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sergey [27]
3 years ago
13

Write the net ionic equation for the reaction between hydrocyanic acid and potassium hydroxide. Do not include states such as (a

q) or (s).
Chemistry
1 answer:
eimsori [14]3 years ago
4 0

Answer:

The net ionic equation is as follows:

HCN(aq) + OH-(aq) ----> H20(l) + CN-(aq)

Explanation:

The reaction between Hydrocyanic acid, HCN, and sodium hydroxide is a neutralization reaction between a weak acid and a strong base.

Hydrocyanic acid being a weak acid ionizes only slightly, while sodium hydroxide being a strong base ionizes completely. The equation for the reaction is given below:

A. HCN(aq) + NaOH-(aq) ----> NaCN(aq) + H2O(l)

Since Hydrocyanic acid is written in the aqueous form as it ionizes only slightly and the ionic equation is given below:

HCN(aq) + Na+(aq)+OH-(aq) ----> Na+(aq)+CN-(aq) + H2O(l)

Na+ being a spectator ion is removed from the net ionic equation given below:

HCN(aq) + OH-(aq) ----> H20(l) + CN-(aq)

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Type the correct answer in the box. Spell all words correctly. In which system are both energy and matter exchanged with the sur
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Explanation:

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3 years ago
Complete and balance the reaction in acidic solution. equation: ZnS + NO_{3}^{-} -> Zn^{2+} + S + NO ZnS+NO−3⟶Zn2++S+NO Which
kotykmax [81]

Answer:

Balanced reaction: 3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O

S is oxidized and N is reduced.

NO_{3}^{-} is the oxidizing agent and ZnS is the reducing agent.

Explanation:

Reaction: ZnS+NO_{3}^{-}\rightarrow Zn^{2+}+S+NO

\Rightarrow Oxidation: ZnS\rightarrow Zn^{2+}+S

Balance charge: ZnS-2e^{-}\rightarrow Zn^{2+}+S  ...............(1)

\Rightarrow Reduction: NO_{3}^{-}\rightarrow NO

Balance H and O in acidic medium : NO_{3}^{-}+4H^{+}\rightarrow NO+2H_{2}O

Balance charge: NO_{3}^{-}+4H^{+}+3e^{-}\rightarrow NO+2H_{2}O ...............(2)

[3\timesEquation-(1)] + [ 2\timesEquation-(2)]:

3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O

Oxidation number of S increases from (-2) to (0) for the conversion of ZnS to S. Therefore S is oxidized.

Oxidation number of N decreases from (+5) to (+2) for the conversion of NO_{3}^{-} to NO. Therefore N is reduced.

NO_{3}^{-} consumes electron from ZnS. Therefore NO_{3}^{-} is the oxidizing agent and ZnS is the reducing agent.

7 0
4 years ago
The highest energy occupied molecular orbital in the f-f bond of the f2 molecule is _____
Morgarella [4.7K]
The basis of finding the answer to this problem is to know the electronic configuration of Fluorine. That would be: <span>[He] 2s</span>²<span> 2p</span>⁵. The valence electrons, which are the outermost electrons of the atom, are the ones that participate in bonding. <em>Since the highest orbital for F is 2p, that means the highest energy occupied would be 2.</em>
3 0
3 years ago
A 30.0g sample of O2 at standard temperature and pressure (STP) would occupy what volume in liters
Murrr4er [49]

Answer:

Explanation:

A 30.0g sample of O2 at standard temperature and pressure (STP) would occupy what volume in liters

PV =nRT

at STP  P= 1atm. T= 273 K

n is the number of moles.  O2 has a molar mass of 32.

30 gm of O2 is 30/32= 0.94 =n

PV = nRT

at STP: P= 1 atm, T=273 K, R is always 0.082 for P in atm and T in K

SO

1 X V = 0.94 X 0.082 X 273

using high school freshman algebra,

V= 0.94 X 0.082 X 273 = 21L

using high school algebra I,

V=

6 0
2 years ago
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