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3 years ago

If you have 8 g of a radioactive element there will be Of the radioactive element left after one half life

Chemistry

1 answer:

marta [7]3 years ago

3 0

Answer:

4g of the radioactive element present

Explanation:

Radioactive elements are atoms that are unstable in normal conditions and are converted in another different atoms.

1 half life express the time that is need to decay the half of the initial amount of the radioactive element.

If initially we have 8g of the radioactive element, after 1 life we will have 8g/2 =

4g of the radioactive element present

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A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 12

Serhud [2]

Answer:

$P_{He}=219mmHg$

$m_{He}=0.618gHe$

Explanation:

Hello,

By applying the Dalton's law, we can compute the partial pressure of the helium has:

$P_{tot}=P_{CO_2}+P_{Ar}+P_{O_2}+P_{He}$

Now, solving for the partial pressure of the helium gas we get:

$P_{He}=P_{tot}-P_{CO_2}-P_{Ar}-P_{O_2}=745mmHg-125mmHg-214mmHg-187mmHg\\P_{He}=219mmHg=0.288atm$

On the other hand, the mass of the helium gas is computed via the ideal gas equation in terms of the helium's mass:

$PV=\frac{m_{He}}{M_{He}}RT\\m_{He}=\frac{M_{He}PV}{RT} =\frac{4g/mol*0.288atm*12L}{0.082\frac{atm*L}{mol*K}*273K}\\ m_{He}=0.618gHe$

Best regards.

7 0

3 years ago

Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in and the atom emits a photon of lig

Usimov [2.4K]

Complete Question

Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in n=4 and the atom emits a photon of light with a wavelength of 486 nm. Group of answer choices

Answer:

$n=2$

Explanation:

From the question we are told that:

Wavelength $\lambda=486nm=>486*10^{-9}$

Generally the equation for Atom Transition is mathematically given by

$\frac{1}{\lambda}=R_{\infty }(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where

Rydberg constant $R_{\infty}=1.097*10^7$

Therefore

$\frac{1}{486*10^{-9}}=1.097*10^7*(\frac{1}{n_1^2}-\frac{1}{4^2})$

$(\frac{1}{n_1^2}-\frac{1}{4^2})=\frac{1}{486*10^{-9}*1.097*10^7}$

$n_1^2=3.98$

$n=1.99$

$n=2$

6 0

3 years ago

Draw the structures of organic compounds a and b. omit all byproducts

Georgia [21]

81. There is 1 carbon, 2 chlorine and fluorine atoms in Freon 12. To draw them it forms a cross with C in the middle and Cl and F both on the opposite side.

8 0

3 years ago

For the reaction 2h2+o2>h20, how many grams of water are produced from 70 grams of hydrogen

KonstantinChe [14]

630g

Explanation:

firstly the equation has to be balanced. 2H2+o2>2H2o.

finding the mole of hydrogen?

mole = 70/2 which is equal to 35 mole of hydrogen.

35 mole of hydrogen is equal to the mole of water.

finding the mass of water=?

35=mass/18

mass= 35*18=630g of water

3 0

3 years ago

A 100 gram block is placed on a scale that is known to have a 12% error. What will the scale read?

sesenic [268]

The scale will read 112 grams.

Why?

Since we already know the percent error of the scale (12%) and we also know the mass of the block that is placed on the scale, we can calculate what will be the read of the scale by using the following formula:

$ScaleRead=100g+(100g*\frac{12}{100})=100g+(100g*0.12)=100g+12g\\\\ScaleRead=112g$

We can see that the scale will read 112 grams.

Let's verify that we are right by using the following formula to verify if the obtained value for the scale read will be 12%:

$PercentError=\frac{ActualValue-RealValue}{RealValue}*100=\frac{112g-100g}{100g}*100\\\\PercentError=\frac{12g}{100g}*100=0.12*100=12(percent)$

Hence, we can see that the scale will read 112 grams.

Have a nice day!

7 0

4 years ago