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noname [10]
2 years ago
15

Oxalic acid (h2c2o4) has a ka of 5. 9×10−2, while the hydrogen oxalate anion (hc2o−4) has a ka of 6. 1×10−5. Is an aqueous solut

ion of a hydrogen oxalate salt (e. G. Nahc2o4) acidic, basic, or neutral?
Chemistry
1 answer:
finlep [7]2 years ago
6 0

Resultant aqueous solution of a hydrogen oxalate salt is basic in nature.

<h3>What is pH?</h3>

pH scale is used to determine the acidity, basicity or neutral nature of any solution.

  • Values from 0 to 6.9 shows the acidity.
  • 7 shows the neutrality.
  • And 7.1 to 14 shows the basicity.

Chemical equation for the given salt in an aqueous solution will be shown as:

NaHC₂O₄ + H₂O → NaOH + H₂C₂O₄

From the equation it is clear that reaction produces-

  • NaOH which is a strong base and
  • H₂C₂O₄ oxalic acid which is a weak acid.

So that resultant solution is basic in nature.

Hence the aqueous solution of salt is basic.

To know more about basicity, visit the below link:
brainly.com/question/172153

#SPJ1

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The N2O4−NO2 reversible reaction is found to have the following equilibrium partial pressures at 100∘C. Calculate Kp for the rea
timofeeve [1]

Answer:

K_{p} for the reaction is 18.05

Explanation:

Equilibrium constant in terms of partial pressure (K_{p}) for this reaction can be written as-

                K_{p}=\frac{P_{NO_{2}}^{2}}{P_{N_{2}O_{4}}}

where P_{NO_{2}} and P_{N_{2}O_{4}} are equilibrium partial pressure of NO_{2} and N_{2}O_{4} respectively

Hence K_{p}=\frac{(0.095)^{2}}{(0.0005)} = 18.05

So, K_{p} for the reaction is 18.05

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4 years ago
Why potasium more reactive than lithium and sodium
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5 0
3 years ago
Which of these statements best answers the question: What is art?
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8 0
3 years ago
What happens to the speed of motion of the particles of a gas when a certain volume of the gas is heated at constant pressure? E
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8 0
3 years ago
Read 2 more answers
Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
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