<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:
![\displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})](https://tex.z-dn.net/?f=%5Cdisplaystyle%203.77%20%5C%20mol%20%5C%20K_3N%28%5Cfrac%7B131.31%20%5C%20g%20%5C%20K_3N%7D%7B1%20%5C%20mol%20%5C%20K_3N%7D%29)
- Multiply/Divide:
![\displaystyle 495.039 \ g \ K_3N](https://tex.z-dn.net/?f=%5Cdisplaystyle%20495.039%20%5C%20g%20%5C%20K_3N)
<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
V ( H2SO4) = 35 mL / 1000 => 0.035 L
M ( H2SO4) = ?
V ( NaOH ) = 25 mL / 1000 => 0.025 L
M ( NaOH ) = 0.320 M
number of moles NaOH:
n = M x V
n = 0.025 x 0.320 => 0.008 moles of NaOH
Mole ratio:
<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
2 moles NaOH ---------------------- 1 mole H2SO4
0.008 moles moles NaOH ---------- ??
0.008 x 1 / 2 => 0.004 moles of H2SO4 :
Therefore:
M ( H2SO4) = n / V
M = 0.004 / 0.035
= 0.114 M
hope this helps!
(B), because 1.0 moles would be 6.02 x 10^23 molecules. So you have half a mole.<span>
</span>
Answer:
The correct answer is option B.
Explanation:
As given ,that 30.24 mL of base was neutralize by 86.14 mL of acid which means that moles of base present in 30.24 mL are neutralized by moles of acid present in 86.14 mL.
After dilution of base from 30.24 mL to 50.0 mL .Since, the moles of base are same in the solution as that of the moles in solution before dilution. Moles of acid require to neutralize the base after dilution will same as a that of present moles of acid present in 86.14 mL.
Answer:
Mass: 981.0 g
Density: 5.61 g/cm^3
Hardness: = 2.5 - 3
Unknown material: Chalcocite
Explanation: