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nekit [7.7K]
3 years ago
15

Which type of volcano forms when eruptions contain mostly dush ashes and cinders?

Chemistry
1 answer:
Alisiya [41]3 years ago
6 0
Stratovolcanoes, or composite volcanoes, mostly erupt dust, ash, and cinders.
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How do polar molecules respond when they are placed between two metal plates, one positively charged and one negatively charged?
kakasveta [241]

Gradpoint

The molecules align to the electric field

5 0
3 years ago
Read 2 more answers
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
If the lungs of a child hold 0.11 mol of air in a volume of 2.8 L, then the lungs of an average female adult, with a volume is 4
Aleks [24]

Answer:

Explanation:

A childs lung can hold .11mols/ per 2.8 L so that gives us a molarity of .039M

A adults lungs can hold .18 mols /per 4.6 so that gives us .039M aswell meaining that the lung capacity between the two is not different.

4 0
2 years ago
Relate what you have learned about the word quantum to the Bohr model.​
ss7ja [257]
I forgot what quantum means to be honest, the Bohr model In atomic physics, the Bohr model or Rutherford–Bohr model, presented by Niels Bohr and Ernest Rutherford in 1913, is a system consisting of a small, dense nucleus surrounded by orbiting electrons—similar to the structure of the Solar System, but with attraction provided by electrostatic forces in place of gravity. After the cubical model (1902), the plum pudding model (1904), the Saturnian model (1904), and the Rutherford model (1911) came the Rutherford–Bohr model or just Bohr model for short (1913). The improvement over the 1911 Rutherford model mainly concerned the new quantum physical interpretation.
7 0
2 years ago
A mole equals 6.02 x 10^23 . Answer these questions below.
siniylev [52]

Answer:

1. 1.25 mol ants x 6.02*10^23 ants/1 mol ants = 7.53*10^23 ants

2. 4.92*10^26 pencils x 1 mol pencils/6.02*10^23 pencils = 817 mol pencils

3. 0.26 mol molecules x 6.02*10^23 molecules/1 mol molecules = 1.6*10^23 molecules

4. 3.46*10^19 molecules x 1 mol molecules/6.02*10^23 molecules = 5.75*10^-5 mol molecules

5. 5.3*10^20 atoms x 1 mol atoms/6.02*10^23 atoms = 8.8 mol atoms

6. 0.11 mol atoms x 6.02*10^23 atoms/1 mol atoms = 6.6*10^22 atoms

I would suggest looking into "dimensional analysis" for help with this type of material. Dimensional analysis will stick with you all throughout chemistry, so picking it up will be extremely beneficial.

7 0
2 years ago
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