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3 years ago

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What are the covalent bonds called that form between atoms of different electronegativities?

1 answer:

ZanzabumX [31]3 years ago

5 0

Polar bonds are formed between atoms of different electronegativities.

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When we experience higher than normal low tides and lower than normal high tides we are experiencing?

DanielleElmas [232]

Answer: a full or new moon

Explanation:

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3 years ago

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PLEASE HELP: You have been given 0.507 moles of cesium fluoride (CsF), determine the mass in grams of cesium fluoride that you h

Degger [83]

Answer:

C because 77.0 CsF

Explanation:

That is the correct answer because I have that homework and i got it right

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2 years ago

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Volgvan

Answer:

The first one is air the second is decreases the third is water the fourth is gas and the last is liquid.

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3 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.

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3 years ago

Oxidation-reduction reactions (often called "redox" for short) are reactions that involve the transfer of electrons from one spe

Arturiano [62]

A. Fe2O3 + 3CO= 2Fe+3CO2
Here element oxidised is CO or Carbon Monoxide, since oxygen is added.

B. 2HCl+2KMnO4+3H2C2O4=6CO2+2MnO2+2KCl+4H2O
Here Element reduced is 3H2C2O4, since Hydrogen is being added. Also KMnO4 is reduced, since Oxygen is removed.

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4 years ago