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3 years ago

14

Please help will mark brainliest!

1 answer:

Llana [10]3 years ago

8 0

Answer:

1. desert

2. grasslands

3. rainforest

4. tundra

5. temperate deciduous forest

6. coniferous forest

Explanation:

educated guess, good luck

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What does relative abundance mean

fgiga [73]

Relative<span> species </span>abundance<span> is a component of biodiversity and refers to how common or rare a species is </span>relative<span> to other species in a defined location or community. </span>Relative abundance<span> is the percent composition of an organism of a particular kind </span>relative<span> to the total number of organisms in the area.

I don't know if your looking for this exact definition so i'm sorry if i'm wrong

</span>

4 0

3 years ago

An erlenmeyer flask contains 15.00mL of 0.030 M HCI before titration. 5.00 mL of 0.050 of M NaOH is added to the HCI in the flas

e-lub [12.9K]

Answer:

1.1

0.050*0.005

0.00025

Explanation:

8 0

3 years ago

1. If temperature is increased, the number of<br> collisions per second

andrey2020 [161]

Answer:

Increases

Explanation:

Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions.

4 0

3 years ago

The molar solubilities of the following compounds (in mol/L) are:

IceJOKER [234]

<u>Answer:</u> The decreasing order of $K_{sp}$ is $AgSCN>AgBr>AgCN$

<u>Explanation:</u>

<u>For AgBr:</u>

The balanced equilibrium reaction for the ionization of silver bromide follows:

$AgBr\rightleftharpoons Ag^{+}+Br^-$

s s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][Br^-]$

We are given:

$s=7.3\times 10^{-7}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.3\times 10{-7})^2=5.33\times 10^{-13}$

Solubility product of AgBr = $5.33\times 10^{-13}$

<u>For AgCN:</u>

The balanced equilibrium reaction for the ionization of silver cyanide follows:

$AgCN\rightleftharpoons Ag^{+}+CN^-$

s s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][CN^-]$

We are given:

$s=7.7\times 10^{-9}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.7\times 10{-9})^2=5.93\times 10^{-17}$

Solubility product of AgCN = $5.33\times 10^{-17}$

<u>For AgSCN:</u>

The balanced equilibrium reaction for the ionization of silver thiocyanate follows:

$AgSCN\rightleftharpoons Ag^{+}+SCN^-$

s s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][SCN^-]$

We are given:

$s=1.0\times 10^{-6}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(1.0\times 10{-6})^2=1.0\times 10^{-12}$

Solubility product of AgSCN = $1.0\times 10^{-12}$

The decreasing order of $K_{sp}$ follows:

$AgSCN>AgBr>AgCN$

4 0

4 years ago

Which of the following is a physical change?

MissTica

Answer:

boiling water

Explanation:

6 0

3 years ago