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alina1380 [7]
3 years ago
6

Ted used a balance scale to weigh a 2.233 g sample of copper sulfate. Which of these measurements made by Ted is the most accura

te?
Chemistry
1 answer:
Vika [28.1K]3 years ago
6 0
There’s only one measurement the question doesn’t make sense
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This green solution of chromium(III) can further be reduced by zinc metal to a blue solution of chromium(II) ions. Write the bal
Contact [7]

Answer:

Half-reactions:

Cr³⁺ + 1e⁻ → Cr²⁺; Zn → Zn²⁺ + 2e⁻

Net ionic equation:

2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺

Explanation:

The Cr³⁺ is reduced to Cr²⁺:

<h3>Cr³⁺ + 1e⁻ → Cr²⁺ -Half-reaction 1-</h3>

Zn is oxidized to Zn²⁺:

<h3>Zn → Zn²⁺ + 2e⁻ -Half-reaction 2-</h3>

Twice the reduction of Cr:

2Cr³⁺ + 2e⁻ → 2Cr²⁺

Now this reaction + Oxidation of Zn:

2Cr³⁺ + 2e⁻ + Zn → 2Cr²⁺ + Zn²⁺ + 2e⁻

<h3>2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ - Net ionic equation</h3>

6 0
2 years ago
What is the mass number of an atom which is made up of 27 protons,33 neutrons and 27 electrons?
Gelneren [198K]
27 is the mass number

7 0
3 years ago
Which of the following solutions would have the lowest freezing point? 1.0 nacl 3.0 nacl 2.0 nacl or 5.0 nacl
Scilla [17]
The solution that will have the lowest freezing point is 5.0 SODIUM CHLORIDE. 
Adding solute to solvents usually result in the depression of the freezing point. The higher the quantity of the solute that is added, the lower the freezing point of the solution.
5 0
3 years ago
Read 2 more answers
Is anyone else's brainly not working or is mine the only one that is tweaking?
omeli [17]
Mine tweaking as well
5 0
3 years ago
If one mole of a substance has a mass of 56.0 g, what is the mass of 11 nanomoles of the substance? Express your answer in nanog
9966 [12]

Answer:

616,0 ng is the right answer.

Explanation:

You should know that 1 mole = 1 .10^9 nanomoles

Get the rule of three.

1 .10^9 nanomoles ...................... 56.0 gr

11 nanomoles .....................

(11 x 56) / 1 .10^9 nanomoles = 6.16 x 10^-7 gr

Let's convert

6.16 x 10^-7 gr x 1 .10^9 = 616 ngr

8 0
3 years ago
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