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Naya [18.7K]
3 years ago
9

In the compound KHSO4, there is an ionic bond between the(1) KH+ and SO42- ions(2) KHSO3+ and O2- ions(3) K+ and HS- ions

Chemistry
2 answers:
anygoal [31]3 years ago
4 0
In an ionic compound, we have a reaction between a negative and a positive atom. 

Therefore, we automatically can conclude that our answer is (4) K+ and HSO4- ions
Hope I helped. :)
wolverine [178]3 years ago
3 0
The answer is (4) K+ and HSO4- ions, because the two radical ions bond to create KHSO4. 

It also makes sense chemically speaking, because KH+ and SO4^2- cannot create a non-charged molecule, and naturally speaking all compounds that are not ions want to balance out their charges.
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A sample with a molar mass of 34.00 g/mol is found to consist of 5.9783% Hydrogen and 94.0217% oxygen. Find it’s molecular formu
kozerog [31]

Answer:

The formula is H202 (hydrogen peroxide, known as hydrogen peroxide)

Explanation:

100%----34g

5, 9783%---X= (5, 9783%x 34g)/100% =2 g

1g---1 atom of H

2g----x= 2g x 1 atom of H/1g = 2 atom of H

100%----34g

94, 0217%---X= (94,0217%x 34g)/100% =32 g

16g---1 atom of 0

32g----x= 32g x 1 atom of 0/16g = 2 atom of 0

5 0
3 years ago
In response to action potentials arriving along the transverse tubules, the sarcoplasmic reticulum releasesA) acetylcholine.B) s
stealth61 [152]

Answer:

Calcium ions.

Explanation:

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Thus, the correct answer is option (D).

7 0
3 years ago
Very urgent/easy chem question easy points
aev [14]
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8 0
3 years ago
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replacing standard incandescent lightbulbs with energy-efficient compact fluorescent lightbulbs can save a lot of energy. Calcul
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3 years ago
The rate constant k for a certain reaction is measured at two different temperatures:
bogdanovich [222]

Answer:

Ea=5.5 Kcal/mole

Explanation:

Let rate constant are K_1  and K_2  at temperature T_1  and T_2

By using Arrhenius equation at two different two different temperature,

Log K_1/K_2 =E_a/2.303R*(1/T_2 -1/T_2 );T_1=273+376=649K  ;K_1=4.8*10^8;T_2=273+280=553K  ;K_2=2.3*10^8;R=2 cal/(mole.K);Log (4.8*10^8)/(2.3*10^8 )=E_a/2.303R*(1/553K-1/649); Log 4.8/2.3=E_a/2.303R*96/358897 ;0.32=E_a/2.303R*96/358897;E_a=(0.32*2.303R*358897)/96;  

By putting value of R=2 cal/mole.K

E_a=5510.265cal/mole;

By rounding off upto 2 significant figure;

E_a=5.5Kcal/mole;

8 0
3 years ago
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