Answer:
Explanation:
Nitrogen gas in nitrogen cylinder is in gaseous form only . So when gas is released , pressure decreases because moles of gas in the cylinder decreases .
Propane gas in cylinder is in liquified form . The gas is partially in gaseous and liquid form . So when gas is released outside , pressure decreases inside but due to low pressure , some liquid form is converted into gaseous form which restores the pressure of gas inside cylinder . Hence , the pressure of gas remains unchanged inside cylinder.
Answer:
2MnO2 + H2O => Mn2O3 + 2OH is the correct balanced equation.
Answer:

Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble the data in one place.
2Mg + O₂ ⟶ 2MgO
n/mol: 2 5
Calculate the moles of MgO we can obtain from each reactant.
From Mg:
The molar ratio of MgO:Mg is 2:2

From O₂:
The molar ratio of MgO:O₂ is 2:1.

Answer:
Total energy required to raise the temperature of 425 g of tin from 298.15 K to 505.05 K and to melt the tin at 505.05 K is 45.249 kiloJoules.
Explanation:
Mass of the tin ,m= 425 g
Heat capacity of the tin ,c= 0.227 J/g K
Initial temperature of the tin ,
= 25.0 °C = 298.15 K
Final temperature of the tin,
= 231.9 °C = 505.05 K
Let the heat required to change the temperature of tin from 298.15 K to 505.05 K be Q.


Heat required to melt tin at 505.05 K be Q'
The heat of fusion of tin metal =

Total energy required to raise the temperature of 425 g of tin from 298.15 K to 505.05 K and to melt the tin at 505.05 K is:
= Q+Q' = 19.961 kJ + 25.288 kJ = 45.249 kJ
Answer:
2.7 g/cm³
Explanation:
Step 1: Calculate the mass of kerosene
The mass of the full beaker (mFB) is equal to the sum of the masses of the empty beaker (mEB) and the mass of the kerosene (mK).
mFB = mEB + mK
mK = mFB - mEB
mK = 60 g - 20 g = 40 g
Step 2: Calculate the density of kerosene
Density (ρ) is an intrinsic property of matter. It can be calculated as the quotient between the mass of kerosene and its volume.
ρ = m/V
ρ = 40 g/15 cm³ = 2.7 g/cm³