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Len [333]
3 years ago
9

Convert 250 g O2 to mol O2​

Chemistry
1 answer:
Nezavi [6.7K]3 years ago
8 0

Answer:

250 g converted to mol O2 is O16.

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calculate the volume occupied by 10g of propane gas, under normal conditions of temperature and pressure
andriy [413]

Answer:

5.5 L

Explanation:

First we <u>convert 10 g of propane gas</u> (C₃H₈) to moles, using its <em>molar mass</em>:

  • 10 g ÷ 44 g/mol = 0.23 mol

Then we <u>use the PV=nRT formula</u>, where:

  • P = 1 atm & T = 293 K (This are normal conditions of T and P)
  • n = 0.23 mol
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • V = ?

1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K

  • V = 5.5 L
3 0
2 years ago
Both protons and neutrons (and their anti-particles) froze out:
VladimirAG [237]
I think Both protons and neutrons (and their anti-particles) froze out at 1013 K, about 0.0001 seconds after the Big Bang. Protons and neutrons are sub atomic particles of an atom that are found in the nucleus of an atom. Proton is the positively charge particle while the neutron has no charge. The proton positive charge accounts for the positive nuclear charge. 
6 0
3 years ago
Can somebody help me with this question please
Tasya [4]

Answer:

concave lens

Explanation:

it's concave lens because it diverges the ray/beam of light.

7 0
3 years ago
State ten important of osmosis?​
algol [13]

Explanation:

Osmosis is the movement of water molecules, across a semipermeable membrane, from a hypotonic solution to a hypertonic solution

Osmosis is critical in the maintenance of the right amount of water in the cells an organism. Water is a universal solvent and is especially important for biochemical activities of a cell. Osmosis maintains the right water homeostasis in cells.

Osmosis is useful in the absorption of water by the roots of the plants and the maintenance of transpirational pull that ensures herbecious plants remain upright.

Osmosis enables the opening and closing of plant stomata hence regulating the loss of water by the plants.  

Learn More:

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5 0
2 years ago
An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the
san4es73 [151]

Answer:

the ionic radius of the anion r^- = 312.52 \ pm

Explanation:

From the diagram shown below :

The anion Cl^- is located at the corners

The cation Cs^+ is located at the body center

The Body diagonal length =  \sqrt{3 \ a }

∴ 2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a}  \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a

Given that :

\frac{r^+}{r^-} =0.84   (i.e the  ratio of the ionic radius of the cation to the ionic radius of

                 the anion )

0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a  \\ \\  1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a

Also ; a =  664 pm

Then :

r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm

Therefore,  the ionic radius of the anion r^- = 312.52 \ pm

4 0
3 years ago
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