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vekshin1
3 years ago
11

What is the most threatening, long term,aspect of nuclear fallout?

Chemistry
2 answers:
Nana76 [90]3 years ago
7 0
D) Radioactive contamination

natima [27]3 years ago
4 0

Answer: Radioactive contamination

Explanation: Radioactive wastes can be defined as the wastes that is leaked out due to the nuclear fallout.

The radioactive contamination can affect the huge population of animals that will drink the contaminated water. It is a way more dangerous if consumed by the human beings.

The radioactive contamination remains in the water for years and affects the health of the individual.

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Which material is a mixture 1 water 2 air 3 methane 4 magnesium
amm1812

Answer: Air

Explanation: Pure air is a mixture of several gases that are invisible and odorless. Consists of 78% nitrogen, 21% oxygen, and less than 1% of argon, carbon dioxide.

8 0
3 years ago
Who uses holmium? I need to know for a science project.
Tanzania [10]

Answer:

Holmium can absorb neutrons, so it is used in nuclear reactors to keep a chain reaction under control. Its alloys are used in some magnets. Holmium has no known biological role, and is non-toxic. Holmium is found as a minor component of the minerals monazite and bastnaesite.

Explanation:

this is basically used in industries

7 0
3 years ago
112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.
dolphi86 [110]

<u>Answer:</u> 4.999 moles of excess reactant will be left over.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       .....(1)

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol

For the given chemical reaction:

2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = \frac{12}{2}\times 0.778=4.668 mol of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

8 0
2 years ago
Stannum has a body centered tetragonal with lattice constant, a = b = 5.83A and c = 3.18A. If the atomic radius is 0.145 nm, det
Fed [463]

Answer:

the atomic packing factor of Sn is 0.24

Explanation:

a = b = 5.83A and c = 3.18A.

Volume of unit cell = a²c

= (5.83)² *  3.18 * 10⁻²⁴ cm³

= 1.08 * 10⁻²²cm³

Volume of atoms =

2 \times  \frac{4}{3} \pi r^3

(∴ BCC, effective number of atom is 2)

Volume of atoms =

2 * \frac{4}{3} *3.14*(0.145*10^-^7cm)^3

= 2.55*10⁻²³cm³

\text {Atomic packing factor}=\frac{\text {volume occupied by atom}}{\text {volume of unit cell }}

=\frac{2.55*10^-^2^3}{1.08*10^-^2^2} \\\\=0.24

<h3>therefore, the atomic packing factor of Sn is 0.24</h3>
4 0
3 years ago
A tank in the shape of a rectangular prism measures 10 dm by 4 dm by 6 dm. the tank is completely filled with 8908.8 l of liquid
nevsk [136]
1) The question contains an unknown unit


The number 8908.8 has to be in units of mass: for example, kg or grams.


Here you indicated L.


I am going to work assuming that L is a mass unit. So you can see the way to solve the problem, but you have to verifiy the real unit of the statement and substitute with it.



With that in mind you can find the density of the liquid from:


density  = mass / volume


2) Calculate the volume.


The volume of the liquid is the volume of the vessel, because it is filled.


The volume of the vessel is calculated from the formula of volume for a rectantular prism.


Volume of a rectangular prism = area of the base * height = side * side * height


=> Volume = 10 dm * 4 dm * 6 dm = 240 dm^3 = 240 liter


3) Calculate the density:


density = mass /volume = 8,908.8 L / 240 liter   =  37.12 L / dm^3



Answer: 37.12 L / dm^3


 
6 0
3 years ago
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