Answer:
Mass of chemical = 1.5 mg
Explanation:
Step 1: First calculate the concentration of the stock solution required to make the final solution.
Using C1V1 = C2V2
C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution
C1 = C2V2/V1
C1 = (6 * 25)/ 0.1
C1 = 1500 ng/μL = 1.5 μg/μL
Step 2: Mass of chemical added:
Mass of sample = concentration * volume
Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL
Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg
Therefore, mass of sample = 1.5 mg
Answer:
The symbol is the right answer.
Explanation:
The “ Symbol” is the correct answer because chemist uses the letters of the alphabet to denote the element. For instance, the element oxygen is denoted by the letter of the alphabet “O”, the hydrogen is denoted by the letter of alphabet “H”, Boron is denoted by the letter of alphabet “B”, etc. Here these are the examples that use one letter but there are other elements that use more than 1 letter as the symbol. For example, the Chlorine is represented by the Cl.
Answer:
Water will boil at
.
Explanation:
According to clausius-clapeyron equation for liquid-vapor equilibrium:
![ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%29%3D%5Cfrac%7B-%5CDelta%20H_%7Bvap%7D%5E%7B0%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D-%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
where,
and
are vapor pressures of liquid at
(in kelvin) and
(in kelvin) temperatures respectively.
Here,
= 760.0 mm Hg,
= 373 K,
= 314.0 mm Hg
Plug-in all the given values in the above equation:
![ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B314.0%7D%7B760.0%7D%29%3D%5Cfrac%7B-40.7%5Ctimes%2010%5E%7B3%7D%5Cfrac%7BJ%7D%7Bmol%7D%7D%7B8.314%5Cfrac%7BJ%7D%7Bmol.K%7D%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D-%5Cfrac%7B1%7D%7B373K%7D%5D)
or, 
So, 
Hence, at base camp, water will boil at 
<h3>
Answer:</h3>
138 g SO₂
<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>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 2.16 moles SO₂
[Solve] grams (mass) SO₂
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of S - 32.07 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of SO₂ - 32.07 + 2(16.00) = 64.07 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
138.391 g SO₂ ≈ 138 g SO₂
Answer:
MARK AS THE BRAINLIEST ANSWER
Explanation:
BRAINLIEST