Answer:
The answer to your question is M = 36.49 g
Explanation:
Data
mass = 8.21 g
volume = 4.8064 L
Temperature = 200°C
Pressure = 1.816 atm
M = ?
Process
1.- Convert temperature to °K
°K = 273 + 200
°K = 473
2.- Calculate the number of moles
n = (PV)/RT
n = (1.816)(4.8064)/(0.082)(473)
n = 0.225
3.- Calculate the molar mass
M --------------- 1 mol
8.21 g ---------- 0.225 moles
M = (1 x 8.21)/0.225
M = 36.49 g
Reactivity is a chemical
property of a substance. According to EPA regulations, it is normally unstable
and readily
undergoes violent change without
detonating. it can explode or violently react when exposed to water, when
heated, or under STP.
Answer:
The given atom is of Ca.
Explanation:
Given data:
Speed of atom = 1% of speed of light
De-broglie wavelength = 3.31×10⁻³ pm (3.31×10⁻³ / 10¹² = 3.31×10⁻¹⁵ m)
What is element = ?
Solution:
Formula:
m = h/λv
m = mass of particle
h = planks constant
v = speed of particle
λ = wavelength
Now we will put the values in formula.
m = h/λv
m = 6.63×10⁻³⁴kg. m².s⁻¹/3.31×10⁻¹⁵ m ×( 1/100)×3×10⁸ m/s
m = 6.63×10⁻³⁴kg. m².s⁻¹/ 0.099×10⁻⁷m²/s
m = 66.97×10⁻²⁷ Kg/atom
or
6.69×10⁻²⁶ Kg/atom
Now here we will use the Avogadro number.
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
Now in given problem,
6.69×10⁻²⁶ Kg/atom × 6.022 × 10²³ atoms/ mol × 1000 g/ 1kg
40.3×10⁻³×10³g/mol
40.3 g/mol
So the given atom is of Ca.
Answer:
6) solid- dots close together
liquid- a little bit farther apart than solid
gas- far apart
7) the movement of anything from a higher area of concentration to an area of lower concentration.
8) B. evaporation
9) The solute is the thing being dissolved, the solvent is the thing dissolving it, and the solution is the product of the solute and solvent.
10) D. regular brewed coffee
*I'm not exactly sure on #10 but I hope I could've helped a little at least
