Answer:
The new volume of the balloon is 600cm³
Explanation:
Given parameters:
Initial Volume, which is V₁ = 200cm³
Initial temperature, T₁ = 60°C
Final temperature T₂ = 180°C
Final Volume V₂ =?
To solve this kind of problem, we apply one of the gas laws that shows the relationship between volume and temperature.
This law is the Charles law, it states that " the volume of a fixed mass of a gas is directly proportional to its absolute temperature if pressure is constant".
It is simply expressed as:
V₁/T₁ = V₂/T₂
Since our unknown is V₂, we make it the subject of the expression given above:
V₂ = V₁T₂/T₁
Now input the corresponding values and solve:
V₂ = 200 x 180 / 60
V₂ = 36000/60
V₂ = 600cm³
The new volume of the balloon is 600cm³
Density = mass / volume
D = 500 g / 50 cm³
D = 10 g/cm³
hope this helps!
Answer:
1.1 × 10² g
Explanation:
First, we will convert 1.0 L to cubic centimeters.
1.0 L × (10³ mL/1 L) × (1 cm³/ 1 mL) = 1.0 × 10³ cm³
The density of water is 1.0 g/cm³. The mass corresponding to 1.0 × 10³ cm³ is:
1.0 × 10³ cm³ × (1.0 g/cm³) = 1.0 × 10³ g
1 mole of water (H₂O) has a mass of 18 g, consisting of 2 g of H and 16 g of O. The mass of Hydrogen in 1.0 × 10³ g of water is:
1.0 × 10³ g H₂O × (2 g H/18 g H₂O) = 1.1 × 10² g
Answer:
44.8 L of N₂.
Explanation:
We'll begin by writing the balanced equation for the. This is illustrated below:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
1 mole of N₂ reacted to produce 2 moles of NH₃.
Therefore, Xmol of N₂ will react to produce 4 moles of NH₃ i.e
Xmol of N₂ = (1 × 4)/2
Xmol of N₂ = 2 moles
Thus, 2 moles of N₂ reacted to produce 4 moles of NH₃.
Finally, we shall determine the volume of N₂ required for the reaction. This can be obtained as follow:
1 mole of N₂ occupies 22.4 L at STP.
Therefore, 2 moles of N₂ will occupy = (2 × 22.4) = 44.8 L
Thus, 44.8 L of N₂ is needed to produce 4 moles of NH₃.