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Anettt [7]
3 years ago
10

What is the identity of the atom shown?

Chemistry
2 answers:
posledela3 years ago
3 0

Answer:

A

Explanation:

Argon has 17 protons and 18 neutron

atomic number is the same as the protons number.

sesenic [268]3 years ago
3 0

Answer:

it a

Explanation:

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Which is a diatomic molecule?<br><br> A)Ar<br><br> B)CO<br><br> C)CO2<br><br> D)NaCl
LUCKY_DIMON [66]

Answer:

C. Co2

Explanation:

Oxygen is a diatomic molecule because 2 atoms are conisistebof a normal molecule.

5 0
3 years ago
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Which of the following symbols represents a chlorine ion with a stable arrangement of eight valence electrons?
frozen [14]

A stable arrangement of eight valence electrons : ³⁵Cl⁻¹

<h3>Further explanation</h3>

Chlorine is a halogen gas, located in group 17, p block

Chlorine has an atomic number of 17 and an atomic mass of 35

Electron configuration: [Ne] 3s²3p⁵

If we look at the electron configuration, then Cl will bind 1 more electron so that the configuration is stable like Argon (atomic number 18)

So by binding this one electron, chlorine forms negative ions (anions)

³⁵Cl⁻¹

B. Cl⁻² binds 2 electrons, exceeding the octet rule

C. Cl⁺¹, releases 1 electron, remains unstable

D. Cl, the neutral form of Cl, is still unstable with a 7-electron valence configuration

3 0
3 years ago
Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

3 0
3 years ago
Scientists must be careful not to use inductive reasoning, because it can lead to faulty conclusions true or false?
LekaFEV [45]
True. I think it's true but I could be wrong
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Where is ocean water the densest?
Vika [28.1K]

Answer:

your answer should be the bottom

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Explanation:

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