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satela [25.4K]
3 years ago
6

What are valence electrons of an atom used for?

Chemistry
1 answer:
Murrr4er [49]3 years ago
5 0

In chemistry and physics, a valence electron is an outer shell electron that is associated with an atom, and that can participate in the formation of a chemical bond if the outer shell is not closed; in a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair.
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g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the
weqwewe [10]

Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

5 0
2 years ago
A sample of 10K gold contains the following: 10.0g gold, 4.0g silver, 5.0g copper, and 5.0g nickel. What is the percent gold in
stellarik [79]

Answer:

oook

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Explanation:

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6 0
2 years ago
Read 2 more answers
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Lera25 [3.4K]

Answer:

NH3

Explanation:

2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)

So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.

n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol

n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol

From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2

4 0
3 years ago
Hey I need help with this element, compound, and mixtures worksheet
Alika [10]

We should describe a little bit the legend.

A - Element - we should have circles with same color and not bonded together (argon gas).

B - Compound - here we may have circles with same or different color bonded together (water or oxygen which is a diatomic molecule).

C - Mixture of elements - circles with different colors not not bonded together (mixture of noble gases).

D - Mixture of compounds - circles with same or different color bonded together but we should see two or more types of connectivity between circles (mixture of water and ethanol).

E - Mixture of elements and compounds - circles with same or different color bonded together mixed with circles with same color and not bonded together (a mixture between oxygen which is a diatomic molecule and noble gas like argon).

Now we may answer the question:

1) B

2) C

3) D

4) D

5) A

6) B

7) B

8) E

9) E

10) D

11) B

12) D

13) D

14) D

15) D

7 0
3 years ago
What is the molarity of 70.6 g C2H6O in 2.25 L of solution
myrzilka [38]

Hey there!

Molar mass C2H6O = 46.0684 g/mol

Number of moles:

n = mass of solute / molar mass

n = 70.6 / 46.0684

n = 1.532 moles

Therefore:

M = number of moles / volume ( L )

M = 1.532 / 2.25

= 0.680 M

Hope that helps!

7 0
3 years ago
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