Answer:
a. 3.72 [atm]
Explanation:
For a gas at constant temperature, (with no change in number of molecules of the gas), we can apply Boyle's Law: ![P_1V_1=P_2V_2](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2)
![(1.556[atm])(268.5[mL])=P_2(112.4[mL])](https://tex.z-dn.net/?f=%281.556%5Batm%5D%29%28268.5%5BmL%5D%29%3DP_2%28112.4%5BmL%5D%29)
![\dfrac{(1.556[atm])(268.5[mL\!\!\!\!\!\!\!\!{--}])}{112.4[mL \!\!\!\!\!\!\!\!{--}]}=\dfrac{P_2(112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})}{112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%281.556%5Batm%5D%29%28268.5%5BmL%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B--%7D%5D%29%7D%7B112.4%5BmL%20%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B--%7D%5D%7D%3D%5Cdfrac%7BP_2%28112.4%5BmL%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B-----%7D%29%7D%7B112.4%5BmL%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B-----%7D%7D)
![3.716957[atm]=P_2](https://tex.z-dn.net/?f=3.716957%5Batm%5D%3DP_2)
It seems like the answer should have 4 significant figures since all of the other quantities have 4 significant figures, but the closest answer choice of those provided is a. 3.72
Answer:
15 mL of the solute
Explanation:
From the question given above, the following data were obtained:
Solution = 50 mL
Solvent = 35 mL
Solute =?
Solution is simply defined as:
Solution = solute + solvent
With the above formula, we can easily obtain the solute in the solution as follow:
Solution = 50 mL
Solvent = 35 mL
Solute =?
Solution = solute + solvent.
50 = solute + 35
Collect like terms
50 – 35 = solute
15 = solute
Solute = 15 mL
Therefore, 15 mL of the solute is required.
<h3>
Answer:</h3>
0.024 kg CaO
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Aqueous Solutions</u>
- Molarity = moles of solute / liters of solution
<u>Atomic Structure</u>
- Reading a Periodic Tables
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
0.41 mol CaO
2.5 M Solution
<u>Step 2: Identify Conversions</u>
1000 g = 1 kg
Molar Mass of Ca - 40.08 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of CaO - 40.08 + 16.00 = 56.08 g/mol
<u>Step 3: Convert</u>
- Set up:
![\displaystyle 0.43 \ mol \ CaO(\frac{56.08 \ g \ Cao}{1 \ mol \ CaO})(\frac{1 \ kg \ CaO}{1000 \ g \ CaO})](https://tex.z-dn.net/?f=%5Cdisplaystyle%200.43%20%5C%20mol%20%5C%20CaO%28%5Cfrac%7B56.08%20%5C%20g%20%5C%20Cao%7D%7B1%20%5C%20mol%20%5C%20CaO%7D%29%28%5Cfrac%7B1%20%5C%20kg%20%5C%20CaO%7D%7B1000%20%5C%20g%20%5C%20CaO%7D%29)
- Multiply:
![\displaystyle 0.024114 \ kg \ CaO](https://tex.z-dn.net/?f=%5Cdisplaystyle%200.024114%20%5C%20kg%20%5C%20CaO)
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>
0.024114 kg CaO ≈ 0.024 kg CaO
The information that a foliated metamorphic rock provide about the conditions it was formed under are:(for many diffrent rocks like Phyllite and Amphibolite)
you can tell by its features like if its smooth and sleek or if its rough and bumpy those physical features.