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3 years ago

Which of the following is a proton donor. a. Bronsted-Lowry Base 6 Bronsted-Lowry Acid c. Lewis Acid d. Lewis Base

Chemistry

1 answer:

Margarita [4]3 years ago

3 0

Answer:

Bronsted-Lowry acid

Explanation:

Bronsted-Lowry Base -

According to Bronsted-Lowry theory of acids and bases:

Acids are those substances which donate proton in the solution whereas bases are those substances which accept proton in the solution.

Consider an example,

$HCO_3^{-} + HOH\; \rightarrow H_2CO_3 + OH^{-}\\$

HCO3- behave as base as it accepts proton in the reaction whereas HOH behaves as acid as it releases proton.

According to Lewis theory of acids and bases:

Acids are electron acceptor and bases are electron donar. Acids are generally electron deficient species whereas Lewis are chemical species having lone pair of electrons.

For example, AlCl3 is behave as Lewis acid because of being electron deficient species.

NH3 is a Lewis base because of presence of lone pair of electrons.

Therefore, among given options, Bronsted-Lowry acid is the correct option.

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3 years ago

a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re

Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

moles of $H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61$

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

$M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL$

Putting values in above equation, we get:

$11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL$

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

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3 years ago

In the rock cycle, the process of weathering and erosion happens between:

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Sedimentary and metamorphic :)

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2 years ago

What is the concentration of the unknown h3po4 solution? the neutralization reaction ish3po4(aq)+3naoh(aq)→3h2o(l)+na3po4(aq) -g

hjlf

First, we need to get moles of NaOH:

when moles NaOH = volume * molarity

= 0.02573L * 0.11 M

= 0.0028 moles

from the reaction equation:

H3PO4(aq) + 3NaOH → 3 H2O(l) + Na3PO4(aq)

we can see that when 1 mol H3PO4 reacts with→ 3 mol NaOH

∴ X mol H3PO4 reacts with → 0.0028 moles NaOH

∴ moles H3PO4 = 0.0028 mol / 3 = 9.4 x 10^-4 mol

now we can get the concentration of H3PO4:

∴[H3PO4] = moles H2PO4 / volume

= 9.4 x 10^-4 / 0.034 L

= 0.028 M

6 0

3 years ago

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