Answer:
4,38%
small molecular volumes
Decrease
Explanation:
The percent difference between the ideal and real gas is:
(47,8atm - 45,7 atm) / 47,8 atm × 100 = 4,39% ≈ <em>4,38%</em>
This difference is considered significant, and is best explained because argon atoms have relatively <em>small molecular volumes. </em>That produce an increasing in intermolecular forces deviating the system of ideal gas behavior.
Therefore, an increasing in volume will produce an ideal gas behavior. Thus:
If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to <em>decrease</em>
<em />
I hope it helps!
Answer:
D
Explanation:
If the pressure remains constant then the temperature and Volume are all that you have to consider.
Givens
T1 = 19oC = 19 + 273 = 292o K
T2 = 60oC = 60 + 273 = 333oK
V1 = 250 mL
V2 = x
Formula
V1/T1 = V2/T2
250/292 = x/333
Solution.
The solves rather neatly. Multiplly both sides by 333
250*333 / 292 = 333 *x / 333
Do the multiplication
250 * 333 / 292 = x
83250 / 292 = x
Divide by 292
x = 285.1 mL
The answer is D
For this question, assume that you have 1 compound. This compound is divided in half once, so you are left with 0.5. That 0.5 that remains is divided in half again, this is the second half-life, and you are left with 0.25. The final half life involves dividing 0.25 in half, which means you are left with 0.125. For the answer to make sense, you need to know your conversions between decimals and fractions. To make it simple, if you have 0.125 and you times it by 8, you are left with your initial value of 1. Therefore, after three half-lives, you are left with 1/8th of the compound.
Answer:
a. 1.12 L
Explanation:
Step 1: Write the balanced equation for the photosynthesis
6 CO₂(g) + 6 H₂O(l) ⇒ C₆H₁₂O₆(s) + 6 O₂(g)
Step 2: Calculate the moles corresponding to 2.20 g of CO₂
The molar mass of CO₂ is 44.01 g/mol.
2.20 g × 1 mol/44.01 g = 0.0500 mol
Step 3: Calculate the moles of O₂ produced
The molar ratio of CO₂ to O₂ is 6:6. The moles of O₂ produced are 6/6 × 0.0500 mol = 0.0500 mol
Step 4: Calculate the volume occupied by 0.0500 moles of O₂ at STP
At STP, 1 mole of O₂ occupies 22.4 L.
0.0500 mol × 22.4 L/1 mol = 1.12 L
