1. What do they have in common?
As mentioned in the problem, these gases are present in equal amounts. So, that would infer that they are common in terms of their mass. Also, it is specified that the temperature is 25°C. Connected to that is the average kinetic energy, which is directly proportional. Hence, they are also common in temperature and average kinetic energy.
2. What are the differences?
They differ in type, of course. Also, they differ in average velocities which is a factor of temperature of molar mass. Since they are 3 different types of gases with different molar masses, they would also differ in their average velocities.
Answer:
Molarity of NaOH = 1.8 M.
Explanation:
From the question given above, the following data were obtained:
Mass of NaOH = 36 g
Molar mass of NaOH = 40 g/mol
Volume = 500 mL
Molarity of NaOH =?
Next, we shall determine the number of mole in 36 g of NaOH. This can be obtained as follow:
Mass of NaOH = 36 g
Molar mass of NaOH = 40 g/mol
Mole of NaOH =?
Mole = mass / molar mass
Mole of NaOH = 36 / 40
Mole of NaOH = 0.9 mole
Next, we shall convert 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Finally, we shall determine the molarity of NaOH. This can be obtained as follow:
Mole of NaOH = 0.9 mole
Volume = 0.5 L
Molarity of NaOH =?
Molarity = mole / Volume
Molarity of NaOH = 0.9 / 0.5
Molarity of NaOH = 1.8 M
Answer:
Explanation:
q= mc theta
where,
Q = heat gained
m = mass of the substance = 670g
c = heat capacity of water= 4.1 J/g°C
theta =Change in temperature=(
66-25.7)
Now put all the given values in the above formula, we get the amount of heat needed.
q= mctheta
q=670*4.1*(66-25.7)
=670*4.1*40.3
=110704.1
Lets organise the data given in the question
[ClO₂] (m) [OH⁻] (m) initial rate (m/s)
<span>0.060 0.030 0.0248
</span><span> 0.020 0.030 0.00276
</span><span> 0.020 0.090 0.00828
rate equation as follows
rate = k [</span>ClO₂]ᵃ [OH⁻]ᵇ
where k - rate constant
we need to find order with respect to ClO₂ therefore lets take the 2 equations where OH⁻ is constant.
1) 0.00276 = k [0.020]ᵃ[0.030]ᵇ
2) 0.0248 = k [0.060]ᵃ[0.030]ᵇ
divide first equation from the second
0.0248/0.00276 = [0.060/0.020]ᵇ
8.99 = 3ᵇ
8.99 rounded off to 9
9 = 3ᵇ
b = 2
order with respect to ClO₂ is 2
