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antoniya [11.8K]

2 years ago

10

2. The table below shows the atomic numbers of elements of the periodic table represented by letters I to Q The letters are not

the actual chemical symbols for the elements. Element J Atomic number www I 且 } T H 3 Lithiur K 7 Witrogen 8 Select two elements which belong ta Olygin The same period of the periodic table Jk The same group of the period table. Select the element which: twitt form a divalent andon Reacts murst vigorously with water. 9 N Home 10 (Ink (1mk bºr (Ink 12 May Nashy 13 Aluminian C 14 St Veer

1 answer:

Komok [63]2 years ago

4 0

Explanation:

cudhdbsieehe eirbruwnee8ebeuw eodjsuww8

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Temperate grasslands have trees and shrubs <br> True or false

laila [671]

it should be True unless if I’m wrong

5 0

3 years ago

Read 2 more answers

WHO IS GOOD AT CHEMISTRY EQUATIONS? I NEED A QUESTION ANSWERED BY TONIGHT! ITS THE ONLY ONE IM STUCK ON:

sattari [20]

Answer:

374 atoms would be present in 22 units of aluminum sulfate.

Explanation:

Formula for Aluminium Sulfate : Al₂(SO₄)₃

Al = 2

S = 1

O = 4

(SO) = 3

∴ Number of atom in one unit is : 2 + (1+4)3 ==> 17

If there are 17 atoms in one unit, therefore;

Number of atoms in 22 units = 17 x 22

= 374 atoms

4 0

3 years ago

Why carbon nanotubes are so strong? Quick please urgent

Goryan [66]

Answer: Carbon nanotubes, especially multi walled carbon nanotubes, are so strong because they are a single chain of unbroken covalent carbon-carbon bonds.

Explanation:

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4 years ago

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0

3 years ago

Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t

Rasek [7]

Answer:

$4.6*10^{-14}$ M

Explanation:

Concentration of $Cu^{2+} = [Cu(NO_3)_2]$ = 0.020 M

Constructing an ICE table;we have:

$Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}$

Initial (M) 0.020 0.40 0

Change (M) - x - 4 x x

Equilibrium (M) 0.020 -x 0.40 - 4 x x

Given that: $K_f =1.7*10^{13}$

$K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}$

$1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}$

Since x is so small; 0.40 -4x = 0.40

Then:

$1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}$

$1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}$

$1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x$

$8.704*10^9-4.352*10^{11}x =x$

$8.704*10^9 = 4.352*10^{11}x$

$x = \frac{8.704*10^9}{4.352*10^{11}}$

$x = 0.0199999999999540$

$Cu^{2+}= 0.020 - 0.019999999999954$

$Cu^{2+} = 4.6*10^{-14}$ M

8 0

3 years ago