Answer:
2.28 × 10^-3 mol/L
Explanation:
The equation for the equilibrium is
CN^- + H2O ⇌ HCN + OH^-
Ka = 4.9 × 10^-10
KaKb = Kw
4.9 × 10^-10 Kb = 1.00 × 10^-14
Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5
Now, we can set up an ICE table
CN^- + H2O ⇌ HCN + OH^-
I/(mol/L) 0.255 0 0
C/(mol/L) -x +x +x
E/(mol/L) 0.255 - x x x
Ka = x^2/(0.255 - x) = 2.05 × 10^-5
Check for negligibility
0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255
x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6
x = sqrt(5.20 × 10^-6) = 2.28 × 10^-3
[OH^-] = x mol/L = 2.28 × 10^-3 mol/L
There are 19.5 g Na in 71.4 g NaHCO₃
Calculate the <em>molecular mass of NaHCO₃</em>.
1 Na = 1 × 22.99 u = 22.99 u
1 H = 1 × 1.008 u = 1.008 u
1 C = 1 × 12.01 u = 12.01 u
3 O = 3 × 16.00 u = <u>48.00 u
</u>
TOTAL = 84.008 u
So, there are 22.99 g of Na in 84.008 g NaHCO₃.
∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na
Explanation:
Mass of the organic compound = 200g
Mass of carbon = 83.884g
Mass of hydrogen = 10.486g
Mass of oxygen = 18.640g
The mass of nitrogen = mass of organic compound - (mass of carbon + mass of hydrogen + mass of oxygen)
Mass of nitrogen = 200 - (83.884 + 10.486 + 18.64) = 200 - 113.01
Mass of nitrogen = 86.99g
The empirical formula of a compound is its simplest formula.
It is derived as shown below;
C H O N
Mass 83.884 10.486 18.64 86.99
molar
mass 12 1 16 14
Moles 83.884/12 10.486/1 18.64/16 86.99/14
6.99 10.49 1.17 6.21
Divide
by
lowest 6.99/1.17 10.49/1.17 1.17/1.17 6.21/1.17
6 9 1 5
Empirical formula C₆H₉ON₅
learn more:
Empirical formula brainly.com/question/2790794
#learnwithBrainly
Answer:
4.6 mol Si
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 2.8 × 10²⁴ atoms Si
[Solve] moles Si
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide [Cancel like units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
4.64962 mol Si ≈ 4.6 mol Si
Pyruvic acid is the end product of the Link reaction in respiration.
Hope that helps
