All categories

4 years ago

Which atom will gain electrons during ionic bonding? a) Li b) F c) Mg d) K

Chemistry

1 answer:

Sphinxa [80]4 years ago

4 0

Answer:

b) F

Explanation:

Only non-metals can gain electrons during ionic bonding.

b) F

Li, Mg and K are metals.

You might be interested in

What scientist do that is the basis

UNO [17]

What does this question mean

5 0

3 years ago

Read 2 more answers

Is melting sugar a chemical or physical change? Please provide an explanation

liraira [26]

Answer:

Hola Amigo! Here's Ur Answer :D

Explanation:

The chemical reaction we are most familiar with is that of melting: sugar decomposes at a temperature ranging between 184 and 186°C. This is a very recent discovery we owe to a team of researchers in Illinois. Basically, when we heat sucrose gently, this produces a phenomenon known as “ apparent melting ”.

Happy to Help! :D

3 0

3 years ago

Read 2 more answers

The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti

lyudmila [28]

$NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}$

Here the base is a benzoate ion, which is a weak base and reacts with water.

$C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)$

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or $[OH^{-}] = 10^{^{-pOH}}$

$[OH^{-}] = 10^{^{-4.96}}$

$[OH^{-}] = 1.1\times 10^{-5}$

The base dissociation equation kb = $\frac{Product}{Reactant}$

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = $1.6\times 10^{-10}$

And value of [OH-] we have calculated as $1.1\times 10^{-5}$ and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

$1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}$

$[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}$

$[C6H5COO^{-}] = 0.76 M$ or $0.76\frac{mol}{L}$

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = $0.76(\frac{mol}{L}) \times (0.50L)$

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

8 0

3 years ago

Read 2 more answers

If the volame ofa gas coetainer at 32°C changes froem 1.55 L to 753 ml, what will the final temperature be? Assume the pressure

motikmotik

Answer : The final temperature of gas will be, 149 K

Explanation :

Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

$V\propto T$