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Angelina_Jolie [31]
3 years ago
9

(25points) Explain in your own words the difference between atomic mass and atomic number. Where can you find this information o

n the periodic table?
Chemistry
2 answers:
Fiesta28 [93]3 years ago
8 0

Answer:

No time for it Ill help soon qw q

Explanation: And yes you know that 2332423 .

Eduardwww [97]3 years ago
6 0

Answer:

Atomic mass= The number of protons and neutrons in the atom

Atomic Number= Just the number of protons

Explanation:

You might be interested in
B) Name three substances which can undergo sublimation and deposition respectively​
SpyIntel [72]

solid carbon dioxide, iodine, arsenic, and naphthalene

Explanation:

Examples of substances that undergo sublimation

Examples of solids that sublime are dry ice (solid carbon dioxide), iodine, arsenic, and naphthalene (the stuff mothballs are made of).

3 0
3 years ago
Find the number of moles of sodium hydroxide in 25cm3 solution of concentration 0,1 mol/dm3​
kozerog [31]

Answer:

0.0025  moles

Explanation:

25cm^3 = 25/1000 dm^3

Conc = mol/dm^3

Mol = conc * dm^3

Mol = 0.1 * 25/1000

3 0
2 years ago
2 A certain gas of 25 g at 25°c and 0.65 atm occupies a volume of 23.52L Determine the molecule mass of the gas.​
denpristay [2]

Answer:

{ \bf{PV= \frac{m}{M} RT}} \\  \\ { \tt{(0.65  \times 23.52)  =  \frac{25}{M}  \times 0.081 \times (25 + 273)}} \\  \\ M = 39.5 \: g

8 0
3 years ago
An undergraduate weighed out 20grams of sodium hydroxide pellets. If Na =23, O = 16 andH = 1, What is the mole of this sodium hy
tatuchka [14]
The answer would be .5 mols because you take the total amount of grams, which is 20, and you had up the molar mass of sodium hydroxide, which would be 40. After you have this you would set this up as a stochiometry equation. With 1 mol on top you dived 20/40 to cancel out your grams. This leaves you with .5 mols

3 0
3 years ago
The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
maw [93]
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
                       CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial                Y                   -                 -
Change           -X                  +X              +X
Equilibrium      Y-X                 X                X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

                Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
                    X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
                                                     = 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
                                                 = 5.79 x 10⁻³ g/L

7 0
3 years ago
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