All categories

3 years ago

. how many lone pairs of electrons are present in the lewis structure of calcium sulfide?

Chemistry

1 answer:

miv72 [106K]3 years ago

7 0

Answer:
<u>Four</u>

Explanation:
Calcium Sulfide is an ionic compound made up of Ca²⁺ and S²⁻.
Ca²⁺ is formed as,

Ca → Ca²⁺ + 2 e⁻

These two electrons are accepte by Sulfur as,

S + 2 e⁻ → S²⁻

So, before accepting 2 electrons S was having six valence electrons, after accepting two electrons from Ca it has 8 electrons which are present in four pairs as shown below,

You might be interested in

Suppose the volume of a container is decreased, but the same number of gas molecules is present at the same temperature. What wi

charle [14.2K]

Answer: Option (C) is the correct answer.

Explanation:

When the volume of a container is decreased and same number of gas molecules are present inside the container at the same temperature then it means that more number of gas molecules are spread per unit volume.

As earlier when the volume of container was more then the molecules were widely spread inside the container but on decreasing the volume of container, molecules come closer to each other and hence there will be increase in molecules per unit volume.

6 0

2 years ago

Given the following equation: 2 KCIO, +2 KCl + 30, how many grams of O, can be produced by letting 5.90 moles of KCIO, react ?

Finger [1]

Answer:

12.o of 02

Explanation:

8 0

3 years ago

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.

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.

$N_o=180 mg$

$t_{1/2}= 30 yeras$

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

I’m about to die of cuteness overload <br> Naruto, sasuke, sakura, and Kakashi-sensei

Gnoma [55]

Answer:

dfadskjfkhasdljkfasdhfaljksfh so cuttttteeee :3

;w;""'

7 0

2 years ago

Read 2 more answers

A beaker contains 0.125 L of a 3.00 M solution. If the volume goes up to 0.325 L, what is the new molarity?

igor_vitrenko [27]

Answer:

1.15 M

Explanation:

Step 1: Given data

Initial volume (V₁): 0.125 L

Initial concentration (C₁): 3.00 M

Final volume (V₂): 0.325 L

Final concentration (C₂): ?

Step 2: Calculate the final concentration of the solution

We want to prepare a dilute solution from a concentrated one by adding water. We can calculate the concentration of the dilute solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂

C₂ = 3.00 M × 0.125 L/0.325 L = 1.15 M

8 0

2 years ago