All categories

3 years ago

Write a balanced nuclear equation for the following: The nuclide astatine-218 undergoes alpha emission to give bismuth-214

Chemistry

1 answer:

galina1969 [7]3 years ago

7 0

Answer:

$^{218}_{85}At\rightarrow ^{214}_{83}Bi+^{4}_{2}He$

Explanation:

Hello there!

In this case, since the radioactive reaction for the alpha emission of astatine-218 to bismith-214 involve the release of a helium atom as shown below:

$^{218}_{85}At\rightarrow ^{214}_{83}Bi+^{4}_{2}He$

Whereas the atomic number decreases by 2 and the mass number by 4 in agreement to the release of the Helium atom.

Regards!

You might be interested in

_______ is a media that consists of natural or synthetic mineral or plant substances that deposit crystalline or splinter-like f

diamong [38]

Answer : Option A) Dry media

Explanation : i) Dry media is a media that deposits graphite, charcoal, pastels, ordinary pencils, then moistened with a wet brush to get various painterly effects on the paper fibers.

ii) Silverpoint - Silverpoint media uses red chalk or traces of black pencil on white-coated paper usually the manuscripts,

iii) Ink Painting - Uses various inks on the paper.

iv) Liquid media - uses liquid pigments suspended in some base which in liquid state that soaks the paper.

3 0

3 years ago

Read 2 more answers

What is the approximate time of day, based on the position of the Sun in the sky?

Ivenika [448]

Answer:

sunrise

Explanation:

The sun rises from the east as there is a time difference the further east you go.

3 0

4 years ago

Read 2 more answers

A substance is dissolved in water, and the temperature of the surroundings decreases. Is this process exothermic or endothermic?

EleoNora [17]

Answer:

endothermic reaction

Explanation:

This is because it absorb heat from the surrounding which is required to break the bond hold it together hence as it absorb heat from the surrounding the temp of surrounding decrease.

5 0

3 years ago

A pharmacy intern is asked to prepare 3 L of a 30% w/v solution. T he pharmacy stocks the active ingredient in 8-ounce bottles o

MariettaO [177]

Answer: The number of bottles that will be needed are 6

Explanation:

We are given:

Amount of solution, the intern is asked to prepare = 3 L = 3000 mL (Conversion factor: 1 L = 1000 mL)

Strength of solution needed = 30 % (w/v)

This means that in 100 mL of solution, the solute present is 30 grams

So, in 3000 mL of solution, the solute present will be = $\frac{30}{100}\times 3000=900g$

Active ingredient present in 1 bottle = 8 ounce of 70 % (w/v)

Conversion factor used: 1 ounce = 29.57 mL

So, $8ounce\times \frac{29.57mL}{1ounce}=236.6mL$

Amount of active ingredient present in 1 bottle = $236.6\times \frac{70}{100}=165.6g$

To calculate the number of bottles, we need to divide the total amount of solution needed by the amount of active ingredient present in 1 bottle, we get:

$\text{Number of bottles}=\frac{\text{Amount of solution to prepare}}{\text{Amount of active ingredient in 1 bottle}}$

Putting values in above equation, we get:

$\text{Number of bottles}=\frac{900g}{165.6g}\\\\\text{Number of bottles}=5.43\approx 6$

Hence, the number of bottles that will be needed are 6

6 0

3 years ago

A sample of an unknown gas takes 222 s to diffuse through a porous plug at a given temperature. At the same temperature, N2(g) t

gulaghasi [49]

Answer:

$\large \boxed{\text{45.1 g/mol}}$

Explanation:

Graham’s Law applies to the diffusion of gases:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r \propto \dfrac{1}{\sqrt{M}}$

If you have two gases, the ratio of their rates of diffusion is

$\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}$

The time for diffusion is inversely proportional to the rate.

$\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}$

Data:

t₂ = 222 s

t₁ = 175 s

M₁ = 28.01

Calculation :

$\begin{array}{rcl}\dfrac{222}{175} & = & \sqrt{\dfrac{M_{2}}{28.01}}\\\\1.269 & = & \sqrt{\dfrac{M_{2}}{28.01}}\\\\1.609 & = & \dfrac{M_{2}}{28.01}\\\\M_{2} & = & 1.609 \times 28.01\\ & = & \textbf{45.1 g/mol}\\\end{array}\\\text{The molar mass of the unknown gas is $\large \boxed{\textbf{45.1 g/mol}}$}$

3 0

3 years ago