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4 years ago

Which substance is amphiprotic?

Chemistry

1 answer:

Aleonysh [2.5K]4 years ago

4 0

Amphiprotic substances

An amphiprotic substance is a substance that can donate or accept a proton, H+

For a substance to be amphiprotic it must

<span>contain a hydrogen atom which is able to be donated to another chemical species.be able to accept a hydrogen ion from another species.</span>

Examples of amphiprotic species include H2O, HCO3-, HSO4-, HPO42- and H2PO4<span>-</span>

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3 years ago

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The ph of a solution for which [OH]= 1.0 x 10^-4 is?<br><br> A 10<br> B 14<br> C 11<br> D 9<br> E 7

zhenek [66]

To find pH, use the following formula ---> pH= - log [H+]

so first we need to calculate the [H+] concentration using the OH concentration. to do this, we need to use this formula--> 1.0x10-14= [H+] X [OH-], so we solve for H+ and plug in

[H+]= 1.0X10-14/[OH-]---> 1.0 x 10-14/ 1.0 x 10-4= 1.0 x 10-10

now that we have the H+ concentration, we can solve of pH

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3 years ago

7.The _____ theory of matter states that all particles of matter are in constant motion.

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4 years ago

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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3 years ago

A jar is filled with water, and the lid is closed tightly. The jar is kept at constant temperature. What happens at the surface

Katarina [22]

<h2>Hello!</h2>

The correct answer is A: Water molecules evaporate and condense at the same rate.

Evaporation is defined as the physical change from liquid to gas, and Condensation is the physical change from gas to liquid.

At any given temperature, these two processes occur at once, in a dynamic equilibrium.

When the lid is closed, evaporation occurs faster than condensation, and pressure increases. Then, when pressure reaches a critical value, condensation starts to occur faster than evaporation, until an equilibrium is reached, and the pressure of the water molecules in the gas phase is maximum for that temperature. The pressure at that point is called Vapor Pressure.

Have a nice day!

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3 years ago