If 0.400 moles CO and 0.400 moles O2 completely react, 17.604 grams of CO2 would be produced.
First, let us look at the balanced equation of reaction:
According to the equation, the mole ratio of CO and O2 is 2:1. But in reality, the mole ratio supplied is 1:1. Thus, CO is the limiting reactant while O2 is in excess.
Also from the equation, the ratio of CO consumed to that of CO2 produced is 1:1. Thus, 0.400 moles of CO2 would also be produced from 0.400 moles of CO.
Recall that: mole = mass/molar mass
Therefore, the mass in grams of CO2 that would be produced from 0.400 moles can be calculated as:
Mass = mole x molar mass
= 0.400 x 44.01
= 17.604 grams
More on calculating mass from number of moles can be found here: brainly.com/question/12513822
The number of atoms in one mole is same in both which is 6 x 10^23 ^23 means power 23
Answer:
3.1 kg
Explanation:
Step 1: Write the balanced combustion equation
C₈H₁₈ + 12.5 O₂ ⇒ 8 CO₂ + 9 H₂O
Step 2: Calculate the moles corresponding to 1.0 kg of C₈H₁₈.
The molar mass of C₈H₁₈ is 114.23 g/mol.
1.0 × 10³ g × 1 mol/114.23 g = 8.8 mol
Step 3: Calculate the moles of CO₂ produced from 8.8 moles of C₈H₁₈
The molar ratio of C₈H₁₈ to CO₂ is 1:8. The moles of CO₂ produced are 8/1 × 8.8 mol = 70 mol.
Step 4: Calculate the mass corresponding to 70 moles of CO₂
The molar mass of CO₂ is 44.01 g/mol.
70 mol × 44.01 g/mol = 3.1 × 10³ g = 3.1 kg
Just use the Heisenberg Uncertainty principle:
<span>ΔpΔx = h/2*pi </span>
<span>Δp = the uncertainty in momentum </span>
<span>Δx = the uncertainty in position </span>
<span>h = 6.626e-34 J s (plank's constant) </span>
<span>Hint: </span>
<span>to calculate Δp use the fact that the uncertainty in the momentum is 1% (0.01) so that </span>
<span>Δp = mv*(0.01) </span>
<span>m = mass of electron </span>
<span>v = velocity of electron </span>
<span>Solve for Δx </span>
<span>Δx = h/(2*pi*Δp) </span>
<span>And that is the uncertainty in position. </span>
