All categories

2 years ago

The isotope 208Tl undergoes Î² decay with a half-life of 3.1 min.

Chemistry

1 answer:

melamori03 [73]2 years ago

8 0

Answer:

a. Pb 208

b. About 21.7 minutes

c. only a trace amount

Explanation:

It under goes beta decay.

There should be virtually nothing after an hour

You might be interested in

I need help please WOTH this

barxatty [35]

Answer:

Kenma I'll help you! it's B.

Explanation:

I hope I helped you Kenma

7 0

2 years ago

Why must reagent bottles be closed when not in use?

Katen [24]

Answer to your question.

7 0

2 years ago

Magda has the two magnets shown below what will happen if magda tries to push the north poles of the two magnets toward one anot

ryzh [129]

They will repel one another because opposites attract, but like charges will repel.

8 0

2 years ago

If the pressure changes from 1 atm to 3 atm, what does the volume of 30 L change to

DiKsa [7]

Answer: 10L

Explanation:

Given that:

Initial pressure P1 = 1 atm

New pressure P2 = 3 atm

Initial volume V2 = 30 L

New volume V2 = ?

Since pressure and volume are involved, apply the formula for Boyle's law

P1V1 = P2V2

1 atm x 30L = 3 atm x V2

30 atm L = 3 atm x V2

V2 = (30 atm L / 3 atm)

V2 = 10L

Thus, volume changed to 10 liters

5 0

2 years ago

When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H

Taya2010 [7]

Answer:

- Empirical:

$C_3H_7$

- Molecular:

$C_6H_{14}$

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

$n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC$

Moreover, the moles of hydrogen:

$n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH$

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

$C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}$

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

$C_3H_7$

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

$C_6H_{14}$

Which is hexane.

Best regards.

6 0

2 years ago