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USPshnik [31]
3 years ago
13

A sphere has a mass of 64 g, and a density of 2.0 g/cm3 . what is the radius of the sphere

Chemistry
1 answer:
slega [8]3 years ago
4 0
The size v=masse/density
v= 4*pi*R^3
R=(3*masse/(4*pi*density))^(1/3)
R=1.9695 cm
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Calculate the value of q (kj) in this exothermic reaction when 5.40 g of hydrogen peroxide decomposes at constant pressure?
Gelneren [198K]
Hydrogen peroxide decomposes to yield water and oxygen gas 
That is; H2O2 (l) = H2O (l) + O2(g)
The standard heat of formation; H2O2 (l) = -187.6 kJ/mol; H2O(l) = -285.8 kJ/mol
1 mole of hydrogen peroxide contains 34 g
Thus, 5.4 g contains 5.4/34 = 0.1588 moles 
The moles of water produced will also be equivalent to 0.1588 moles 
Heat = heat of formation of product - reactant
Therefore; Heat = (0.1588 moles × -285,8 )- (0.1588× -187.6)
                           =  -15.594 kJ
6 0
3 years ago
What type of van der Waals interactions occur between molecules
BigorU [14]

Given what we know, we can confirm that the type of van der Waals interactions that occur between molecules of O2, SCl2, and CH4 in liquids of these substances are the presence of <u>London dispersion forces</u>.

<h3>What are London dispersion forces?</h3>
  • They are a force of attraction between atoms.
  • They are generated by electrostatic attraction.
  • These forces are common between atoms in close proximity and occur often when compounds have a symmetrical distribution of atoms.
  • They are generated by the formation of temporary dipoles.

Therefore, given the symmetry of the atoms disposition in these compounds and the temporary dipoles generated by the atoms being in close proximity, we can confirm that the van der Waals forces present in each compound are London dispersion forces.

To learn more about van deer Waals forces visit:
brainly.com/question/13201335?referrer=searchResults

7 0
2 years ago
When the nuclide bismuth-210 undergoes alpha decay:
pishuonlain [190]

Explanation:

An atom undergoes alpha decay by losing a helium atom.

So when bismuth undergoes alpha decay, we have;

²¹⁰₈₃Bi --> ⁴₂He + X

Mass number;

210 = 4 + x

x = 206

Atomic number;

83 = 2 + x

x = 81

The element is Thallium. The symbol is Ti.

For the second part;

X --> ⁴₂He + ²³⁴₉₀Th

Mass number;

x = 4 + 234 = 238

Atomic Number;

x = 2 + 90 = 92

The balanced nuclear equation is;

²³⁸₉₂U --> ⁴₂He + ²³⁴₉₀Th

7 0
4 years ago
Which of the following properties decrease down a column in the periodic table?
dsp73

Answer:

Effective nuclear charge and ionization energy decreased down in the column.

Explanation:

Along Group:

As we move from top to bottom in group the atomic size increases with increase of atomic number. The electron is added into the next shell hence the valance electrons farther away from the nucleus and hold of nucleus become weaker on the valance electrons. The addition of electrons also increase the shielding and protect the outer electrons from the hold of nucleus. Thus it becomes easier to remove the electron from an atom and less energy is required that's why ionization energy decreases from top to bottom and effective nuclear charge also decrease because of shielding effect.

Along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons and shielding remain constant due to addition of electron in same shell. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases.

3 0
4 years ago
Read 2 more answers
2. Write the formula or name for the following
Tems11 [23]

Answer:

Diphosphorus pentoxide

Carbon dichloride

BCl3

N2H4

Explanation:

These are all covalent compounds. To name covalent compounds, you add prefixes to the beginning of their names depending on what the subscript is of each element. The prefixes are:

1: Mono

2: Di

3: Tri

4: Tetra

5: Penta

6: Hexa

7: Hepta

8: Octa

9: Nona

10: Deca

For example, since the first one is Phopsphorus with a 2 next to it, you add the prefix Di to it.

If the first element in the compound only has one, meaning no number next to it, you do not say mono. This is why we just say "Carbon" for the second one instead of "Monocarbon."

Finally, you always have to end the second element in the compound with "ide." So, "chlorine" becomes "chloride," "oxygen" becomes "oxide," and so on.

7 0
4 years ago
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