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3 years ago

Determine the maximum number of electrons in the 2f designation

1 answer:

4 0

Google said

How many electrons fit in each shell around an atom?

The maximum number of electrons that can occupy a specific energy level can be found using the following formula:

Electron Capacity = 2n2

The variable n represents the Principal Quantum Number, the number of the energy level in question.

Energy Level
(Principal Quantum Number) Shell Letter Electron Capacity
1 K 2
2 L 8
3 M 18
4 N 32
5 O 50
6 P 72
Keep in mind that an energy level need not be completely filled before electrons begin to fill the next level. You should always use the Periodic Table of Elements to check an element's electron configuration table if you need to know exactly how many electrons are in each level.

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A buffer is prepared by mixing 50.3 mL of 0.183 M NaOH with 128.8 mL of 0.231 M acetic acid. What is the pH of this buffer

kvv77 [185]

Answer:

A buffer system can be made by mixing a soluble compound that contains the conjugate ... 10.0 grams of sodium acetate in 200.0 mL of 1.00 M acetic acid.

Explanation:

7 0

2 years ago

In a hospital laboratory, a 10.0 mL sample of gastric juice (predominantly HCl), obtained several hours after a meal, was titrat

LiRa [457]

Answer: 1.14

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL$

Putting values in above equation, we get:

$1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M$

To calculate pH of gastric juice:

molarity of $H^+$ = 0.072

$pH=-log[H^+]$

$pH=-log(0.072)=1.14$

Thus the pH of the gastric juice is 1.14

3 0

3 years ago

Is heat a from of kinetic or potential energy?

Leni [432]

Answer:

its made up of both but I would probably say kinetic

Explanation:

3 0

2 years ago

Read 2 more answers

Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3

krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction $(\Delta S^o)$ .

$\Delta S^o=S_{product}-S_{reactant}$

$\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0{(NH_3)}$ = standard entropy of $NH_3$

$\Delta S^0{(H_2)}$ = standard entropy of $H_2$

$\Delta S^0{(N_2)}$ = standard entropy of $N_2$

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]$

$\Delta S^o=-198.3J/K$

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0

2 years ago

When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a

AnnZ [28]

Answer:

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

Explanation:

Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.

H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

We will balance it using the trial and error method.

First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

Finally, we will get the balanced equation by multiplying H₂O by 6.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

3 0

2 years ago