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2 years ago

HELPPP PLEASE!!

Chemistry

1 answer:

Rzqust [24]2 years ago

6 0

Answer:

Herbivores would increase causing producers to decrease.

Explanation:

If the carnivores are gone, herbivores wouldn't get eaten and increase, and more herbivores mean less producers available.

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Which of the fallowing elements is the most reactive?

Alexxx [7]

Which elements are most reactive?

The alkali metals, found in group 1 of the periodic table (formerly known as group IA), are very reactive metals that do not occur freely in nature. These metals have only one electron in their outer shell. Therefore, they are ready to lose that one electron in ionic bonding with other elements.

Fluorine is the most reactive

5 0

4 years ago

The diagram below shows changes to the nucleus of an atom.

anzhelika [568]

Answer:

maybe 1 or 3 im not sure

Explanation:

i didn't study it yet sorry for not helping but try asking someone else

8 0

3 years ago

Read 2 more answers

The combustion of one mole of liquid ethanol, CH3CH2OH, produces 1367 kJ of heat. Calculate how much heat is produced when 235.0

deff fn [24]

Answer:- 6984 kJ of heat is produced.

Solution:- From given information, 1367 kJ of heat is produced by the combustion of 1 mole of ethanol. We are asked to calculate the heat produced by the combustion of 235.0 g of ethanol.

Let's convert given grams to moles and multiply by the heat produced by one mole of ethanol to get the total heat produced. Molar mass of ethanol is 46 grams per mole. The set will be:

$235.0g(\frac{1mole}{46g})(\frac{1367 kJ}{1mole})$

= 6984 kJ

So, 6984 kJ of heat is produced by the combustion of 235.0 g of liquid ethanol.

6 0

4 years ago

Draw a line graph showing the population of beetles and healthy trees in the forest for five years. Round the numbers to the nea

Sergio [31]

how do i draw a line ghraph on a computer

8 0

3 years ago

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Consider the insoluble compound silver bromide , AgBr . The silver ion also forms a complex with ammonia . Write a balanced net

Degger [83]

Answer:

- $AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)$

- $K=1.2x10^{-5}$

Explanation:

Hello,

In this case, by considering the dissolution of silver bromide:

$AgBr(s)\rightleftharpoons Ag^+(aq)+Br^-(aq) \ \ \ Ksp=[Ag^+][Br^-]=7.7x10^{-13}$

And the formation of the complex:

$Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)\ \ \ Kf=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.6x10^7$

We obtain the balanced net ionic equation by adding the aforementioned equations:

$AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)+Ag^+(aq)\\\\AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)$

Now, the equilibrium constant is obtained by writing the law of mass action for the non-simplified net ionic equation:

$AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-+Ag^+\\\\K=[Ag^+][Br^-]*\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}$

So we notice that the equilibrium constant contains the solubility constant and formation constant for the initial reactions:

$K=Ksp*Kf=7.7x10^{-13}*1.6x10^{7}\\\\K=1.2x10^{-5}$

Best regards.

4 0

3 years ago