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Stolb23 [73]
3 years ago
5

Answer as many as you can please, 40 POINTS

Chemistry
1 answer:
ICE Princess25 [194]3 years ago
6 0
It would be to see your awnsers

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How can you write tht in mg ??
inessss [21]
Rris is how u right it Mg=? Because it ask u a question and then you put the equal sign and then you put the question mark because you don't know it yet
3 0
3 years ago
Read 2 more answers
Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.
lianna [129]

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

4 0
3 years ago
¿Qué es un orbital s, un orbital p, un orbital d, un orbital f y cómo se representa cada uno?
irina1246 [14]

Answer:

Cada uno de esos orbitals sons los differentes grupos en la tabla periodica.

Explanation:

3 0
3 years ago
Read 2 more answers
A buffer consists of 0.45 M CH3COOH (acetic acid) and 0.35 M CH3COONa. The Ka of acetic acid is 1.8 x 10-5 . a) Calculate the pH
Zigmanuir [339]

Answer:

A) pH of Buffer solution = 4.59

B) pH after 5.0 ml of 2.0 M NaOH have been added to 400 ml of the original    buffer solution = 4.65

Explanation:

This  is the Henderson-Hasselbalch Equation:

 pH = pKa + log\frac{[conjugate base]}{[acid]}

to calculate the pH of the following Buffer solutions.

8 0
3 years ago
_Fe2O3 + 2CO —> _Fe + _CO2
soldi70 [24.7K]
<h3>Answer:</h3>

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

<h3>Explanation:</h3>

Concept tested: Balancing of chemical equations

  • A chemical equation is balanced by putting appropriate coefficients on the products and reactants of the equation.
  • Balancing chemical equations ensures that chemical equations obey law of conservation of mass.
  • In this case; to balance the above equation we put the coefficients, 1, 3, 2, and 3 on the reactants and products.
  • Therefore; the balanced chemical equation for the reaction is;

      Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

8 0
3 years ago
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