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3 years ago

What is POH of a solution containing hydrogen ion concentration of 0.008

Chemistry

1 answer:

kotykmax [81]3 years ago

6 0

Given-

Hydrogen ion concentration pH = 0.008

Now,

we know that pH + pOH = 14

pOH = 14-0.008

pOH = 13.992

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3 years ago

Anyway, which is a spectator ion involved in the reaction of k2cro4(aq) and ba(no3)2(aq)?

marissa [1.9K]

Nitrate ions $\text{NO}_3^{-}$ and potassium ions $\text{K}^{+}$ .

<h3>Explanation</h3>

Potassium dichromate undergoes a double-decomposition reaction with barium nitrate to produce barium dichromate and potassium nitrate. The reaction is possible due to the low solubility of barium dichromate that precipitate out of the solution readily after its production.

Start with the balanced chemical equation for this process:

$\text{K}_2\text{CrO}_4 \; (aq) + \text{Ba}(\text{NO}_3)_2 \; (aq) \to \text{Ba}(\text{CrO}_4)_2 \; (s) + 2 \; \text{KNO}_3 \; (aq)$

Rewrite the chemical equation as an ionic one; express all soluble salts- those with state symbol (aq)- as their constituting ions while leaving the insoluble (s) intact.

$2 \; \text{K}^{+}\; (aq)+ 2\; \text{CrO}_4^{-} \; (aq) + \text{Ba}^{2+} \; (aq) + 2 \; \text{NO}_3^{-} \; (aq) \\ \to \text{Ba}(\text{CrO}_4)_2 \; (s) + 2 \; \text{K}^{+} \; (aq) + 2 \; \text{NO}_3^{-} \; (aq)$

$\text{K}^{+}$ and $\text{CrO}_4^{-}$ are found on both sides of the equation by the same quantity. The two ions thus took no part in the net reaction and act as spectator ions.

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3 years ago

Calculate the energy released in each of the following fusion reactions. Give your answers in MeV. (a) 2H + 3H → 4He + n 17.54 C

Vikentia [17]

Answer:

a) E = 17.55 MeV

b) E = 18.99 MeV

c) E = 3.29 MeV

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) E = 4.075 MeV

Explanation:

Energy Released, $E = \triangle M * 931.5$

$\triangle M = \sum M_{product} - \sum M_{reactant}$

Mass of 1H, $M_{H} = 1.007823$

Mass of 2H, $M_{2H} = 2.0141u$

Mass of 3H, $M_{3H} = 3.016 u$

Mass of Helium, $M_{4He} = 4.002602u$

Mass of Beryllium, $M_{7Be} = 7.01693 u$

Mass of neutron, $M_{n} = 1.008664 u$

a) $2H + 3H \rightarrow 4He + n$

$\triangle M = (4M_{He} + M_{n} ) - (2M_{H} + 3 M_{H} )\\\triangle M = ( 4.0026 + 1.008664) - (2.0141 + 3.016 )\\\triangle M = -0.01884u$

Energy released,

$E = -0.01884 * 931.5\\E = -17.55 Mev$

Energy released = 17.55 MeV

b) $4He + 4He \rightarrow 7Be + n$

$\triangle M = (M_{7Be} + M_{n} ) - (M_{4He} + M_{4He} )\\\triangle M = ( 7.01693 + 1.008664) - (4.002602 + 4.002602 )\\\triangle M = 0.02039 u$

Energy released,

$E = 0.02039 * 931.5\\E = 18.99 Mev$

c) $2H + 2H \rightarrow 3 He$ + n

$\triangle M = (M_{3He} + M_{n} ) - (M_{2H} + M_{2H} )\\\triangle M = ( 3.016 + 1.008664) - (2.0141 + 2.0141 )\\\triangle M = -0.003536 u$

Energy released,

$E = -0.003536 * 931.5\\E = -3.29 Mev$

E = 3.29 MeV(Energy is released)

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) $2H + 2H \rightarrow 3H + 1H$

$\triangle M = (M_{3H} + M_{1H} ) - (M_{2H} + M_{2H} )\\\triangle M = ( 3.016 + 1.007825) - (2.0141 + 2.0141 )\\\triangle M = -0.00435 u$

$E = -0.00435 * 931.5\\E =-4.075 Mev$

E = 4.075 MeV ( Energy is released)

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3 years ago

to balance a chemical equation, only the subscript of reactant and product can be changed. true or false

Lostsunrise [7]

Answer:

False

Explanation:

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3 years ago

Explain why is sodium less reactive than potassium

leva [86]

Answer:

As potassium is larger than sodium, potassium's valence electron is at a greater distance from the attractive nucleus and is so removed more easily than sodium's valence electron. As it is removed more easily, it requires less energy, and can be said to be more reactive.

Explanation:

<em>Hope you're having a splendiferous day</em><em>.</em>

<em>Just a bored kid willing to help...</em>

8 0

3 years ago

What is POH of a solution containing hydrogen ion concentration of 0.008​

What is POH of a solution containing hydrogen ion concentration of 0.008