Answer:
B. A precipitate will form since Q > Ksp for calcium oxalate
Explanation:
Ksp of CaC₂O₄ is:
CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻
Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:
Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹
In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.
Replacing in Ksp formula:
[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.
If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.
Thus, right answer is:
<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>
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Carbon source is the answer for ur question
Hi Lara
The answer is : C
Is equal to
I hope that's help!
If you would draw the Lewis structures of these atoms, you would see that A has 2 electron pairs and 2 lone electrons (that can bond). For B you’d see that you only have 1 electron that can form a bond. This means that 1 atom of A (2 lone electrons) can bond with 2 atoms of B. To know the kind of bond you have to know wether or not there will be a ‘donation’ of an electron from one atom to another. This happens when the number of electrons on one atoms is equal to the number of electrons another atom needs to reach the noble gas structure. As you can see, this is not the case here. This means that you get an AB2 structure with covalent character.
Answer:
133 g
Explanation:
Step 1: Write the balanced equation
2 Al(s) + 3 Br₂(l) ⇒ 2 AlBr₃(s)
Step 2: Calculate the moles corresponding to 15.0 g of Al
The molar mass of aluminum is 26.98 g/mol. The moles corresponding to 15.0 g of Al are:
Step 3: Calculate the moles of Br₂ that react with 0.556 moles of Al
The molar ratio of Al to Br₂ is 2:3. The moles of bromine that react with 0.556 moles of aluminum are:
Step 4: Calculate the mass corresponding to 0.834 moles of Br₂
The molar mass of bromine is 159.81 g/mol. The mass corresponding to 0.834 moles of Br₂ is:
