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3 years ago

How many electrons are in p orbitals in Si (Silicon)?

Chemistry

1 answer:

insens350 [35]3 years ago

8 0

Answer:

6 electrons

Explanation:

The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s.

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A 0.227 mol chunk of dry ice (solid CO2) changes to gas. What is the volume of that gas measured at 27 °C and 740 mmHg?

jeka57 [31]

Answer:

Explanation:3.2 ft 2 fti2 ft 4 ft ft2

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2 years ago

You are on a flight to Los Angeles, California. As the plane gets ready to land, it flies out over a large body of water. You lo

sergey [27]

What ocean is next to California on google maps? That is your answer.

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3 years ago

Read 2 more answers

Magnesium combines with oxygen to form magnesium oxide. If 10.57 g of

Phantasy [73]

The total mass is 10.57+6.96=17.53 g.

So, the percent by mass of magnesium is (10.57)/(17.53) * 100 = 60.3%

The percent by mass of oxygen is (6.96)/(17.53) * 100 = 39.7%

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2 years ago

Consider the reaction. 2 HBr(g) ¡ H2(g) + Br2(g) a. Express the rate of the reaction in terms of the change in concentration of

Studentka2010 [4]

Answer :

(A) The rate expression will be:

$Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}$

(B) The average rate of the reaction during this time interval is, 0.00176 M/s

(C) The amount of Br₂ (in moles) formed is, 0.0396 mol

Explanation :

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The given rate of reaction is,

$2HBr(g)\rightarrow H_2(g)+Br_2(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}$

Part A:

The rate expression will be:

$Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}$

Part B:

$\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Average rate}=-\frac{1}{2}\frac{(0.512-0.600)M}{(25.0-0.0)s}$

$\text{Average rate}=0.00176M/s$

The average rate of the reaction during this time interval is, 0.00176 M/s

Part C:

As we are given that the volume of the reaction vessel is 1.50 L.

$\frac{d[Br_2]}{dt}=0.00176M/s$

$\frac{d[Br_2]}{15.0s}=0.00176M/s$

$[Br_2]=0.00176M/s\times 15.0s$

$[Br_2]=0.0264M$

Now we have to determine the amount of Br₂ (in moles).

$\text{Moles of }Br_2=\text{Concentration of }Br_2\times \text{Volume of solution}$

$\text{Moles of }Br_2=0.0264M\times 1.50L$

$\text{Moles of }Br_2=0.0396mol$

The amount of Br₂ (in moles) formed is, 0.0396 mol

8 0

3 years ago

The Ksp for BaCO3 is 5.1x10^-9. how many grams of BaCO3 will dissolve in 1000ml of water

Amanda [17]

Answer:

= 0.014 g of BaCO3

Explanation:

Let x = mol/L of BaCO3 that dissolve.

This will give;

x mol/L Ba2+ and x mol/L CO32-

But;

Ksp = 5.1x10^-9.

Therefore;

Ksp = 5.1 x 10^-9 = (x)(x)

Thus;

x = molar solubility

= √ (5.1 x 10^-9)

= 7.1 x 10^-5 M

Therefore;

Mass BaCO3 = 7.1 x 10^-5 M x 1 L x 197.34 g/mol

= 0.014 g

3 0

3 years ago