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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

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Is pentacosane a molecular compound?

jekas [21]

Answer:

Yes

Explanation:

A molecular compound is that one which is composed of molecules formed by atoms of two or different elements. A pentacosane is an alkaline made of unbranched chain of 25 carbons atoms. It has applications in plant metabolite. The molecular formula of pentacosane is C₂₅H₅₂

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2 years ago

What are the limitations of bohr's model of atom

nataly862011 [7]

Answer:

The Bohr Model is very limited in terms of size. Poor spectral predictions are obtained when larger atoms are in question. It cannot predict the relative intensities of spectral lines. It does not explain the Zeeman Effect, when the spectral line is split into several components in the presence of a magnetic field.

Explanation:

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2 years ago

Read 2 more answers

Calculate a reasonable amount (mass in g) of your unknown acid to use for a titration. You will want about 30 mL of titrant to g

Vinil7 [7]

Answer:

"0.60 g" is the appropriate solution.

Explanation:

The given values are:

Volume of base,

= 30 ml

Molarity of base,

= 0.05 m

Molar mass of acid,

= 400 g/mol

As we know,

⇒ $Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}$

On substituting the values, we get

⇒ $0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}$

⇒ $Number \ of \ moles \ of \ base=0.05\times 30\times 10^{-3}$

⇒ $=1.5\times 10^{-3}$

hence,

⇒ $Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}$

On substituting the values, we get

⇒ $1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}$

⇒ $Mass \ of \ acid=1.5\times 10^{-3}\times 400$

⇒ $=0.60 \ g$

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2 years ago

If two substances travel up the chromatography paper this implies that

Colt1911 [192]

Answer:

They are seperating

Explanation:

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2 years ago

The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abb

Nimfa-mama [501]

Answer:

$\boxed{\text{254 g}}$

Explanation:

$\begin{array}{rcl}\text{\% m/V} & = & \dfrac{\text{Mass of sucrose}}{\text{Volume of solution}}\\\\\text{Let m}& = &\text{mass of sucrose}\\\dfrac{\text{35.0 g}}{\text{100 mL}}& = & \dfrac{m}{\text{725 mL}}\\\\m & = &\dfrac{\text{35.0 g}\times 725}{100}\\\\ & = &\textbf{254 g}\\\end{array}\\\text{You need $\boxed{\textbf{254 g}}$ of sucrose}$

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3 years ago