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3 years ago

Before it is used in the reactor, nuclear fuel is it not radioactive ?

Chemistry

1 answer:

puteri [66]3 years ago

3 0

Answer:

This is false as nuclear fuel is radioactive even before its use in a reactor.

A nuclear material has a half life that means in that time half of its radioactivity is gone

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Megan prepares a pitcher of lemonade by adding a quarter cup of granular sugar to the mixture. Which action should she take so t

REY [17]

Stir the water continuously, this is the only logical answer. Adding powdered sugar, decreasing the volume, increasing the amount of suga, cooling, all don't make the sugar dissolve quicker.

5 0

3 years ago

Read 2 more answers

According to the graph, if you are in a biome where the annual precipitation is about 300 cm and the average temperature is arou

alina1380 [7]

Answer:

We are in Tropical Rainforests.

Explanation:

Biome is basically an ecological region of earth that has specific type of flora and fauna (plants and animals). Every biome has specific abiotic factors like specific climate, temperature, geology, vegetation, soils and relief.

There are total 8 biomes in the world

Desert
Grassland
Temperate Boreal Forest
Chaparral
Tropical Savanna
Tropical Rainforest
Tundra
Temperate Deciduous Forest

Now in our question, we have been given some conditions and are asked to identify the biome. The annual precipitation is about 300 cm and the average temperature is around 25 degrees Celsius.

Now, if we study the characteristics of all these biomes, it is very simple to identify that the biome is Tropical Rainforest.

Important characteristics of Tropical Rainforest:

They receive more than 200 cm of rain each year.
The temperature of the biome is between 20 degree Celsius and 25 degree Celsius during the whole year.
Around 50 percent of the animal specie of the world are found here.

<em>Note: </em><em>You can study the characteristics of world’s other biomes in the link : http://www.physicalgeography.net/fundamentals/9k.html </em>

Hope it helps! :)

8 0

2 years ago

sashaice [31]

I think it's DNA forms a double helix and RNA consists a single strand.
I may be wrong...

3 0

2 years ago

Read 2 more answers

Write a nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90.

rusak2 [61]

Answer:

235/92U+10n→144/54Xe+90/38Sr+2/10n

Explanation:

The nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90 is represented by;

235/92U+10n→144/54Xe+90/38Sr+2/10n

In nuclear fission reactions a heavy nuclide is split into two light nuclides and is coupled by the release of energy.

3 0

3 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago