Answer:
Single Displacement reaction
In a displacement reaction, a more reactive element replaces a less reactive element from a compound.
Change in colour takes place with no precipitate forms.
Metals react with the salt solution of another metal.
Examples:
2KI + Cl2 → 2KCl + I2
CuSO4 + Zn → ZnSO4 + Cu
Double displacement reaction
In a double displacement reaction, two atoms or a group of atoms switch places to form new compounds.
Precipitate is formed.
Salt solutions of two different metals react with each other.
Examples:
Na2SO4 + BaCl2 → BaSO4 + 2NaCl
2KBr + BaCl2 → 2KCl + BaBr2
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Answer : The value of 
at this temperature is 66.7
Explanation : Given,
Pressure of 
at equilibrium = 0.348 atm
Pressure of 
at equilibrium = 0.441 atm
Pressure of 
at equilibrium = 10.24 atm
The balanced equilibrium reaction is,
The expression of equilibrium constant for the reaction will be:
Now put all the values in this expression, we get :
Therefore, the value of at this temperature is 66.7
The number of proton in the nucleus of an atom is its Atomic Number.
Answer:
108 kPa
Step-by-step explanation:
To solve this problem, we can use the <em>Combined Gas Laws</em>:
p₁V₁/T₁ = p₂V₂/T₂ Multiply each side by T₁
p₁V₁ = p₂V₂ × T₁/T₂ Divide each side by V₁
p₁ = p₂ × V₂/V₁ × T₁/T₂
Data:
p₁ = ?; V₁ = 34.3 L; T₁ = 31.5 °C
p₂ = 122.2 kPa; V₂ = 29.2 L; T₂ = 21.0 °C
Calculations:
(a) Convert temperatures to <em>kelvins
</em>
T₁ = (31.5 + 273.15) K = 304.65 K
T₂ = (21.0 + 273.15) K = 294.15 K
(b) Calculate the <em>pressure
</em>
p₁ = 122.2 kPa × (29.2/34.3) × (304.65/294.15)
= 122.2 kPa × 0.8542 × 1.0357
= 108 kPa
