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sashaice [31]
2 years ago
9

How many grams are in 3.45 moles of potassium oxide?

Chemistry
2 answers:
mote1985 [20]2 years ago
6 0

your answer would be B) 325 g

Aleksandr-060686 [28]2 years ago
6 0

1 mole= 94.196g

94.196g x 3.45g = 324. 9762g

So if you round that you will get 325g

So the answer you seek is B) 325g

Hope this helps. Have a great day.

Brainliest please

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Simplify: <br>(100 m)/(26 s)
podryga [215]

100m/26s=50m/13s

50m/13s=3.846m/s

4 0
2 years ago
Read 2 more answers
Describe a single measurement that you could make that shows that a kettle is less than 100% efficient (providing evidence that
alexgriva [62]

Answer:

As you haven't explained what measurements you took before solving this problem, I will explain the general procedure to evaluate the efficiency of a kettle. I hope it helps you. I´ll send an attachement file with the full answer, since I couldn't write it here.

I assume that the material that is going to be heated in the kettle is water.

1- You have to boil water in it and take the time it takes to its boiling point (in seconds).  

2- You have to evaluate the amount of energy the water absorbed Q with the efficiency formula which I explain in the attachement file.

3- Divide Q by the time it took to bring the water to boiling so you can have the power it consumed.

4- You divide the last value you obtained by the Kettles's power rating.  

5- Multiply the last value by 100 to obtain a percentage value of efficiency.

Explanation:

Efficiency is the ration of a machine's useful work, in this case how much energy the water absorbed to get to its boiling point divided by the time it took to get to this point, and the total energy expended, in this case the kettles's power rating.

7 0
2 years ago
Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such
8_murik_8 [283]

Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

  • 1 A = 1 C/s
  • The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
  • 1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
  • The molar mass of silver is 107.87 g/mol

The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g

5 0
2 years ago
The statement that percent yield can never be greater than theoretical yield is another example of the ________.
Gnom [1K]
We can rephrase the statement with a little more specificity in order to understand the answer here.

The mass of the products can never be more than the The mass that is expected.
3 0
2 years ago
6K + B2O3 → 3K2O + 2B
atroni [7]

Answer:

104.84 moles

Explanation:

Given data:

Moles of Boron produced = ?

Mass of B₂O₃ = 3650 g

Solution:

Chemical equation:

6K + B₂O₃    →    3K₂O + 2B

Number of moles of B₂O₃:

Number of moles = mass/ molar mass

Number of moles = 3650 g/ 69.63 g/mol

Number of moles = 52.42 mol

Now we will compare the moles of  B₂O₃ with B from balance chemical equation:

                 B₂O₃          :          B

                    1              :          2

                52.42         :        2×52.42 = 104.84

Thus from 3650 g of  B₂O₃  104.84 moles of boron will produced.

6 0
2 years ago
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