Explanation:
but.......you havent attached a picture?
Explanation:
This energy comes from the food we eat. Our bodies digest the food we eat by mixing it with fluids (acids and enzymes) in the stomach. When the stomach digests food, the carbohydrate (sugars and starches) in the food breaks down into another type of sugar, called glucose.
Answer:
Radius of the bubble reaches the surface of the water = 4.68 cm
Explanation:
We know Ideal gas equation PV = nRT
Since amount of gas and temperature remains constant
so p₁v₁ = p₂v₂
Volume of Spherical bubble with a radius of 3.0 cm
⇒ v₁=
⇒ v₁ = x 3.14 x
Pressure at the depth p₁ = 4 atm
Volume of bubble when it reaches the surface of water =
Pressure p₂ = 1 atm
p₁v₁ = p₂v₂
⇒ 4 x x 3.14 x = 1 x
⇒ r = 4.68 cm
Answer:
The heat needed to convert 1 kg of feed water at 20°C into dry saturated steam at a pressure of 9 bar is 2690.19 kJ/kg.
Explanation:
Step 1 : Obtain the enthalpy of staurated steam and enthalpy of evaporation at 9 bar pressure from the steam table:
From steam table, at 9 bar, the enthalpy of saturated water = 742.83 kJ /kg
enthalpy of evaporation = 2031.1 kJ /kg
Step 2: Calculate the enthalpy of dry saturated steam:
Enthalpy of dry saturated steam = enthalpy of saturated water + enthalpy of evaporation
= 742.83 + 2031.1 = 2773. 93 kJ/ kg
Step 3: Calculate the enthalpy of 1 kg of feed water at 20°C
Enthalpy of 1 kg of feed water = c * ( T2 -T1)
= 4.187 * (20-0)
= 83.74 kJ /kg
Step 4 : calculate the heat needed by 1 kg of feed water at 20°C to be converted to dry saturated steam at 9 bar:
Heat needed = Enthalpy of dry saturated steam - enthalpy of feed water
Heat needed = 2773.93 kJ/kg - 83.74 kJ/kg
Heat = 2690.19 kJ/kg
Answer:
Option a. 4 g of He
Explanation:
6.02×10²³ is the Avogadro's number. The amount of units that corresponds to 1 mol, in this case the units are molecules.
Let's find out which mass corresponds to 1 mol (mass / molar mass)
a. 4g / 4g/mol = 1 mol He
b. 14g / 6.94 g/mol = 2.01 mol Li
c. 80g /159.8 g/mol = 0.5 moles Br₂
d. 4g / 2g/mol = 2 mol H₂