Answer:
A.
Explanation:
Physical properties are characteristics of a substance that can be observed or measured without changing the composition of the substance. Some examples of physical properties are its volume and its density. There are two types of physical properties, intensive physical properties and extensive physical properties.
Answer:
For oxygen: mass % O = (mass of 1 mol of oxygen/mass of 1 mol of CO2) x 100. mass % O = (32.00 g / 44.01 g) x 100. mass % O = 72.71 %
Explanation:
Answer is: 56 gallons of
70% antifreeze and 84 gallons of 95% antifreeze.
ω₁ = 70% ÷ 100% = 0.7; 70% pure antifreeze.
ω₂ = 95% ÷ 100% = 0.95.
ω₃<span> = 85% ÷ 100% = 0.85.
V</span>₁ = ?; volume of 70% antifreeze.
V₂ = ?; volume of 95% antifreeze.<span>
V</span>₃ = V₁ + V₂<span>.
V</span>₃ = 140 gal.
V₁ = 140 gal - V₂<span>.
ω</span>₁ · V₁ + ω₂ ·V₂ = ω₃ · V₃.
0.70 · (140 gal -
V₂) + 0.95 · V₂ = 0.85 · 140 gal.
98 gal - 0.7V₂ + 0.95V₂ = 119 gal.
0.25V₂ = 21 gal.
V₂ = 21 gal ÷ 0.25.
V₂ = 84 gal.
V₁ = 140 gal - 84 gal.
V₁ = 56 gal.
Should this not be in biology? But the answer is Endoplasmic reticulum
Answer:
0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.
Explanation:
Step 1: Data given
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = Shows how much the boiling point increases
⇒i = the van't Hoff factor: Says in how many particles the compound will dissociate
⇒ Since all are aqueous solutions Kb for all solutions is the same (0.512 °C/m)
⇒m = the molality
Step 2:
0.20 m glucose
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for glucose = 1
⇒ Kb = 0.512 °C/m
⇒m = 0.20 m
ΔT = 1*0.512 * 0.20
<u>ΔT = 0.1024 °C</u>
0.30 m BaCl2
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for BaCl2 = Ba^2+ + 2Cl- : i = 3
⇒ Kb = 0.512 °C/m
⇒m = 0.30 m
ΔT = 3*0.512 * 0.30
<u>ΔT = 0.4608 °C</u>
0.40 m NaCl
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for NaCl = Na+ + Cl- : i = 2
⇒ Kb = 0.512 °C/m
⇒m = 0.40 m
ΔT = 2*0.512 * 0.40
<u>ΔT = 0.4096 °C</u>
0.50 m Na2SO4.
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for Na2SO4 = 2Na+ + SO4^2- : i =3
⇒ Kb = 0.512 °C/m
⇒m = 0.50 m
ΔT = 3*0.512 * 0.50
<u>ΔT = 0.768 °C</u>
0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.
