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MrMuchimi
2 years ago
13

Why does the atomic radius decrease across a period

Chemistry
1 answer:
IgorC [24]2 years ago
4 0
Because the nuclear charge increases across a period and so it has a stronger pull on the outer electrons and will pull in the radius
You might be interested in
Need help on last 3 questions
Olin [163]

Answer:

Explanation:

1)

Given data:

Initial volume of balloon = 0.8 L

Initial temperature = 12°C ( 12+273= 285 K)

Final temperature = 300°C (300+273 = 573 K)

Final volume = ?

Solution:

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 0.8 L .573 K / 285 K

V₂ = 458.4 L / 285

V₂ = 1.61 L

2)

Initial pressure = 204 kpa

Initial temperature = 29°C ( 29 + 273 = 302 K)

Final temperature = ?

Final pressure = 300 kpa

Solution:

P₁/T₁ = P₂/T₂

T₂ = T₁P₂/P₁  

T₂ = 302 K . 300 kpa / 204 kpa

T₂ = 90600 K/ 204

T₂ = 444.12 K

3)

Given data:

Initial volume = 14 L

Initial pressure = 2.1 atm

Initial temperature = 100 K

Final temperature = 450 K

Final volume = ?

Final pressure = 1.2 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 2.1 atm × 14 L × 450 K / 100 K × 1.2 atm  

V₂ = 13230 L / 120

V₂ = 110.25 L

5 0
3 years ago
Three 5-l flasks, fixed with pressure gauges and small valves, each contains 4 g of gas at 273 k. flask a contains h2, flask b c
Varvara68 [4.7K]
First, please check the missing part in your question in the attachment.
a) So first, the Rank of pressure:
according to this formula PV = nRT and when n = m/Mw
PV = m/Mw * R*T
when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater 
when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g
∴ Pressure :
 (A) > (B) > (c)

B) The rank of average molecular kinetic energy:
when K = 3/2 KB T
when K is the average kinetic energy per molecule of gas 
and KB is Boltzmann's constant
and T is the temperature (K)
So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:
∴ A = B = C
C) the rank of diffusion rate after the valve is opened:
according to this formula:
R2/R1 = √M1/M2
from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,
when Mw H2 (A) = 2 g & Mw He(B) = 4 g & CH4 (C) = 16 g
∴ the rank of diffusion:
A > B > C

D) The rank of the Total kinetic energy of the molecules:
when we have the Mw different so it will make the no.of molecules differs as when the Mw is low the no.of molecules will be hight, and when the average molecular kinetic energy equals. so the total kinetic energy will depend on no. of molecules 
∵ Mw A < Mw B < Mw C 
∴no .of molecules of A > B >C
∴ the rank of total kinetic energy is:
A > B > C

e) the rank of density:

when ρ = m/ v 
and m is the mass & v is the volume and we have both is the same for A, B, and C
so the density also will be the same, ∴ the rank of the density is:
A = B = C

F) the rank of the collision frequency:
as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.
∴ Collision frequency will only depend on the no.of molecules
we have no.of molecules of A > B > C as Mw A < B < C 
∴the rank of the collision frequency is:
A > B > C 

 



7 0
2 years ago
Annabelle was explaining the carbon cycle to her friend. She said that all the carbon
vagabundo [1.1K]

Answer: O2+6H12O6=CO2+ENERGY(ATP)

I DON'T THINK SHE IS CORRECT

Explanation:

5 0
3 years ago
At standard temperature and pressure. 0.500 mole of xenon gas occupies
ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Mole ratio
  • Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

0.500 mole Xe (g)

<u>Step 2: Convert</u>

  1. [DA] Set up:                                                                                                  \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)
  2. [DA] Evaluate:                                                                                               \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}

Topic: AP Chemistry

Unit: Stoichiometry

3 0
2 years ago
When are conservation efforts most effective? a. Conservation efforts are most effective when multiple groups cooperate. b. Cons
alexandr402 [8]
Conservation efforts most effective if <span>Conservation efforts are most effective when multiple groups cooperate. The answer is letter A. The rest of the choices do not answer the question above</span>
7 0
3 years ago
Read 2 more answers
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