When the temperature of a gas within a rigid container decreases, the particles on average move more slowly and do not collide with one another or the container walls as often. It is the the ideal gas law that applies in this situation and states that a decrease in the temperature of a gas also results in a decrease in the pressure. Thus, automobile tyres which have a colder gas in the winter sometimes need additional air to provide suffiicient pressure.
Answer:
D
Explanation:
We know that the
reaction catalyzing power of a catalyst ∝ surface area exposed by it
Given
volume V1= 10 cm^3
⇒
hence r= 1.545 cm
also, surface area S1= 
now when the sphere is broken down into 8 smaller spheres
S2= 8×4πr'^2
now, equating V1 and V2 ( as the volume must remain same )
and solving we get
r'= r/2
therefore, S2=
S2=
S2= 2S1
hence the correct answer is
. The second run has twice the surface area.
Using stoichiometry:
5.5 L of blood x (1000 mL/1L) x (15 g/100 mL) x (1 kg/1000 g) = 0.825 kg
Answer:
Perflutren
Explanation:
Perflutren is a fluorinated hydrocarbon and gaseous substance used as an imaging contrast agent.
Answer:
Explanation:
- For the balanced reaction:
<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).</em>
It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.
- Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:
no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.
- Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:
<em><u>Using cross multiplication:</u></em>
4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.
0.64 mol of Fe is needed to react with → ??? mol of O₂.
∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.
- Finally, we can get the volume of oxygen using the information:
<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1 mol of O₂ occupies → 22.4 L, at STP conditions.
0.48 mol of O₂ occupies → ??? L.
∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.
