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harkovskaia [24]

2 years ago

6

A 50-gram sample of a radioactive nuclide is left to decay for 10 hours. At the end of 10 hours, only 25 grams of the sample is

left. What is the half-life of the nuclide sample?

1 answer:

Tanzania [10]2 years ago

4 0

The answer is 10 hours I had the same question

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If 0.0106 g of a gas dissolves in 0.792 L of water at 0.321 atm, what quantity of this gas (in grams) will dissolve at 5.73 atm?

Alex73 [517]

Answer:

0.189 g.

Explanation:

This problem is an application on <em>Henry's law.</em>

Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.

Solubility of the gas ∝ partial pressure

If we have different solubility at different pressures, we can express Henry's law as:

<em>S₁/P₁ = S₂/P₂,</em>

S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm

S₂ = ??? g/L and P₂ = 5.73 atm

So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.

<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>

<em></em>

8 0

2 years ago

What is the molarity of a stock solution made from 40 grams of LiOH in 16050mL of water

tekilochka [14]

Answer: heck the chemistry app itll help you i dont know this answer but the app will tell u!

6 0

2 years ago

The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151

vitfil [10]

Answer:- The natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for $^1^5^1_E_u$ and $^1^5^3_E_u$ as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of $^1^5^1_E_u$ as n then the abundance of $^1^5^3_E_u$ would be 1-n .

Let's plug in the values in the formula:

$151.964=150.919860(n)+152.921243(1-n)$

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

$151.964-152.921243=150.919860n-152.921243n$

$-0.957243=-2.001383n$

negative sign is on both sides so it is canceled:

$0.957243=2.001383n$

$n=\frac{0.957243}{2.001383}$

$n=0.478$

The abundance of $^1^5^1_E_u$ is 0.478 which is 47.8%.

The abundance of $^1^5^3_E_u$ is = $1-0.478$

= 0.522 which is 52.2%

Hence, the natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

3 0

2 years ago

I’LL MAKE YOU BRAINLIEST+ FREE POINTS

Pavel [41]

Answer:

The solution's new volume is 1.68 L

Explanation:

Dilution is the procedure to prepare a less concentrated solution from a more concentrated one, and simply consists of adding more solvent. So, in a dilution the amount of solute does not vary, but the volume of the solvent varies.

In summary, a dilution is a lower concentration solution than the original.

The way to do the calculations in a dilution is through the expression:

Ci*Vi=Cf*Vf

where C and V are concentration and volume, respectively; and the i and f subscripts indicate initial and final respectively.

In this case, being:

Ci= 7 M
Vi= 0.60 L
Cf= 2.5 M
Vf=?

Replacing:

7 M*0.60 L= 2.5 M* Vf

Solving:

$Vf=\frac{7 M*0.60 L}{2.5 M}$

Vf= 1.68 L

<u><em>The solution's new volume is 1.68 L</em></u>

4 0

2 years ago

The two boron atoms listed in the table are isotopes of the boron. The two carbon atoms and the two oxygen atoms are also called

Ivan

Answer:

A type of an atom which has a different number of neutrons but the same atomic number, therefore making it the same element. This atom would still have the same properties as well. (Ex: Vanadium-51 is an isotope of Vanadium that has 51 neutrons but still has 23 protons, as its atomic number is 23.)

3 0

1 year ago