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GalinKa [24]
2 years ago
6

Calculate the ph of a 0.20 m solution of iodic acid (hio3, ka = 0.17).

Chemistry
1 answer:
saveliy_v [14]2 years ago
7 0
Iodic acid partially dissociates into H+ and IO3- 
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will  be released as that of H+, its concentration is also X.  The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3]; 
<span>Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; </span>
<span>At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; </span>
<span>so 0.17 = x² / (0.20 - x); </span>
<span>Solving for x using the quadratic formula: </span>
<span>x = [H+] = 0.063 M or pH = - log [H+] = 1.2.</span>
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Boron - 10 emits alpha particles and cesium - 137 emits beta particles. Write balanced nuclear
anzhelika [568]

Answer:

B10  5N +5P= Li6 3N +3P

Cs 137 82N+55P = I 133  80N + 53P

Explanation:

8 0
2 years ago
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How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C
NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0
1 year ago
6.74 g of the monoprotic acid KHP (MW = 204.2 g/mol) is dissolved into water. The sample is titrated with a 0.703 M solution of
Dennis_Churaev [7]

Answer:

Volume of the calcium hydroxide solution used is 0.0235 mL.

Explanation:

2KHP+Ca(OH)_2\rightarrow 2H_2O+Ca(KP)_2

Moles of KHP = \frac{6.74 g}{204.2 g/mol}=0.0330 mol

According to reaction, 2 moles of KHP  with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;

\frac{1}{2}\times 0.0330 mol=0.01650 mol of calcium hydroxide

Molarity of the calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

Molarity=\frac{Moles}{Volume(L)}

0.703 M=\frac{0.01650 M}{V}

V=\frac{0.01650 M}{0.703 M}=0.0235 mL

Volume of the calcium hydroxide solution used is 0.0235 mL.

4 0
2 years ago
2. Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:
Nesterboy [21]

Answer

  • increase in temperature
  • decrease in pressure
  • continuous removal of PH3
  • adding more of P into the system

Explanation:

        In the reaction   P4(g)+6H2(g) ⇌ 4PH3(g);

  • The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:

        heat+A⇌BΔH=+

  • Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.
  • According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.

4 0
1 year ago
How many atoms of zirconium are in 0.3521 mol of zirconium?
lora16 [44]

Answer:

2.12×10²³ atoms.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This simply means that 1 mole of zirconium also 6.02×10²³ atoms.

Thus, we can obtain the number of atoms present in 0.3521 mole of zirconium as follow:

1 mole of zirconium also 6.02×10²³ atoms.

Therefore, 0.3521 mole of zirconium will contain = 0.3521 × 6.02×10²³ = 2.12×10²³ atoms.

Therefore, 0.3521 mole of zirconium contains 2.12×10²³ atoms.

3 0
2 years ago
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