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mojhsa [17]
3 years ago
9

What is the balanced equation when aluminum reacts with copper (ii) sulfate in a single displacement reaction?

Chemistry
1 answer:
Vanyuwa [196]3 years ago
3 0
This will be the balanced equation: <span>2Al + 3CuSO4 -> 3Cu + Al2(SO4)3.</span>
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The term “periodic” helps identify _______ that occur in the table, like electronegativity or atomic radii.
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Explanation:

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Express 749 000 000 in scientific notation,
Dima020 [189]

Answer:

7.49 × 108

Explanation:

Scientific notation is a way to express numbers in a form that makes numbers that are too small or too large more convenient to write. It is commonly used in mathematics, engineering, and science, as it can help simplify arithmetic operations. In scientific notation, numbers are written as a base, b, referred to as the significant, multiplied by 10 raised to an integer exponent, n, which is referred to as the order of magnitude:

8 0
3 years ago
20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc
bezimeni [28]

Answer:

Concentration of the barium ions  = [Ba^{2+}] = 0.4654 M

Concentration of the chloride ions  = [Cl^{-}]=0.9308 M

Explanation:

Moles (n)=Molarity(M)\times Volume (L)

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride  = 0.1355 M

n=0.1355 M\times  0.04389 L=0.005947 mol

Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

\frac{1}{2}\times 0.05947 mol=0.029735 mol barium hydroxide

Moles of barium hydroxide = 0.029735 mol

Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

=1\times 0.029735 mol= 0.029735 mol

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL  = 63.89 mL = 0.06389 L

Concentration of the barium ions =[Ba^{2+}]

[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M

Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M

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