What are the correct coefficients when this equation is balanced? K + Br2 KBr

Which of the following pairs lists a substance that can neutralize H2SO4 and the salt that would be produced from the reaction?

Bogdan [553]

The second option only.

LiOH, Li₂SO₄.

<h3>Explanation</h3>

A base neutralizes an acid when the two reacts to produce water and a salt.

Sulfuric acid H₂SO₄ is the acid here. There are more than one classes of bases that can neutralize H₂SO₄. Among the options, there are:

Metal hydroxides

Ca(OH)₂ and
LiOH.

Metal hydroxides react with sulfuric acid to produce water and the sulfate salt of the metal.

$\text{Ca}(\text{OH})_{\bf 2}+\text{H}_2\text{SO}_4 \to \textbf{Ca}\textbf{SO}_{\bf 4} +{\bf 2}\;\text{H}_2\text{O}$ .

The formula for calcium sulfate $\text{CaSO}_4$ in option A is spelled incorrectly. Why? The charge on each calcium $\text{Ca}^{2+}$ is +2. The charge on each sulfate ion ${\text{SO}_4}^{2-}$ is -2. Unlike $\text{Li}^{+}$ ions, it takes only one $\text{Ca}^{2+}$ ion to balance the charge on each ${\text{SO}_4}^{2-}$ ion. As a result, $\text{Ca}^{2+}$ and ${\text{SO}_4}^{2-}$ ions in calcium sulfate exist on a 1:1 ratio.

$2\;\text{LiOH} +\text{H}_2\text{SO}_4 \to \text{Li}_2\text{SO}_4 + 2\;\text{H}_2\text{O}$ .

Ammonia, NH₃

Ammonia NH₃ can also act as a base and neutralize acids. NH₃ exists as NH₄OH in water:

$\text{NH}_3 + \text{H}_2\text{O} \to \textbf{NH}_{\bf 4}\text{OH}$ .

The ion ${\text{NH}_4}^{+}$ acts like a metal cation. Similarly to the metal hydroxides, NH₃ (or NH₄OH) neutralizes H₂SO₄ to produce water and a salt:

$2\;\textbf{NH}_{\bf 4}\text{OH}+ \text{H}_2\text{SO}_4 \to (\textbf{NH}_{\bf 4})_2\text{SO}_4+2\;\text{H}_2\text{O}$ .

The formula of the salt (NH₄)₂SO₄ in the fourth option spelled the ammonium ion incorrectly.

As part of the salt (NH₄)₂SO₄, the ammonium ion NH₄⁺ is one of the products of this reaction and can't neutralize H₂SO₄ any further.

7 0

3 years ago

What are some examples of lipids?

Anestetic [448]

Saturated fat, milk, cheese, and meat.

3 0

3 years ago

What is used up in and stops a chemical reaction?

CaHeK987 [17]

B - limiting reactant

6 0

3 years ago

Read 2 more answers

what modifications to either or both of the half-cell reactions would kim need to make if she wanted to use them to produce equa

kompoz [17]

We have to add the both half cell equations and eliminate the number of electrons lost/gained.

<h3>What modification must Kim make to the equations?</h3>

The term redox reaction is a type of reaction that occurs when an electron is lost or gained in a reaction system. We can see that in this reaction, zinc looses two electron which are gained by copper.

If we want to obtain the equation 4.9 which is the overall equation of the redox reaction from the various half cell equations then we have to add the both half cell equations and eliminate the number of electrons lost/gained.

Learn kore about redox reaction:brainly.com/question/13293425

#SPJ1

7 0

1 year ago

Calcule la densidad del hidrógeno H2 en g/L a 327 mm Hg y 48ºc

tankabanditka [31]

<h3>The density of H₂ = 0.033 g/L</h3><h3>Further explanation</h3>

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm , N/m²

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m², v= m³)

T = temperature, Kelvin

n = N / No

n = mole

No = Avogadro number (6.02.10²³)

n = m / MW

m = mass

MW = molecular weight

For density , can be formulated :

$\tt \rho=\dfrac{P\times MW}{R\times T}$

P = 327 mmHg = 0,430263 atm

R = 0.082 L.atm / mol K

T = 48 ºC = 321.15 K

MW of H₂ = 2.015 g/mol

The density :

$\rho=\dfrac{0,430263\times 2.015 }{0.082\times 321.15}\\\\\rho=0.033~g/L$

4 0

3 years ago