Boyle's Law
Pressure 1 × Volume 1 = Pressure 2 × Volume2
or P1×V1= P2×V2
Finding Volume 2
(5.75atm)(5.14L) = (2.49atm)(V2)
Then Multiply
29.55atm•L = 2.49atm(V2)
Divided by 2.49
29.56atm•L / 2.49atm = 2.49atm(V2) / 2.49atm
V2= 11.87 L
Hello there.
<span>Which VIIA element doesn't have stable isotopes in nature?
</span>Db. At
<h3>
Answer:</h3>
2.999 mol Br
<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>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
1.806 × 10²⁴ molecules Br
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
<u />
= 2.999 mol Br
<u>Step 4: Check</u>
<em>We are given 4 sig figs. Follow sig fig rules and round.</em>
Our final answer is already in 4 sig figs, so there is no need to round.
Answer: 
Explanation: Hydrocarbons are the compounds which contain only carbon and hydrogen as their constituent elements. These are considered to be as the non polar species as because of the less electronegativity difference between the two components.
Now as it is a non polar species thus it will dissolve only a non polar solute following the principle of 'Like dissolves Like'. Thus the most non polar species given in the options is pentane. Cyclohexane will definitely dissolve this pentane as both are non polar in nature.
, HI and NaBr are polar specie which cannot be dissolved in non polar species.
Answer is: volume of oxygen is 4.63 liters.
Balanced chemical reaction: 2C + O₂ → 2CO.
m(C) = 4.50 g.
n(C) = m(C) ÷ M(C).
n(C) = 4.50 g ÷ 12 g/mol.
n(C) = 0.375 mol.
From chemical reaction: n(C) : n(O₂) = 2 : 1.
n(O₂) = 0.1875 mol.
T = 48°C = 321.15 K.
p = 810 mmHg ÷ 760 mmHg/atm= 1.066 atm.
<span>R = 0.08206
L·atm/mol·K.
Ideal gas law: p·V = n·R·T.</span>
V(O₂) =
n·R·T / p.<span>
V(O₂) =
0.1875 mol · 0.08206 L·atm/mol·K · 321.15 K / 1.066 atm.</span><span>
V(O₂<span>) =
4.63 L.</span></span>