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Sever21 [200]
3 years ago
8

Which can occur during a physical change? (Select all that apply.)

Chemistry
1 answer:
koban [17]3 years ago
7 0

Answer:

the shape of a piece of matter can change

You might be interested in
Vanadium has an atomic mass of 50.9415 amu. It has two common isotopes. One isotope has a mass of 50.9440 amu and a relative abu
grandymaker [24]

Answer:

Average atomic mass of the  vanadium = 50.9415 amu

Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975

Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu

Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025

Atomic mass of Isotope (II) of vanadium ,m' = ?

Average atomic mass of vanadium =

m × abundance of isotope(I) + m' × abundance of isotope (II)

50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025

m'= 49.944 amu

Explanation:

6 0
3 years ago
A match has about 21 milligrams of red phosphorus coating the tip. How many atoms of phosphorus is this?
MAXImum [283]

Answer:

4.083 * 10^20 atoms.

Explanation:

One Mole of phosphorus  contains 6.022 * 10^23 atoms (Avogadros number)'

Since 1 mole of Phosphorus  has a mass of  30.974 grams, 21 milligrams has

6.022 * 10^23  * 0.021 / 30.974

= 0.004083 * 10^23

= 4.083 * 10^20

3 0
3 years ago
Each 5-ml teaspoon of Extra Strength Maalox Plus contains 450 mg of magnesium hydroxide and 500 mg of aluminum hydroxide. How ma
Art [367]

Answer:

0.0347 moles of hydronium ions

Explanation:

The equation of the neutralization reaction between hydroxide and hydronium ions is given below:

H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)

From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.

The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:

Number of moles = mass / molar mass

Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol

Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol

Mass of magnesium hydroxide = 450 g = 0.45 g

Mass of aluminium hydroxide = 500 mg = 0.5 g

Moles of magnesium hydroxide = (0.45/58) moles

Moles of aluminium hydroxide = (0.5/78) moles

Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:

Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)

Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)

Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles

Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles

Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions

Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.

3 0
3 years ago
What would you need to do to calculate the molality of 10 mol of NaCl in 200
Makovka662 [10]
It will be the second one
8 0
2 years ago
What is the boiling point elevation constant, Kb, of diethyl ether if 38.2 g of the nonelectrolyte benzophenone, C6H5COC6H5, dis
kakasveta [241]

Answer: the boiling point elevation constant is 1.73^0C/m

Explanation:

Elevation in boiling point is given by:

\Delta T_b=i\times K_b\times m

\Delta T_b=T_b-T_b^0= = Elevation in boling point

i= vant hoff factor = 1 (for non electrolyte)

K_b =boiling point constant = ?

m= molality

\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (diethylether)= 330 g = 0.33 kg

Molar mass of solute (benzophenone)= 182 g/mol

Mass of solute (benzophenone) = 38.2 g

(35.7-34.6)^0C=1\times K_b\times \frac{38.2g}{182g/mol\times 0.33kg}

K_b=1.73^0C/m

Thus the boiling point elevation constant is 1.73^0C/m

3 0
2 years ago
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