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liberstina [14]
2 years ago
15

A tank containing both hf and hbr gases developed a leak. The ratio of the rate of effusion of hf to the rate of effusion of hbr

is ________.
Chemistry
1 answer:
Pavlova-9 [17]2 years ago
4 0

Answer:

2.01

Explanation:

The effusion is the passage of the molecules by a small hole by a difference of pressure. By Graham's Law, the rate of the effusion is inversely proportional to the square of the molar mass of the compound. Thus,

rateHF/rateHBr = √MHBr /√MHF

MHBr = 81 g/mol

MHF = 20 g/mol

rateHF/rateHBr = √81/√20

rateHF/rateHBr = 2.01

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A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.7 nm.93.7 nm. What is the final state of the hy
sammy [17]

Answer:

5

Explanation:

Given that the formula is;

1/λ= R(1/nf^2 - 1/ni^2)

λ = 93.7 nm or 93.7 * 10^-9 m

R= 1.097 * 10^7 m-1

nf = ?

ni = 1

From;

ΔE = hc/λ

ΔE = 6.63 * 10^-34 * 3* 10^8/93.7 * 10^-9

ΔE = 21 * 10^-19 J

ΔE = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)

21 * 10^-19 J = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)

21 * 10^-19/-2.18 * 10^-18 = (1/nf^2 - 1/1^2)

-0.963 = (1/nf^2 - 1)

-0.963 + 1 = 1/nf^2

0.037 = 1/nf^2

nf^2 = (0.037)^-1

nf^2 = 27

nf = 5

7 0
2 years ago
Franklin was performing an experiment by combining hydrochloric acid and sodium hydroxide. He measured the mass of his reactant
VARVARA [1.3K]

Answer:

B

Explanation:

took the test

4 0
1 year ago
Read 2 more answers
Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample
Dimas [21]

Answer:

<em>3.27·10²³ atoms of O</em>

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is <em>Na₂SO₄, </em>and its molar mass is approximately 142.05\frac{g}{mol}.

We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.

19.3g <em>Na₂SO₄</em> · \frac{1 mol Na^2SO^4}{142.05g Na^2SO^4} · \frac{4 mol O}{1 mol Na^2SO^4} ·\frac{6.022x10^2^3}{1 mol O}

After doing the math for this dimensional analysis, you should get a quantity of approximately <em>3.27·10²³ atoms of O</em>.

3 0
2 years ago
Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha
OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

KH_2PO_4 acts as a weak acid and Na_2HPO_4 as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

7.00=6.86+log(\frac{base}{0.10})

7.00-6.86=log(\frac{base}{0.10})

0.14=log(\frac{base}{0.10})

Taking antilog:

10^0^.^1^4=\frac{base}{0.10}

1.38=\frac{base}{0.10}

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is Na_2HPO_4 required to make the buffer is 0.138M.

5 0
2 years ago
Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t
Dafna11 [192]

Answer:

The plane with aluminium can lift more mass of passangers than the plane of steel.

Explanation:

The total mass the airplane canc lift is:

m_{tot}=m_{fuselage}+m_{passangers}

For aluminium:

m_{tot}=m_{fus-Al}+m_{pas-Al}

m_{fus-Al}=\delta _{Al}*V_{fuselage}

and

V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]

where:

  • L is lenght
  • D is diameter
  • e is thickness

m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}

For steel (same procedure):

m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel

Knowing that the total mass the airplane can lift is constant and that aluminum has a lower density than the steel, we can afirm that the plane with aluminium can lift more mass of passangers.

Also you can estimate an average weight of passanger to estimate a number of passangers it can lift.

5 0
2 years ago
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