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IgorC [24]
2 years ago
12

1. how common are the elements that living systems are made out of?

Chemistry
1 answer:
Bas_tet [7]2 years ago
4 0

<h3>Following are the answers to the questions - </h3>

1. Common elements of the living organisms are - Nitrogen, oxygen, phosphorus, sulphur and carbon. These are the 6 most knowing common elements.

2. It's included more then 95% to 97% mass on a human body . Matter itself has energy and matter can be converted to energy and vice versa. The equal amount of matter can have totally different amount of energy and so show up the different type of matter.

3. Electron can be transferred one atom to another this is the most basic way that outer electron can from atom bond .

To know more about electron atom click

brainly.com/question/23390564

#SPJ4

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SOMEONE PLEASE HELP ME THIS IS DUE TOMORROW!! I WILL MARK YOU AS A BRAINLIEST IF ANSWERED NOW!!
AnnZ [28]
K= [Ar] 4s^1

Pb= [Xe]  6s^2 5d^10 6p^2

Sc= [Ar] 4s^2 3d^1

Ra= [Rn] 7s^2

O= [He] 2s^2 2p^4

Ag= [Kr] 5s^2 4d^9

Ru= [Kr] 5s^1

Ce=  [Xe] 6s^2 5d^1 4f^1

I= [Kr] 5s^2 4d^10 5p^5

F= [He] 2s^2 2p^5



6 0
4 years ago
Find the empirical formula of the following compounds:
Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus \frac{mol phosphorus}{ 30.97 g phosphorus}= 0.0293 mol

Moles bromine 6.99 g bromine\frac{mol bromine}{79.90 g bromine}=0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

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8 0
1 year ago
A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric ac
UkoKoshka [18]

Answer:

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

pH = pKa + log(\frac{[KF]}{[HF]})

pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04  

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq)  +  F⁻(aq) ⇄   HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

\eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol

\eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

\eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles  

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.  

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq)    (1)

The number of moles of HF after the reaction of KF with HCl is:

\eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles

And the concentration of HF after the reaction of KF with HCl is is:

C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L

Now, from the equilibrium of equation (1) we have:

Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]}

Ka = \frac{x^{2}}{0.531 - x}  (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!

8 0
4 years ago
What element does the following orbital notation represent?
Snezhnost [94]
Answer is Ti

..............................
5 0
4 years ago
HELP! <br><br>How many moles of a solute would 3 liters of a 2-molar solution contain? ​
katrin [286]

Answer: If it was 3 mol of solute in 2 L of solution it would be 1.5 mol/L.

However when the solute dissolves in the water creating the solution, the volume increases. So 3 mol of solute in 2 L of water creates more than 2 L of solution.

The correct method for making a 3 mol/L solution would be to place some water into a two liter volume container. Dissolve all 3 mol of the solute into the water. Then add water to the 2 L mark. Now there is 3 mol of solute and 2 L of solution.

Explanation: I hope this helps XDDDD

8 0
3 years ago
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