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melomori [17]
2 years ago
11

A flask contains 1/3 mole of h2 and 2/3 mol of he. Compare the force on the wall per impact of h2 relative to that for he.

Chemistry
2 answers:
densk [106]2 years ago
8 0

The force on the wall is actually the pressure exerted by gas molecules

Higher the pressure more the force exerted on the walls of container

The pressure depends upon the number of molecules of a gas

In a mixture of gas the pressure depends upon the mole fraction of the gas

As given the mole fraction of He is more than that of H2 therefore He will exert more pressure on the wall

The ratio of impact will be

H2 / He = 2/3 / 1/3 = 2: 1

brilliants [131]2 years ago
4 0

Answer:

1/2

Explanation:

By Raoult's Law, the pressure of a gas in a gas mixture, is its mole fraction multiplied by the total pressure of the gas mixture. Thus, calling Pt as the total pressure:

P_{H_{2}} = \frac{1}{3} *Pt

P_{He} = \frac{2}{3} *Pt

The pressure is the force divided by the area, thus:

\frac{F_{H_{2}}}{A} = \frac{1}{3} *Pt

\frac{F_{He}}{A} = \frac{2}{3}*Pt

Because the area is the same:

\frac{F_{H_{2}}}{F_{He}} = \frac{1/3}{2/3}

\frac{F_{H_{2}}}{F_{He}} = \frac{1}{2}

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frosja888 [35]

Answer:

The 9th percentile is 4.464ml

Explanation:

Hello!

Given the variable X: Amount of dye dispensed into a paint can.

With a normal distribution, mean μ= 5ml and standard deviation δ= 0.4ml

You need to calculate the 9th percentile of the distribution.

The percentile is a measure of the position that indicates the number that separates the distribution or data set in a percentage of interest. In this case, the 9th percentile is the value of the distribution that separates the bottom 9% from the top 91%.

Symbolically:

P(X≤x₀)= 0.09

x₀ represents the value of the percentile.

The best way to calculate this value is by using the standard normal distribution since it is already tabulated, and then "translate" the Z-value to a value of the variable.

Under the standard normal distribution you have to look for:

P(Z≤z₀)= 0.09

The value marks the bottom 9% of the distribution, this means that you'll find it in the left tail of it. Remember the mean of the standard normal distribution is zero, so all values under the mean will be negative. Using the left entry of the Z-table you have to look for 0.09 in the body of the table and reach the margins to find the corresponding value:

z₀= -1.34

Now using the formula of the distribution you can "translate" the Z-value in terms of the variable of interest

Z=(X-μ)/ δ ~N(0;1)

z₀=(x₀-μ)/ δ

z₀*δ=x₀-μ

x₀=(z₀*δ)+μ

x₀=(-1.34*0.4)+5

x₀= 4.464

The 9th percentile is 4.464ml

I hope you have a SUPER day!

4 0
2 years ago
Hydroxyapatite, Ca 10 ( PO 4 ) 6 ( OH ) 2 Ca10(PO4)6(OH)2 , has a solubility constant of Ksp = 2.34 × 10 − 59 2.34×10−59 , and d
Margarita [4]

Answer:

1.315x10⁻³M = [Ca²⁺]

Explanation:

Based in the reaction:

Ca₁₀(PO₄)₆(OH)₂(s) ⇄ 10Ca²⁺(aq) + 6PO₄³⁻(aq) + 2OH⁻(aq)

Solubility product, ksp, is defined as:

ksp = [Ca²⁺]¹⁰ [PO₄³⁻]⁶ [OH⁻]²

From 1 mole of hydroxyapatite are produced  10 moles of Ca²⁺ and 6 moles of PO₄³⁻. That means moles of PO₄³⁻ are:

6/10 Ca²⁺ = PO₄³⁻

Replacing in ksp formula:

ksp = [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [OH⁻]²

As [OH⁻] is 2.50x10⁻⁶M and ksp is 2.34x10⁻⁵⁹:

2.34x10⁻⁵⁹ =  [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [2.50x10⁻⁶]²

3.744x10⁻⁴⁸ = 0.046656[Ca²⁺]¹⁶

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<em>1.315x10⁻³M = [Ca²⁺]</em>

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I hope it helps!

5 0
2 years ago
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nikdorinn [45]

Answer:

b . Irradiated food is shown to not be radioactive.

Explanation:

If it can be proven that irradiated food is not radioactive, then it will effective dispute the idea that irradiated food are less safe to eat.

  • An irradiated food is one in which ionizing radiations have been employed to improve food quality.
  • Thus, bacteria and other food spoilers can be exterminated from the food.
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