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3 years ago

What is the net ionic equation for the reaction that is represented by the following total ionic equation?

Chemistry

2 answers:

xeze [42]3 years ago

7 0

Answer:

The net ionic equation is

2PO₄³⁻ + 3Ca²⁺ → Ca₃(PO₄)₂ (s)

Written in words, the net ionic equation is

2 upper P upper O subscript 4 superscript 3 minus, plus 3 upper C a superscript 2 plus right arrow upper C a subscript 3 (upper P upper O subscript 4) subscript 2.

Option B.

Explanation:

The equation described, written down is

6Na⁺ + 2PO₄³⁻ + 3Ca²⁺ + 6Cl⁻ → 6Na⁺ + 6Cl⁻ + Ca₃(PO₄)₂

To obtain the net ionic equation, we examine the reaction and eliminate the ions that occur on both sides of the equation.

On checking this reaction, we see that 6Na⁺ and 6Cl⁻ are the ions that occur on both sides. They are thus, eliminated.

It becomes clear that the overall reaction is a reaction of aqueous Na₃PO₄ with aqueous CaCl₂ to give aqueous NaCl and insoluble Ca₃(PO₄)₂.

The aqueous compounds exist in ionic form in solutions, hence, the net ionic equation of the double displacement reaction after eliminating the ions that remain in solution before and after the reaction, we obtain

2PO₄³⁻ + 3Ca²⁺ → Ca₃(PO₄)₂ (s)

Better written as

3Ca²⁺ + 2PO₄³⁻ → Ca₃(PO₄)₂ (s)

The net ionic equation written in words is

2 upper P upper O subscript 4 superscript 3 minus, plus 3 upper C a superscript 2 plus right arrow upper C a subscript 3 (upper P upper O subscript 4) subscript 2.

Hope this Helps!!!

Goryan [66]3 years ago

4 0

Answer:

The answer is B. on ed

Explanation:

i took the test and got it right

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3 years ago

Question 9 (1 point)

pantera1 [17]

Answer:

8 moles of C

Explanation:

From the question given above, the following equation was obtained:

3A + 2B —> 6C

From the equation above,

3 moles of A reacted to produce 6 moles of C.

Thus, the number of mole of C produced by reacting 4 moles of A can be obtained as follow:

From the equation above,

3 moles of A reacted to produce 6 moles of C.

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3 years ago

A 7.85 × 10-5 mol sample of copper-61 emits 1.47 × 1019 positrons in 90.0 minutes. what is the decay constant for copper-61

Lera25 [3.4K]

4.14x10^-3 per minute
First, calculate how many atoms of Cu-61 we initially started with by
multiplying the number of moles by Avogadro's number.

7.85x10^-5 * 6.0221409x10^23 = 4.7273806065x10^19
Now calculate how many atoms are left after 90.0 minutes by subtracting the
number of decays (as indicated by the positron emission) from the original
count.
4.7273806065x10^19 - 1.47x10^19 = 3.2573806065x10^19
Determine the percentage of Cu-61 left.
3.2573806065x10^19/4.7273806065x10^19 = 0.6890455577
The formula for decay is:
N = N0 e^(-Î»t)
where
N = amount left after time t
N0 = amount starting with at time 0
Î» = decay constant
t = time
Solving for Î»:
N = N0 e^(-Î»t)
N/N0 = e^(-Î»t)
ln(N/N0) = -Î»t
-ln(N/N0)/t = Î»
Now substitute the known values and solve:
-ln(N/N0)/t = Î»
-ln(0.6890455577)/90m = Î»
0.372447889/90m = Î»
0.372447889/90m = Î»
0.00413830987 1/m = Î»
Rounding to 3 significant figures gives 4.14x10^-3 per minute as the decay
constant.

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3 years ago

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ollegr [7]

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