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Mandarinka [93]
2 years ago
12

Name the element in the fourth period of the periodic table with six 4p electrons.

Chemistry
1 answer:
Oxana [17]2 years ago
6 0

Answer:

- Krypton.

- Silicon.

Explanation:

Hello.

In this case, since the first element's last term in the electron configuration is 4p⁶, by looking at the fourth period, the only element having sic 4p electrons is Kr as its electron configuration is:

Kr^{36}:1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^1^0,4p^6

In the second case, the element of the third period having two 3p electrons or 3p^2 the last term in its electron configuration is silicon as shown below:

Si^{14}:1s^2,2s^2,2p^6,3s^2,3p^2

Best regards.

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alina1380 [7]

Answer:

potasium to erbium

Explanation:

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3 years ago
If a car is traveling 100 km/h and comes to a stop in 3 minutes,what is acceleration of a passenger who is using vehicle restrai
Rama09 [41]

We are given –

  • Final velocity of car is, v= 0
  • Initial velocity of car is, u= 100 km/hr
  • Time taken, t is = 3 minutes or 180 sec

Here–

\qquad\pink{\bf \longrightarrow  Initial\:  velocity =  100 \:km/hr}

\qquad\sf \longrightarrow  Initial\:  velocity = \dfrac{ 100 \times 1000}{3600} \:m/s

\qquad\pink{ \bf \longrightarrow  Initial\:  velocity = 27.78\: m/s}

Now –

\qquad____________________________

\qquad\purple{\bf \longrightarrow  Acceleration  = \dfrac{Final\: Velocity -Initial  \:Velocity }{Time}}

\qquad\purple{\bf \longrightarrow  Acceleration  = \dfrac{v -u}{t}}

\qquad\sf \longrightarrow  Acceleration = \dfrac{(0- 27.78)}{1800}

\qquad\sf \longrightarrow  Acceleration =\cancel{ \dfrac{- 27.78}{1800}}

\qquad\purple{\bf \longrightarrow  Acceleration = -0.15 \: m/s^2}

\qquad_______________________________

6 0
2 years ago
Which of the following is true of solids?
Over [174]
<span>Answer: B. Ionic solids have higher melting points than molecular solids.
</span>
This is because the rest are false, as solids are able to melt, and do have melting points. Also, not all solids have the same melting points. 
5 0
2 years ago
The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-
grin007 [14]

Answer:

Total pressure 5.875 atm

Explanation:

The equation for above decomposition is

2N_2O \rightarrow 2N_2 + O_2

rate constant k =  1.94\times 10^{-4} min^{-1}

Half life t_{1/2} = \frac{0.693}{k} = 3572 min

Initial pressure N_2 O = 4.70 atm

Pressure after 3572 min = P

According to first order kinematics

k = \frac{1}{t} ln\frac{4.70}{P}

1.94\times 10^{-4} = \frac{1}{3572} \frac{4.70}{P}

solving for P we get

P = 2.35 atm

2N_2O \rightarrow 2N_2 + O_2

initial           4.70                         0             0

change        -2x                          +2x           +x

final             4.70 -2x                     2x           x

pressure ofO_2 after first half life  = 2.35 = 4.70 - 2x

                                                          x = 1.175

pressure of N_2 after first half life  =  2x = 2(1.175) = 2.35 ATM

Total pressure  = 2.35 + 2.35 + 1.175

                          = 5.875 atm

5 0
3 years ago
Read 2 more answers
1. If I have 3 identical books that all have a length of 25cm, a
Ksenya-84 [330]

Answer:

12,375 cm³

Explanation:

v=l×w×h

Since there are 3 books the equation would then be

v=3(l×w×h)

v=3(25×15×11)

v=3(4,125)

v=12,375 cm³

5 0
3 years ago
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