Answer:
1.67mol/L
Explanation:
Data obtained from the question include:
Mole of solute (K2CO3) = 5.51 moles
Volume of solution = 3.30 L
Molarity =?
Molarity is simply the mole of solute per unit litre of the solution. It can be expressed mathematically as:
Molarity = mole of solute /Volume of solution
Molarity = 5.51 mol/3.30 L
Molarity = 1.67mol/L
Therefore, the molarity of K2CO3 is 1.67mol/L
<u>Answer:</u> The value of 
for the given reaction is 1.435
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
Given mass of 
= 9.2 g
Molar mass of = 92 g/mol
Volume of solution = 0.50 L
Putting values in above equation, we get:
For the given chemical equation:
<u>Initial:</u> 0.20
<u>At eqllm:</u> 0.20-x 2x
We are given:
Equilibrium concentration of = 0.057
Evaluating the value of 'x'
The expression of for above equation follows:
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![[NO_2]_{eq}=2x=(2\times 0.143)=0.286M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.143%29%3D0.286M)
![[N_2O_4]_{eq}=0.057M](https://tex.z-dn.net/?f=%5BN_2O_4%5D_%7Beq%7D%3D0.057M)
Putting values in above expression, we get:
Hence, the value of for the given reaction is 1.435
Recycling!! And also planting more trees.
Hope this helps!
In order to <span>decrease the pressure of a gas inside a closed cubical container, you need to decrease the temperature of the container. The volume of the system is rigid so it means volume is constant. By the ideal gas law, temperature and pressure are directly related. Increasing the temperature, increases the pressure and the opposite to happens.</span>
Answer:
Q = 1360.248 j
Explanation:
Given data:
Mass of brass = 298.3 g
Initial temperature = 30.0°C
Final temperature = 150°C
Specific heat capacity of brass = 0.038 J/g.°C
Heat absorbed = ?
SOLUTION:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 150°C - 30.0°C
ΔT = 120°C
Q = 298.3 g × 0.038 J/g.°C × 120°C
Q = 1360.248 j
