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2 years ago

2. Calculate the de Broglie wavelength of an electron moving at 1/137th the speed

Chemistry

1 answer:

anastassius [24]2 years ago

5 0

Explanation:

speed of light is 3×10^8 m/ s

1/137 of 3 × 10^ 8 m/s ( u get the speed the electron was moving with)

mass of electron 9.109× 10^-31 kg( i think we consider the mass of electron if the relativistic correction of the mass is negligible)

then from de Broglie wavelength= h/ mv

h is planck constant 6.626 × 10^-34

substitute the valies and you get the answer

am sorry I don't have my calculator with me I wuld hv written the answer bt I hope this helps

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A certain gas is present in a 12.0 L cylinder at 4.0 atm pressure. If the pressure is increased to 8.0 atm the volume of the gas

diamong [38]

There are 3 parts in this question:
1) To find the initial Boyle's constant $k_{i}$
2) To find the final Boyle's constant $k_{f}$
3) To verify whether gas is obeying Boyle's law or not

Given data:
The initial volume of the cylinder(in litres) = $V_{i}$ = 12.0 L
The initial pressure(in atmospheric pressure) = $P_{i}$ = 4.0 atm

The final pressure(in atmospheric pressure) = $P_{f}$ = 8.0 atm
The final volume of the cylinder(in litres) = $V_{f}$ = 6.0 L

First you need to know what Boyle's law is:
Boyle's law states that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

The Mathematical form of Boyle's law is:
$P = \frac{k}{V}$

Where,
P = Pressure
V = Volume of the gas
k = Boyle's constant

Now let's solve aforementioned parts one by one:
1.
The initial volume of the cylinder(in litres) = $V_{i}$ = 12.0 L
The initial pressure(in atmospheric pressure) = $P_{i}$ = 4.0 atm
The Boyle's constant = $k_{i}$ = ?

According to the Boyle's law,

$P_{i} = \frac{k_{i}}{V_{i}}$

=> $k_{i}$ = $P_{i}V_{i}$
Plug-in the values in the above equation, you would get:
$k_{i}$ = 4.0 * 12.0 = 48

Ans-1) $k_{i}$ = 48

2.
The final pressure(in atmospheric pressure) = $P_{f}$ = 8.0 atm
The final volume of the cylinder(in litres) = $V_{f}$ = 6.0 L
The Boyle's constant = $k_{f}$ = ?

According to the Boyle's law,

$P_{f} = \frac{k_{f}}{V_{f}}$

=> $k_{f}$ = $P_{f}V_{f}$
Plug-in the values in the above equation, you would get:
$k_{f}$ = 8.0 * 6.0 = 48

Ans-2) $k_{f}$ = 48

3.
In order to verify Boyle's law, the initial Boyle's constant should be EQUAL to the final Boyle's constant, meaning:

$k_{i}$ = $k_{f}$

Since,
$k_{i}$ = 48
$k_{f}$ = 48

Therefore,
48=48.

Ans-3) Hence proved: The gas IS obeying the Boyle's law.

-i

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3 years ago

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Compute 4.659×104−2.14×104. Round the answer appropriately.

pentagon [3]

Answer: 25,200.

Explanation:

1) Given: 4.659 × 10⁴ - 2.14 × 10⁴

2) You have to deal with significant figures.

Since, the powers are the same (10⁴) you can directly subtract the decimal numbers. But first analyze the significant figures and the number of decimal digits.

3) The number 4.659 × 10⁴ has four significant figures (4, 6, 5, and 9), while the number 2.14 × 10⁴ has three significan figures (2, 1, and 4).

4) When you add or subtract numbers with diferent amount of decimal digits, the result must show the same number of decimal digits as the term with the least number of decimal digits.

5) Before subtracting, you must round all the terms to the least number of decimal digits. So, since 2.14 has two decimal digits and 4.659 has three decimal digits, you shall round 4.659 to 4.66.

6) Now you subtract 4.66 - 2.14 = 2.52

7) Multiply by the power of 10: 2.52 × 10⁴ = 25,200. And that is the answer.

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2 years ago

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Sav [38]

Answer: The amount of water that can be formed is 160 moles

Explanation:

We are given:

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The chemical equation for the reaction of hydrogen gas and oxygen gas follows:

$2H_2+O_2\rightarrow 2H_2O$

By Stoichiometry of the reaction:

1 mole of oxygen gas reacts with 2 moles of hydrogen gas

So, 80 moles of oxygen gas will react with = $\frac{2}{1}\times 80=160mol$ of hydrogen gas

As, given amount of hydrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of oxygen gas produces 2 moles of water

So, 80 moles of oxygen gas will produce = $\frac{2}{1}\times 80=160moles$ of water

Hence, the amount of water that can be formed is 160 moles

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2 years ago