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Anton [14]
3 years ago
12

What will most likely happen when stress is applied to an equilibrium reaction? The system will not respond to the stress. The s

ystem will use catalysts to change its equilibrium. The equilibrium position will shift to increase the applied stress. The system will change its concentration to shift to a new equilibrium position.
Chemistry
2 answers:
steposvetlana [31]3 years ago
8 0

Answer:

The system will change its concentration to shift to a new equilibrium position.

Explanation:

For example  in the Haber Process

N2  +  3H2  ⇄  2NH3

If the pressure is increased the  process will move to the right - to have more NH3 and less of the nitrogen and hydrogen.

andrey2020 [161]3 years ago
4 0

Answer:

The system will change its concentration to shift to a new equilibrium position.

Explanation:

According to Le Chatelier’s principle when a stress is applied to a system, the equilibrium shifts to counteract the applied stress, such that a new equilibrium position is formed.  

For example, ice and water is taken in a closed container. Let the temperature of the closed container be raised. The equilibrium of the system will shift to the product side, it will use up the added energy and more water will be formed from the ice.  

ice + energy --> water

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Balance <br><br> _Mg + _HCL = _MgCl2 + H2
Pavel [41]
_Mg + _HCL = _MgCl2 + H2
Separate the terms on each side: 

_Mg + _HCl = _MgCl2 + H2

Mg- 1                 Mg-1
               
H-1                    H-2

Cl-1                    Cl-2

Mg is balanced on both sides so move on to the next (put a 1 in the space). 

1Mg

There are two H's and two Cl's on the results side, so to balance the equation put a 2 as a coefficient for HCl and it'll all balance out.

2HCl

Balamced equation will be: 

 1Mg + 2HCL = 1MgCl2 + H2
5 0
3 years ago
Read 2 more answers
What is the percentage composition when 10g of magnesium combines with 4g of nitrogen?
Ivanshal [37]
The %  composition   when 10g of magnesium combine   with  4g of   nitrogen  is  71.43%   magnesium   and  28.57 %  nitrogen

               calculation

%
  composition  =  mass  of an element  / total mass  x100
mass  of magnesium = 10 g
mass of nitrogen  =  4g

calculate  the  total  mass  used

=  10g of  Magnesium  + 4 g of  nitrogen = 14 grams

%   composition for  magnesium  is therefore  =  10/14  x100 = 71.43 %

%  
composition  for  nitrogen  is therefore = 4 /14  x100  =   28.57 %
7 0
3 years ago
How many grams will be produced
Anni [7]

Answer:

\Large \boxed{\sf 34.2 \ g}

Explanation:

Mole ratio

9:4

1.75:x

Moles of CO₂

\displaystyle \frac{4 \times 1.75}{9} =0.78

Use formula to find mass

\displaystyle moles=\frac{mass}{M_r}

Relative molecular mass of CO₂ = 12 + 16 × 2 = 44

\displaystyle 0.78=\frac{mass}{44}

mass=34.2

4 0
3 years ago
Which is not a type of mass movement
lana66690 [7]

Answer:

options?

Explanation:

3 0
3 years ago
Read 2 more answers
Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa
drek231 [11]

Answer:

1.  the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c (\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} })

Explanation:

To find  -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 117 kPa

⇒Absolute pressure ( p1 )  = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 212 kPa

⇒Absolute pressure (p2)  = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. V_{1} = V_{2}

As we know that, P ∝ M

⇒ \frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }

⇒\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }

⇒\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378 ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒\frac{P}{T} ∝ M

⇒\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }

So, The correct relation is c option.

6 0
3 years ago
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