Answer:
The density of acetic acid at 30°C = 1.0354_g/mL
Explanation:
specific gravity of acetic acid = (Density of acetic acid at 30°C) ÷ (Density of water at 30°C)
Therefore, the density of acetic acid at 30°C = (Density of water at 30°C) × (Specific gravity of acetic acid at 30°C)
= 0.9956 g/mL × 1.040
= 1.0354_g/mL
Specific gravity, which is also known as relative density, is the ratio of the density of a substance to the density of a specified standard substance.
Generally the standard substance of to which other solid and liquid substances are compared is water which has a density of 1.0 kg per litre or 62.4 pounds/cubic foot at 4 °C (39.2 °F) while gases are normally compared with dry air, with a density of 1.29 grams/litre or 1.29 ounces/cubic foot under standard conditions of a temperature of 0 °C and one standard atmospheric pressure
Answer:
<em><u>The three-dimensional region of space that indicates where there is a high probability of finding an electron.</u></em>
The metalloids are mostly concentrated in groups 14, 15, and 16. (Some simpler charts will show them as 4A, 5A, and 6A - take a look at the top of the periodic table your class uses to double-check).
If you like my answer, please vote me a 'brainliest' - trying to improve my rank :-)
Answer:
Head loss in turbulent flow is varying as square of velocity.
Explanation:
As we know that head loss in turbulent flow given as
Where
F is the friction factor.
L is the length of pipe
V is the flow velocity
D is the diameter of pipe.
So from above equation we can say that
It means that head loss in turbulent flow is varying as square of velocity.
We know that loss in flow are of two types
1.Major loss :Due to surface property of pipe
2.Minor loss :Due to change in momentum of fluid.
Answer:
V₂ = 22.23 mL
Explanation:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Given data:
Initial volume = 25 mL
Initial pressure = 725 mmHg (725/760 =0.954 atm)
Initial temperature = 20 °C (20 +273 = 293 K)
Final pressure = standard = 1 atm
Final temperature = standard = 273.15 K
Final volume = ?
Solution:
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 0.954 atm × 25 mL × 273.15 K / 293 K × 1 atm
V₂ = 6514.63 mL . atm . K / 293 K . atm
V₂ = 22.23 mL
