Answer:
You must add 48.97 mL of water to make the 0.200 M diluted solution.
Explanation:
In chemistry, dilution is the reduction in concentration of a chemical in a solution. In other words, it is the process of reducing the concentration of solute in solution, simply adding more solvent to the solution.
In a dilution, the quantity or mass of the solute is not changed but only that of the solvent. As only solvent is being added, by not increasing the amount of solute the concentration of the solute decreases.
The expression for the dilution calculations is:
Cinitial* Vinitial = Cfinal* Vfinal
In this case:
- Cinitial= 12 M
- Vinitial= 0.830 mL
- Cfinal= 0.200 M
- Vfinal= ?
Replacing:
12 M*0.830 mL= 0.200 M*Vfinal
Solving:
Vfinal= 49.8 mL
Since 0.830 mL is the volume you initially have of HCl, the amount of water you must add is:
49.8 mL - 0.830 mL= 48.97 mL
<u><em>You must add 48.97 mL of water to make the 0.200 M diluted solution.</em></u>
Answer: The 3rd and 6th bullet point is the quantitative data.
Explanation: Quantitative data is expressed by NUMBERS and Qualitative data is expressed by WORDS. The 3rd and 6th one is correct because they both use numbers to compare how much time hummingbirds spent feeding on nectar.
Answer:
P₂ = 130.18 kPa
Explanation:
In this case, we need to apply the Gay-Lussack's law assuming that the volume of the container remains constant. If that's the case, then:
P₁/T₁ = P₂/T₂ (1)
From here, we can solve for the Pressure at 273 K:
P₂ = P₁ * T₂ / T₁ (2)
Now, all we need to do is replace the given data and solve for P₂:
P₂ = 340 * 273 / 713
<h2>
P₂ = 130.18 kPa</h2>
Hope this helps
Answer:
There is a mass of 154 Grams of Carbon Dioxide.
Explanation:
One mole is equal to 6.02 × 10^23 particles.
This means we have 1.05 X 10^24 total particles of Ethane.
Each ethane particle contains 2 carbon atoms.
If every particle of ethane is burned, we will end up with 2.10 x 10^24 molecules of Carbon Dioxide (Particles of Methane x 2, since each Methane particle contains 2 carbon atoms)
Carbon Dioxide has a molar mass of 44.01 g/mol
So if we take our amount of Carbon Dioxide molecules and divide it by 1 mole, ((2.10 x 10^24)/(6.02 x 10^23) = 3.49) we find that we have 3.49 moles of Carbon Dioxide.
Now all we need to do is multiply our moles of carbon dioxide(3.49) by it's molar mass(44.01) while accounting for significant digits.
What you should end up with is 154 Grams of Carbon Dioxide.
Hope this helps (And more importantly I hope I didn't make any errors in my math lol)
As a side note this is all assuming that this takes place at STP conditions.
