.
But, 2 moles of sodium reacts to form 2 moles of NaCl
Therefore;
Moles of NaCl = Moles of Na
= 6.0 moles
Thus, moles of sodium chloride produced is 6.0 moles
Explanation:
3 little bones in ear takes that vibration and empowers them. Then they transfer the vibrations to next part of ear.
Answer:
0.7561 g.
Explanation:
- The hydrogen than can be prepared from Al according to the balanced equation:
<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>
It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.
- Firstly, we need to calculate the no. of moles of (6.8 g) of Al:
no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.
<em>Using cross multiplication:</em>
2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.
0.252 mol of Al need to react → ??? mol of H₂.
∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.
- Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:
mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.
