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maksim [4K]
2 years ago
11

If 280cm3 of hydrogen diffuses in 40secs, how long will it take for 490cm3 of a gas x, whose vapor density is 25 to diffuse unde

r the same condition (RMM hydrogen =2)​
Chemistry
1 answer:
BaLLatris [955]2 years ago
6 0

The time taken by gas x : 350 s

<h3>Further explanation  </h3>

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or  

the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

Molar mass = 2 x vapour density(VD)

r₁=rate Hydrogen = 280 cm³ / 40 s = 7 cm³/s

M₁=molar mass Hydrogen=2 g/mol

M₂=molar mass gas x=

\tt 2\times VD=2\times 25=50~g/mol

\tt \dfrac{7}{r_2}=\sqrt{\dfrac{50}{2} }\\\\\dfrac{7}{r_2}=5\\\\r_2=1.4~cm^3/s

\tt r_2=\dfrac{V_2}{t_2}\\\\t_2=\dfrac{V_2}{r_2}\\\\t_2=\dfrac{490~cm^3}{1.4~cm^3/s}\\\\t_2=350~s

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Buck's Turf Care Company mowed 233 lawns over a 11 week period. What is the average weekly rate of mowing lawns?
masya89 [10]

Answer: 21 lawns per week

Explanation:

The average weekly rate refers to how many lawns were mowed per week given that 233 were done in 11 weeks.

Rate will be given by;

= Lawns mowed / Weeks taken

= 233 / 11

= 21 lawns per week

6 0
3 years ago
a thermometer containing 8.3g of mercury has broken. if mercury ha a density of 13.6g/mL. what volume is spilled?
scoray [572]
D = m / V

13.6 = 8.3 / V

V = 8.3 / 13.6

V = 0.610 mL

hope this helps!
4 0
3 years ago
Read 2 more answers
Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

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