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3 years ago

Which nuclei is not radioactive? A. Am-241 B. Mg-24 C. Pu-241 D. U-238

Physics

1 answer:

Helen [10]3 years ago

6 0

Hello!

Which nuclei is NOT radioactive?

A) Am-241 B) Mg-24 C) Pu-241 D) U-238

Solving:

It is noteworthy that chemical elements located on the periodic table in the lanthanide and actinide groups are radioactive.

Am-241 (americium) belongs to the group of actinides and is a heavy and radioactive metal.

Mg-24 (magnesium) is an essential element for the body, mainly for the nervous system, in addition to synthesizing proteins and serves for hormonal control, belongs to the group of alkaline earth metals and is a non-radioactive nucleus.

Pu-241 (plutonium) is an element that is isotope of fission by plutonium, belongs to the group of actinides and is a heavy and radioactive metal.

U-238 (uranium) is an element that is isotope of non-fission uranium, belongs to the group of actinides and is a heavy and radioactive metal.

Answer:

B) Mg-24

_______________________

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A professor drives off with his car (mass 870 kg), but forgot to take his coffee mug (mass 0.47 kg) off the roof. The coefficien

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Answer: $6.86 m/s^2$

Explanation:

Given

mas of car=870 kg

coffee mug mass=0.47 kg

coefficient of static friction between mug and roof $\mu _s= 0.7$

Coefficient of kinetic Friction $\mu _k=0.5$

maximum car acceleration is $\mu \times g$

here coefficient of static friction comes in to action because mug is placed over car . If mug is moving relative to car then \mu _k is come into effect

$a_{max}=0.7\times 9.8=6.86 m/s^2$

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3 years ago

The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25°C. How many grams of chlorophyll, C55H72MgN4O5, a nonvolatile,

mixer [17]

Answer:

29.4855 grams of chlorophyll

Explanation:

From Raoult's law

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 457.45 mmHg ÷ 463.57 mmHg = 0.987

Mass of solvent (diethyl ether) = 187.4 g

MW of diethyl ether (C2H5OC2H5) = 74 g/mol

Number of moles of solvent = mass/MW = 187.4/74 = 2.532 mol

Let the moles of solute (chlorophyll) be y

Total moles of solution = moles of solute + moles of solvent = (y + 2.532) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.987 = 2.532/(y + 2.532)

y + 2.532 = 2.532/0.987

y + 2.532 = 2.565

y = 2.565 - 2.532 = 0.033

Moles of solute (chlorophyll) = 0.033 mol

Mass of chlorophyll = moles of chlorophyll × MW = 0.033 × 893.5 = 29.4855 grams

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3 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

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