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3 years ago

Uranium-235 and uranium-238 are considered isotopes of one another

1 answer:

5 0

Indeed they are. The uranium atoms have separate numbers of neutrons in their nuclei, but still have the same number of protons.

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Which is the least dense?

AVprozaik [17]

1000 g of feathers is the least dense

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3 years ago

Explain how you would test the presence of oxygen and hydrogen gases.

Vedmedyk [2.9K]

Place a burning splint near the opening of a test tube. If a popping noise occurs, it's probably hydrogen. Place a glowing splint in the test tube, and if it reignites, it could be oxygen. Place a burning splint into a test tube, and if it goes out, it could be carbon dioxide.

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3 years ago

Jupiter revolves around the sun in a predictable pattern because

Misha Larkins [42]

Jupiter revolves around the Sun in a predictable pattern because the force of gravity between Jupiter and the Sun maintains the orbit.

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2 years ago

A sample of Xe gas is observed to effuse through a pourous barrier in 4.83 minutes. Under the same conditions, the same number o

Solnce55 [7]

Answer:

28.93 g/mol

Explanation:

This is an extension of Graham's Law of Effusion where $\frac{R1}{R2} = \sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$

We're only talking about molar mass and time (t) here so we'll just concentrate on $\sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$ . Notice how the molar mass and time are on the same position, recall effusion is when gas escapes from a container through a small hole. The time it takes it to leave depends on the molar mass. If the gas is heavy, like Xe, it would take a longer time (4.83 minutes). If it was light it would leave in less time, that gives us somewhat an idea what our element could be, we know that it's atleast an element before Xenon.

Let's plug everything in and solve for M2. I chose M2 to be the unknown here because it's easier to have it basically as a whole number already.

$\sqrt{\frac{M2}{131} } = \frac{2.29}{4.83}$

The square root is easier to deal with if you take it out in the first step, so let's remove it by squaring each side by 2, the opposite of square root essentially.

$(\sqrt{\frac{M2}{131} } )^2= (\frac{2.29}{4.83})^2$

${\frac{M2}{131} } = (0.47)^2$

${\frac{M2}{131} } = 0.22$

M2= 0.22 x 131

M2= 28.93 g/mol

8 0

2 years ago

Using the volumes of sodium thiosulfate solution you just entered, the mass of bleach sample, and the average molarity of the so

dedylja [7]

Answer:

3.18 (w/w) %

Explanation:

In the problem, you can find mass of NaClO knowing the reaction of NaClO with Na₂S₂O₃ is:

NaClO + 2Na₂S₂O₃ + H₂O → NaCl + Na₂S₄O₆ +2NaOH + NaCl

<em>Where 1 mole of NaClO reacts with 2 moles of Na₂S₂O₃</em>

<em> </em>Moles of thiosulfate in the titration are:

0.0101L ₓ (0.042mol / L) = 4.242x10⁻⁴ moles of Na₂S₂O₃

Thus, moles of NaClO in the initial solution are:

4.242x10⁻⁴ moles of Na₂S₂O₃ ₓ (1mol NaClO / 2 mol Na₂S₂O₃) = 2.121x10⁻⁴ moles NaClO

As molar mass of NaClO is 74.44g/mol, mass of 2.121x10⁻⁴ moles are:

2.121x10⁻⁴ moles ₓ (74.44g / mol) = <em>0.0158g of NaClO</em>

As mass of bleach is 0.496g, mass percent is:

0.0158g NaClO / 0.496g bleach ₓ 100 =

4 0

3 years ago