Which type of community will have more competition than cooperation?

In a chemical formula the right side is called?

Ghella [55]

Answer:

Product

Explanation:

It is written at the right side of an arrow

8 0

4 years ago

Calculate the new molarity that results when 250.mL of water is added to each of the following solutions.

sergiy2304 [10]

Answer:

The answer to your question is 1) 0.037 M 2) 0.32 M 3) 0.096 M

Explanation:

a) 125 ml of 0.251 M HCl

-Calculate the moles of HCl

Molarity = moles/volume

-Solve for moles

moles = Molarity x volume

-Substitution

moles = 0.251 x 0.125

= 0.0314

-Calculate the new molarity

Molarity = 0.0314/ (0.125 + 0.250)

-Simplification

Molarity = 0.014/0.375

-Result

Molarity = 0.037 M

2.-

445 ml of 0.499 M of H₂SO₄

-Calculate the number of moles

moles = 0.499 x 0.445

moles = 0.222

-Calculate the new molarity

Molarity = 0.222/(0.445 + 0.25)

Molarity = 0.222/0.695

Molarity = 0.32

3)

5.25 l of HCO₃ 0.101 M

Calculate the number of moles

moles = 0.101 x 5.25

moles = 0.53

-Calculate the Molarity

Molarity = 0.53 / (0.25 + 5.25)

Molarity = 0.53 / 5.5

Molarity = 0.096

8 0

3 years ago

H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles

abruzzese [7]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$H_2(g)+F_2(g)\rightarrow 2HF(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]$

We are given:

$\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = $\frac{-14.1}{1}\times 2.20=-31.02J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

7 0

3 years ago

Write a balanced chemical equation for the neutralization reaction between nitric acid and strontium hydroxide

Ilia_Sergeevich [38]

Hope this helps as well!!!!!

8 0

3 years ago

When did the level of oxygen in Earth's atmosphere become high enough to sustain aerobic respiration? When did the level of oxyg

Alexeev081 [22]

Answer:

2.4 billion years ago

Explanation:

The formation of the tertiary atmosphere, with the re-arrangements of the plates, took that influenced the long term evolution of the planet that transferred the carbon dioxide to the large continental stores and the surplus oxygen came into formation about 2.4 billion years ago during the massive oxygenation event that leads to the appearance at the end of the banded iron formations.

The process of photosynthesis also added and reduced the material that reduces the oxidation of the atmosphere and kept ion place till 15% of the Precambrian period had ended.

7 0

3 years ago