Ask question

Ask question

All categories

3 years ago

14

The half-life of f-20 is 11.4 seconds. when there are 200 atoms of f-20 in a sample, a scientist starts a stopwatch. the scienti

st stops the stopwatch at 34.2 seconds; there are _____ atoms of f-20 remaining.

1 answer:

prisoha [69]3 years ago

7 0

There should only be 25 atoms remaining

You might be interested in

A diffusion coefficient KG was determined for the diffusion of a gas A through a stagnant film B [KG = 0.88 mol//h-ft2-atm].

Angelina_Jolie [31]

Explanation:

Transfer of mass A into stagnant film B depends on the availability of driving force.

Whereas driving force is the pressure difference at the surface of A and the bulk.

As, $N_{A} \propto (P_{A1} - P_{A2})$

$N_{A} = K_{G} \times (P_{A1} - P_{A2})$

Therefore, putting the given values into the above formula as follows.

$N_{A} = K_{G} \times (P_{A1} - P_{A2})$

= $0.88 \times (0.2 atm - 0.05 atm)$

= 0.132 $mol/h.ft^{2}$

Thus, we can conclude that the flux of A from a surface into a mixture of A and B is 0.132 $mol/h.ft^{2}$

4 0

4 years ago

Consider the following reaction, equilibrium concentrations, and equilibrium constant at

REY [17]

Answer:

Equilibrium concentration of $H_{2}O$ is 12.5 M

Explanation:

Given reaction: $C_{2}H_{4}+H_{2}O\rightleftharpoons C_{2}H_{5}OH$

Here, $K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}$

where $K_{c}$ represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations

Here, $[C_{2}H_{4}]=0.015M$ , $[C_{2}H_{5}OH]=1.69M$ and $K_{c}=9.0$

So, $[H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M$

Hence equilibrium concentration of $H_{2}O$ is 12.5 M

5 0

3 years ago

How many grams of i2 are needed to react with 30.1 g of n2h4?

Nina [5.8K]

The reaction is,

H2S + I2 --------------> 2 HI +S

Molar weight of H2S = 34 g per mol

Molar weight of HI =128 g per mol

Molar weight of I2 =254 g per mol

Moles of H2S in 49.2 g = 49.2 /34 mol = 1.447 mol

So according to stoichiometry of the reaction, number of I2 mols needed

= 1.447 mol

The mass of I2 needed = 1.447 mol x 254 g

8 0

3 years ago

Read 2 more answers

Which incorrect aspect of Rutherford’s model was fixed by Bohr’s model?

dolphi86 [110]

Answer:

Some things that were wrong with Rutherford's model were that the orbiting electrons should give off energy and eventually spiral down into the nucleus, making the atom collapse. Bohr proposed his quantized shell model of the atom to explain how electrons can have stable orbits around the nucleus. To remedy the stability problem, Bohr modified the Rutherford model by requiring that the electrons move in orbits of fixed size and energy.

Explanation:

7 0

3 years ago

Could someone please care to explain what the answer is

polet [3.4K]

Calculate the percentage composition of Ca(MnO4)2.

<u>Firstly, we see how many atoms are there for each element in the formula</u>.

Ca= 1 atom

Mn= 2 atoms

O= 8 atoms

<u>Next, we are going to consult our periodic table for the atomic mass of each element.</u>

Ca= 40

Mn= 55

O= 16

Then, we have to find the molar mass for the compound..

Here is the formula for calculating molar mass of an element:

Molar Mass= ( no. of atoms of the element × atomic mass of the element)

Now, we have to calculate the atomic mass of the compound. So using the molar mass formula for an element, we calculate the molar mass for each element then we sum up their molar masses to get the compounds molar mass.

Molar mass (Ca)= 1× 40

(Ca)= 40

Molar mass (Mn)=2×55

(Mn)= 110

Molar mass (O)= 8×16

(O)= 128

Now: Molar mass( compound)= (Ca)+(Mn)+(O)

= 40+ 110 128

= 278

This is everything we need to calculate our percentage composition for each element..

* The example says to find the percentage composition for Ca. So we only find for Ca, Which is already done using the formula and the answer is 14.39%.

To prove that your answer is correct, find the percentage composition for Mn and O as well. Then you add up their percentage compositions.

If you do and you get 100 as your answer, then your percentage compositions are correct.

Why don't you try finding the percentage composition for Mn and O, then add up all the three percentage compositions. If you 100 as their sum, then your percentage composition for each of the elements are correct.

6 0

3 years ago