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3 years ago

Is cobalt 60 more dangerous than bananas ( potassium 40) and why ?

Chemistry

1 answer:

maxonik [38]3 years ago

7 0

Answer:

To the best of my knowledge, it is because of the amount of gamma rays is given off.

Explanation:

While both are isotopes, Potassium 40 gives off fewer gamma rays compared to Cobalt 60. Potassium 40 isn't really harmful to humans, but Cobalt 60 (I believe) is used in chemotherapy.

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For the reaction shown, calculate how many grams of each product form when the following amounts of reactant completely react to

stira [4]

This may help you
You need to use some stoichiometry here. The only way to do that is if you're working in moles. Since you're given grams of Al, you can convert that moles by dividing by the molar mass. Then from looking at the coefficients in your equation, you can see that for however many moles of Al react, the same numbers of moles of Fe will be produced, but only half as many moles of Al2O3 will be produced. To go back to grams, multiply the moles of each product that you get by their molar masses!

7 0

3 years ago

PLEASE HELP!!! ON A TIMER, WILL MARK BRAINLIEST

Nadusha1986 [10]

Answer:

Well, Your best answer will be is 2H+ + 2OH- -> 2H2O but you have to reduce it to H+ + OH- -> H2O. Good Luck!

8 0

3 years ago

Read 2 more answers

1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.

Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

$\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}$

$\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}$

$\Delta \ S = {(1*37.7)In \dfrac{263}{273}$

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0

2 years ago

For the reaction 6 Li + N2 → 2 Li3N , what is the maximum amount of Li3N (34.8297 g/mol) which could be formed from 14.18 mol (6

Yuki888 [10]

Moles Li = 3.50 g / 6.941 g/mol= 0.504
the ratio between Li and N2 is 6 : 1
moles N2 required = 0.504 /6=0.0840
we have 3.50 g / 28.0134 g/mol=0.125 moles of N2 so N2 is in excess
the ratio between Li and Li3N is 6 : 2
moles Li3N = 0.504 x 2 /6=0.168
mass Li3N = 0.168 mol x 34.8297 g/mol=5.85 g

4 0

3 years ago

HELP ME ASAP PLEASE AND THANK YOU

mafiozo [28]

F I think since the one on the worst line is an E so I’m pretty sure it’s F :p

7 0

3 years ago