<h3>
Answer:</h3>
495 g K₃N
<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>
3.77 mol K₃N
<u>Step 2: Identify Conversions</u>
Molar Mass of K - 39.10 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
495.039 g K₃N ≈ 495 g K₃N
Answer:
Erosion
Explanation:
Erosion is not a cause of metamorphism
The causes are mostly temperature, pressure, heat from the metamorphic rock
<span> weather is the condition of the climate in a particular place at a particular
time.</span>
Answer:
Volume will goes to increase.
Explanation:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
So when the temperature goes to increase the volume of gas also increase. Higher temperature increase the kinetic energy and molecules move randomly every where in given space so volume increase.
Now we will put the suppose values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 4.5 L × 348 K / 298 k
V₂ = 1566 L.K / 298 K
V₂ = 5.3 L
Hence prove that volume increase by increasing the temperature.
Answer:
The overview of the subject is outlined underneath in the summary tab.
Explanation:
- The molar ratio seems to be essentially a balanced chemical equilibrium coefficient that implies or serves as a conversion factor for the product-related reactants.
- This ratio just says the reactant proportion which reacts, but not the exact quantity of the reacting product. Consequently, the molar ratio should only be used to provide theoretical instead of just a definite mass ratio.
