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TEA [102]
3 years ago
6

If the relative rate of diffusion of ozone as compared to chlorine is 6:3 and further if the density of chlorine is 36 Find out

the density of ozone.​
Chemistry
1 answer:
Alexandra [31]3 years ago
6 0

Answer:

The density of ozone is 4.24.

Explanation:

The relation between the relative rate of diffusion and density is given by :

r\propto \dfrac{1}{\sqrt d}

The given ratio of the relative rate of diffusion of ozone as compared to chlorine is 6:3.

Let the density of ozone is d₂.

\dfrac{r_1}{r_2}=\sqrt{\dfrac{d_2}{d_1}} \\\\\dfrac{6}{3}=\sqrt{\dfrac{d_2}{36}} \\\\3=\dfrac{\sqrt{d_2}}{6}\\\\d_2=\sqrt{18} \\\\d_2=4.24

So, the density of ozone is 4.24.

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Vanyuwa [196]

Answer:

1.3 × 10⁻¹¹ M

Explanation:

We are going to do 4 successive dilutions. In each dilution, we will apply the dilution rule.

C₁.V₁=C₂.V₂

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C₁ and V₁ are concentration and volume of the initial state

C₂ and V₂ are concentration and volume of the final state

<u>First dilution</u>

C₁ = 3.1 × 10⁻⁵ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL

C_{2}=\frac{C_{1}.V_{1}}{V_{2}} =\frac{3.1\times 10^{-5}M \times 1.00mL }{40.00mL} =7.8 \times 10^{-7}M

<u>Second dilution</u>

C₁ = 7.8 × 10⁻⁷ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL

C_{2}=\frac{C_{1}.V_{1}}{V_{2}} =\frac{7.8 \times 10^{-7}M \times 1.00mL }{40.00mL} =2.0 \times 10^{-8}M

<u>Third dilution</u>

C₁ = 2.0 × 10⁻⁸ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL

C_{2}=\frac{C_{1}.V_{1}}{V_{2}} =\frac{2.0 \times 10^{-8}M \times 1.00mL }{40.00mL} =5.0 \times 10^{-10}M

<u>Fourth dilution</u>

C₁ = 5.0 × 10⁻¹⁰ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL

C_{2}=\frac{C_{1}.V_{1}}{V_{2}} =\frac{5.0 \times 10^{-10}M \times 1.00mL }{40.00mL} =1.3 \times 10^{-11}M

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3 years ago
Suppose 3.52 g of calcium chloride is completely dissolved in a beaker of water. what would be the number of chloride ions that
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Answer:

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3 0
3 years ago
The value of ΔG° at 181.0 °C for the formation of calcium chloride from calcium metal and chlorine gas is ________ kJ/mol. At 25
aleksandr82 [10.1K]

Answer:

\Delta G^{0} at 181.0 ^{0}\textrm{C} is -723.3 kJ/mol.

Explanation:

We know, \Delta G^{0}=\Delta H^{0}-T\Delta S^{0}

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Let's assume \Delta H^{0} and \Delta S^{0} does not change in the temperature range 25.0 ^{0}\textrm{C}  - 181.0 ^{0}\textrm{C}.

181.0^{0}\textrm{C} = (273+181.0) K = 454.0 K

Hence, at 181.0 ^{0}\textrm{C}, \Delta G^{0}=(-795.8kJ/mol)-[(454.0 K)\times (-159.8\times 10^{-3}kJ/K.mol)]

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creativ13 [48]
The answer is E. coli
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Answer:

HCl conc.= 6.0mol/L

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From the dissociation of HCl= 1 mole H+ and 1mol Cl-, which is equivalent stoichiometrically in concentration to that of 1 mol HCl,

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