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laila [671]
3 years ago
10

The atomic number of carbon is 6. how many electrons does it contain in its outer valence shell?

Chemistry
1 answer:
Oksi-84 [34.3K]3 years ago
6 0
There are 6 electrons in the atom of a carbon.
The first atomic level will be occupied by 2 electrons while the second will be occupied by 4 electrons.

Based on this, the outer valence shell of carbon (which is L shell) will be occupied by 4 electrons.
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200mL of 4.98M of sodium chloride solution is added to an additional 532 ml of water what is the final molarity?
ELEN [110]

Answer:

M₂ = 1.9 M

Explanation:

Given data;

Volume of sodium chloride = 200 mL

Molarity of sodium chloride = 4.98 M

Volume of water = 532 mL

Final Molarity = ?

Solution:

M₁V₁ = M₂V₂

M₂ =  M₁V₁ /V₂

M₂ = 4.98 M × 200 mL / 532 mL

M₂ = 996 mL. M /532 mL

M₂ = 1.9 M

4 0
3 years ago
Proteins are large molecules. They enter a cell through
Lady bird [3.3K]

The answer is; active transport


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7 0
3 years ago
Why is the deuterium-tritium reaction the most promising nuclear fusion reaction for future energy production?
stepan [7]
The reactions produces an enormous amount of energy per unit mass compared to nuclear fission. Hope this helped
6 0
3 years ago
Read 2 more answers
The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be
Stels [109]

Explanation:

1) Initial mass of the Cesium-137=N_o= 180 mg

Mass of Cesium after time t = N

Formula used :

N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}

Half life of the cesium-137 = t_{1/2}=30 years[p/tex]where,
[tex]N_o = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

t_{\frac{1}{2}} = half life of the isotope

\lambda = rate constant

N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}

Now put all the given values in this formula, we get

N=180mg\times e^{-\frac{0.693}{30 years}\times t}

Mass that remains after t years.

N=180 mg\times e^{0.0231 year^{-1}\times t}

Therefore, the parent isotope remain after one half life will be, 100 grams.

2)

t = 70 years

N_o=180 mg

t_{1/2}= 30 yeras

N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

3)

N_o=180 mg

t_{1/2}= 30 yeras

N = 1 mg

t = ?

1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}

\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0
3 years ago
Can anyone answer this?? Thanks in advance <br><br> AlCl3 + Cu = ?
Kazeer [188]
Reaction won't occur
Because the activity of Cu is less than Al
3 0
3 years ago
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