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3 years ago

Which two atoms could be the "before'' and ''after'' of an alpha ejection

1 answer:

4 0

The answer would be uranium and thorium. When an alpha ejects a particle, it will create a new atom. So, when uranium ejects an alpha particle, it will produce thorium. They call this process as the alpha decay. Alpha decay often happens on atoms that are abundant nuclei such as uranium, radium, and thorium.

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Give an example of gravitational potential

larisa86 [58]

A raised weight.

Water that is behind a dam.

Have a nice day!

4 0

3 years ago

Read 2 more answers

What is the percentage strength (w/w) of a solution made by dissolving 62.5 g of potassium chloride in 187.5 ml of water?

kiruha [24]

Percent strength (% w/w) of a solution is defined as the amount of solute present in 100 g of the solution.

Given data:

Mass of the solute, potassium chloride = 62.5 g

Volume of water (solution) = 187.5 ml

We know that the density of water = 1 g/ml

Therefore, the mass corresponding to the given volume of water

= 187.5 ml * 1 g/1 ml = 187.5 g

We have a solution of 62.5 g of potassium chloride in 187.5 g water

Therefore, amount of solute in 100 g of water= 62.5 * 100/187.5 = 33.33

The percentage strength = 33.33 %

8 0

3 years ago

The radius of a vanadium atom is 130 pm. How many vanadium atoms would have to be laid side by side to span a distance of 1.30 m

monitta

Answer:

5 000 000 (5 million atoms)

Explanation:

Let us assume that a vanadium atom has a spherical shape.

diameter of a sphere = 2 x radius of the sphere

Thus,

Radius of a vanadium atom = 130 pm

= 130 x $10^{-12}$ m

The diameter of a vanadium atom = 2 x radius

= 2 x 130 x $10^{-12}$

= 260 x $10^{-12}$ m

Given a distance of 1.30 mm = 1.30 x $10^{-3}$ m,

The number of vanadium atoms required to span the distance = $\frac{1.3*10^{-3} }{260*10^{-12} }$

= 5000000

Therefore, the number of vanadium atom that would span a distance of 1.30 mm is 5 million.

3 0

3 years ago

Perform the following unit conversions:

Y_Kistochka [10]

Answer :

(a) The 60 feet of water is equal to 179344.2 Pa.

(b) The 220 psi is equal to $31680lbf/ft^2$

(c) The 120 torr is equal to 15998.6 Pa.

(d) The 1.0 atm to inches of glycerin is equal to 323.07 inches.

(e) The 1050 mm Hg is equal to $376.95lbf/ft^2$

Explanation :

(a) The conversion used from feet to pascal is:

$1\text{ feet}=2989.07Pa$

As we are given that 60 feet of water. Now we have to convert into Pa.

As, $1\text{ feet}=2989.07Pa$

So, $60\text{ feet}=\frac{60\text{ feet}}{1\text{ feet}}\times 2989.07Pa=179344.2Pa$

The 60 feet of water is equal to 179344.2 Pa.

(b) The conversion used from $psi$ to $lbf/ft^2$ is:

$1\text{ psi}=144lbf/ft^2$

As we are given that 220 psi. Now we have to convert into $lbf/ft^2$ .

As, $1\text{ psi}=144lbf/ft^2$

So, $220\text{ psi}=\frac{220\text{ psi}}{1\text{ psi}}\times 144lbf/ft^2=31680lbf/ft^2$

The 220 psi is equal to $31680lbf/ft^2$

(c) The conversion used from torr to pascal is:

$1\text{ torr}=133.322Pa$

As we are given that 120 torr. Now we have to convert into Pa.

As, $1\text{ torr}=133.322Pa$

So, $120\text{ torr}=\frac{120\text{ torr}}{1\text{ torr}}\times 133.322Pa=15998.6Pa$

The 120 torr is equal to 15998.6 Pa.

(d) 1.0 atm to inches of glycerin

Formula used : $P=\rho gh$

where,

P = pressure of glycerin = 1.0 atm = 101325 Pa

g = acceleration due to gravity = $9.8m/s^2$

$\rho$ = density of glycerin = $1260kg/m^3$

Now put all the given values in above formula, we get:

$101325Pa=(1260kg/m^3)\times (9.8m/s^2)\times h$

$h=8.206\frac{Pa.m^2s^2}{kg}=8.206m=323.07\text{ inches}$

Conversion used :

$1Pa=\frac{kg}{ms^2}\\\\1m=39.37inches$

The 1.0 atm to inches of glycerin is equal to 323.07 inches.

(e) The conversion used from $psi$ to $lbf/ft^2$ is:

$1\text{ mmHg}=0.359lbf/ft^2$

As we are given that 1050 mmHg. Now we have to convert into $lbf/ft^2$ .

As, $1\text{ mmHg}=0.359lbf/ft^2$

So, $1050\text{ mmHg}=\frac{1050\text{ mmHg}}{1\text{ mmHg}}\times 0.359lbf/ft^2=376.95lbf/ft^2$

The 1050 mm Hg is equal to $376.95lbf/ft^2$

4 0

3 years ago

Connect each of these circumstances to the rate of reaction. Explain how the change in circumstances could be implemented in a l

Marat540 [252]

Answer:

poop

Explanation:

poop

3 0

3 years ago