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hodyreva [135]
3 years ago
10

Identify the outer electron configurations for the (a) alkali metals, (b) alkaline earth metals, (c) halogens, (d) noble gases.

(a) Alkali metal: ns,1 ns2, ns2np1, ns2np2, ns2np3, ns2np4, ns2np5, ns2np6. (b) Alkaline earth metals: ns1, ns2, ns2np1, ns2np2, ns2np3, ns2np4, ns2np5, ns2np6. (c) Halogens: ns1, ns2, ns2np1, ns2np2, ns2np3, ns2np4, ns2np5, ns2np6. (d) Noble gases: ns1, ns2, ns2np1, ns2np2, ns2np3, ns2np4, ns2np5, ns2np6.
Chemistry
1 answer:
ladessa [460]3 years ago
6 0

Answer:

Explanation:

Alkali metals   ------ outermost orbit containing one electron

ns²np¹

Alkaline metals -------- outermost orbit containing two electron

ns²np²

halogens --------------- outermost orbit containing seven electron

ns²np⁵

noble gas --------------- outermost orbit containing eight electron

ns²np⁶.

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When 0.50 liter of a 12m solution is diluted to 1.0 liters, the molarity of the new solution is ____
Akimi4 [234]

Answer:

Final molarity  = 6 M

Explanation:

Given data:

Initial volume = 0.50 L

Initial molarity = 12 M

Final volume = 1 L

Final molarity = ?

Solution:

Formula:

M₁V₁ = M₂V₂

M₁ = Initial molarity

V₁ = Initial volume

M₂ = Final molarity

V₂ = Final volume

Now we will put the values.

12 M × 0.50 L = M₂ ×  1 L

6 M.L = M₂ ×  1 L

M₂ = 6 M.L / 1L

M₂ = 6 M

6 0
4 years ago
Caustic soda is 19.1 M NaOH and is diluted for household use. What is the household concentration if 10 mL of the concentrated s
olga nikolaevna [1]
Hello!

To calculate the household concentration of NaOH we need to use the dilution formula, clearing for M2, as you can see in the equation below:

M1*V1=M2*V2 \\ \\ V2= \frac{M1*V1}{V2}

Now, we input the values from the data we have onto this equation. M1=19,1 M; V1=10 mL; V2=400 mL, and solve the equation to get the result:

V2= \frac{M1*V1}{V2}= \frac{19,1M*10mL}{400 mL}=0,48M

So, the household concentration of NaOH will be 0,48 M

Have a nice day!
4 0
4 years ago
Which sample of gas at STP has the same number of molecules as 6 liters of Cl2(g) at STP?
lara [203]

Answer:

The correct answer is option 2  (6 liters of N2)

Explanation:

The complete question

Which sample of gas at STP has the same number  of molecules as 6 liters of Cl2(g) at STP?

(1) 3 liters of O2(g)

(2) 6 liters of N2(g)

(3) 3 moles of O2(g)

(4) 6 moles of N2(g)

Step 1: Data given

Volume = 6 L

STP = 1 atm and 273 K

Step 2: Calculate moles of Cl2

p*V = n*R*T

n = (p*V)/ (R*T)

⇒with n = the number of moles Cl2 = TO BE DETERMINED

⇒with V = the volume of Cl2 = 6.0 L

⇒with p = the pressure of Cl2 = 1 atm

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 273 K

n = (1.0 * 6.0 ) (0.08206 * 273)

n = 0.2678 moles Cl2

This means we need 0.2678 moles of a gas at STP

Option 3 has 3 moles of O2 ⇒ Not the same number of molecules

Option 4 has 6 moles of N2  ⇒ Not the same number of molecules

For 3 liters of O2 we'll have:

⇒with n = the number of moles O2 = TO BE DETERMINED

⇒with V = the volume of O2 = 3.0 L

⇒with p = the pressure of O2 = 1 atm

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 273 K

n = (1.0 * 3.0 ) (0.08206 * 273)

n = 0,1339 moles ⇒ Not the same number of molecules

For 6 liters of N2 we'll have

⇒with n = the number of moles N2 = TO BE DETERMINED

⇒with V = the volume of N2 = 6.0 L

⇒with p = the pressure of N2 = 1 atm

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 273 K

n = (1.0 * 6.0 ) (0.08206 * 273)

n = 0.2678 moles N2 ⇒ The same number of molecules

The correct answer is option 2  (6 liters of N2)

3 0
3 years ago
A large hamburger sandwich contains 628 kcal and 36 grams of fat. Approximately what percentage of the total energy is contribut
Klio2033 [76]

The percentage of the total energy contributed by fat = 51.59% of fat

<h3>Calculation of total energy in fat</h3>

The quantity of fat in the hamburger = 36grams

But 1 gram of fat = 9 kcal( standard energy value per gram of fat).

That is , 9kcal = 1 gram

X kcal = 36 grams

cross multiply,

X kcal = 9 × 36

= 324kcal

To calculate the percentage of 324kcal of fat in 628 kcal,

\% =  \frac{324}{628}  \times  \frac{100}{1}

\% =  \frac{32400}{628}

% = 51.59% of fat

The percentage of the total energy contributed by fat = 51.59%

Learn more about fats here:

brainly.com/question/1601509

3 0
2 years ago
Which of the following describes an impact of the specific heat of water on the planet? Islands and coastal places have moderate
den301095 [7]

Answer:

Islands and coastal places have moderate pleasant climates.

Explanation:

The specific heat of water is very large as compared to most of the common substances. The result of this characteristic is that water heats slowly and cools slowly. In coastal region and islands, the air above lands gets heated up quickly as compared to air above water. The hot air is lighter and rises. The cool air above water rushes to takes its place. This is known as sea breeze. At night, the wind blows from land towards the sea as land cools down faster than water. land breeze occurs at night. Due to sea breeze and land breeze, moderate pleasant climates characterizes the islands and coastal regions.

3 0
3 years ago
Read 2 more answers
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