Answer:
6.82 kg
Explanation:
Given that the amount of water is 15L and we know that the density of water is ≈ 1kg/L. The mass of water is given by mass = volume x density, i.e,
mass = 15 x 1 = 15 kg. Also the specific heat capacity of water is 4.186 KJ/kg.
The sublimation enthalpy of dry ice is 571 KJ/kg.
Now, the amount of heat lost by water is entirely used up for the sublimation (conversion from soild to gas) of dry ice. And the heat (Q) lost by water is given as : Q = mCΔT, where m is the mass of water, C the specific heat capacity of water and ΔT the change in temperature.
Here, Q = 15 x 4.186 x (90 - 28) = 3892.98 KJ.
This amount of heat is taken up by the dry ice for its sublimation. Also the energy taken by dry ice (Q') for its sublimation is given by: Q' = m'L', where m' is the mass of dry ice, L' is the latent heat of sublimation (i.e, the amount of heat required per kg of a substance to sublime) of dry ice amd L' = 571 KJ/kg.
Now, Q' =m'L' = heat lost by water = 3892.98KJ.
And, m'L' = m' x 571 KJ/kg = 3892.98 KJ. (Dividing with 571)
Therefore, m' = 6.82 kg.
To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy.
The amount of energy required by an isolated, gaseous molecule in the electronic state of the ground to absorb in order to discharge an electron and produce a cation has been known as the ionization energy. The amount of energy required for every atom in a mole to drop one electron is most often given as kJ/mol.
Anything that causes electrically neutral atoms and molecules to gain or lose electrons in order to become electrically charged atoms as well as molecules .
Therefore, the "To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy."
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Answer:
10.6 grams is approximately 0.10 moles. So you would need about 0.10 moles of sulfuric acid. That converts to about 9.80 grams.
Explanation:
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Answer:

Explanation:
Hello!
In this case, we can divide the problem in two steps:
1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

So we solve for C2:

2. Now, since 111 mL of water is added, we compute the final volume, V3:

So, the final concentration of the 139 mL portion is:

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