(01.06 LC)

Hiii pls help me to write out the ionic equation

emmasim [6.3K]

Answer:

STEP I

This is the balanced equation for the given reaction:-

$2KOH_{(aq)} + H_2SO_4{}_{(aq)} \rightarrow K_2SO_4{}_{(aq)} + 2H_2O_{(l)}$

STEP II

The compounds marked with (aq) are soluble ionic compounds. They must be

broken into their respective ions.

see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).

On breaking them into their respective ions :-

2KOH -> 2K+ + 2OH-
H2SO4 -> 2H+ + (SO4)2-
K2SO4 -> 2K+ + (SO4)2-

STEP III

Rewriting these in the form of equation

$\underline{\pmb{2K^+} }+ 2OH^- + 2H^+ + \pmb{\underline{{SO_4{}^{2-}}} \: \rightarrow \: \underline{\pmb{2K^+}}} + \underline{\pmb{SO_4{}^{2-}}} + 2H_2O$

STEP IV

Canceling spectator ions, the ions that appear the same on either side of the equation

(note: in the above step the ions in bold have gotten canceled.)

$\boxed{ \mathfrak{ \red{ 2OH^-{}_{(aq)} + 2H^+{(aq.)} \rightarrow H_2O{}_{(l)}}}}$

This is the net ionic equation.

____________________________

$\\$

$\mathfrak{\underline{\green{ Why\: KOH \:has\: been\: taken\: as\: aqueous ?}}}$

KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.

[Alkali metal hydroxides are the only halides soluble in water ]

4 0

3 years ago

A strontium-90 atom that has a lost two electrons has __________ protons, __________ neutrons, and __________ electrons.

MAXImum [283]

Answer:
A strontium-90 atom that has a lost two electrons has 38 protons, 52 neutrons, and 36 electrons.

Explanation:
Atomic number of Strontium (Sr) is 38.
Atomic number = number of protons
Hence, Strontium has 38 protons.

If the element is in neutral state,
number of protons = number of electrons.
Then, neutral Strontium atom should have 38 electrons.
But the question says Sr has lost 2 electrons. Hence, number of electrons should be 38 - 2 = 36.

Mass number = number of protons + number of neutrons.

The given mass number is 90. Hence, number of neutrons should be 90 - 38 = 52.

4 0

3 years ago

What is the expected value for the heat of sublimation of acetic acid if its heat of fusion is 10.8 kJ/mol and its heat of vapor

Dennis_Churaev [7]

Answer:

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

Explanation:

$CH_3COOH(l)\rightarrow CH_3COOH(g)$ ..[1]

Heat of vaporization of acetic acid = $H^o_{vap}=24.3 kJ/mol$

$CH_3COOH(s)\rightarrow CH_3COOH(l)$ ..[2]

Heat of fusion of acetic acid = $H^o_{fus}=10.8 kJ/mol$

Heat of sublimation of acetic acid = $H^o_{sub}=?$

$CH_3COOH(s)\rightarrow CH_3COOH(g)$ ..[3]

[1] + [2] = [3] (Hess's law)

$H^o_{sub}=H^o_{vap}+H^o_{fus}$

$=24.3 kJ/mol+10.8 kJ/mol=35.1 kJ/mol$

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

5 0

3 years ago

Which three groups of the periodic table contain the most elements classified as metalloids (semimetals)?

attashe74 [19]

The metalloids are mostly concentrated in groups 14, 15, and 16. (Some simpler charts will show them as 4A, 5A, and 6A - take a look at the top of the periodic table your class uses to double-check).

If you like my answer, please vote me a 'brainliest' - trying to improve my rank :-)

8 0

3 years ago

Read 2 more answers

how would you write equation to show how buffers H2CO3 and NaHCO3 behave when (a) HCl is added and (b) NaOH is added.

Vlad1618 [11]

H2CO3 <---> H+ + HCO3-
NaHCO3 <---> Na+ + HCO3-

When acid is added in the buffer, the excess H+ of that acid reacts with HCO3- to form H2CO3, and due to this NaHCO3 dissociates into HCO3- to attain the equilibrium. and hence there is no net effect of H+ due to pH remain almost constant. when a base is added to the buffer, the OH- ion of base react eith H+ ion present in buffer, then to attain equilibrium of H+ ion, the H2CO3 dissociates to produce H+ ion, but now there is the excess of HCO3- due to which Na+ ion react with them to attain equilibrium of HCO3-. hence there is again no net change in H+ ion due to which pH remain constant.....

8 0

3 years ago