<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.
Use the PV = nRT equation T is in Kelvins = 31 + 273 = 304 K
P(0.5) = (2.91)(0.0821)(304)
P(0.5) = 72.6289
P = 145.25 atm or 1.45x10^2 atm
Answer:
A fire extinguisher
Explanation:
A lab station usually contains equipment for the people working at the station to use. However, most rooms will only have one fire extinguisher for the whole room meaning it would most likely be located somewhere that is easy to access by all and not just a singular lab station.
~Hope this Helps!~
Answer:
Milk is essentially a colloidal dispersion of fat in water. ... However, the fact remains that the fat and water components cannot be mixed together from a solution. There are therefore, two distinct immiscible liquid phase's present, which is why it is a heterogeneous mixture.
Answer:
67.1%
Explanation:
Based on the chemical equation, if we determine the moles of sodium carbonate, we can find the moles of NaHCO₃ that reacted and its mass, thus:
<em>Moles Na₂CO₃ - 105.99g/mol-:</em>
6.35g * (1mol / 105.99g) = 0.0599 moles of Na₂CO₃ are produced.
As 1 mole of sodium carbonate is produced when 2 moles of NaHCO₃ reacted, moles of NaHCO₃ that reacted are:
0.0599 moles of Na₂CO₃ * (2 moles NaHCO₃ / 1 mole Na₂CO₃) = 0.1198 moles of NaHCO₃
And the mass of NaHCO₃ in the sample (Molar mass: 84g/mol):
0.1198 moles of NaHCO₃ * (84g / mol) = 10.06g of NaHCO₃ were in the original sample.
And percent of NaHCO₃ in the sample is:
10.06g NaHCO₃ / 15g Sample * 100 =
<h3>67.1%</h3>
