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2 years ago

What are the four most important elements on earth?

Chemistry

2 answers:

Hoochie [10]2 years ago

8 0

Answer:

These four atomic elements are oxygen, carbon, hydrogen, and nitrogen.

Explanation:

hope it helps....

vova2212 [387]2 years ago

5 0

Answer:

Nitrogen/Carbon/Oxygen/Hydrogen

Explanation:

I believe those would be the most important because:

Nitrogen - Composing 78% of earths atmosphere.

Carbon - Building blocks for all known life.

Oxygen - We breath it as well as it composes 21% of our atmosphere.

Hydrogen - Makes up a bunch of compounds (like water) that enable humans to live.

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Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the f

julia-pushkina [17]

Explanation:

As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.

The chemical species which accept proton are called Brønsted-Lowry base.

(a) $HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-$

$HNO_3$ is Bronsted lowry acid and $NO_3^-$ is its conjugate base.

$H_2O$ is Bronsted lowry base and $H_3O^+$ is its conjugate acid.

(b)

$CN^- + H_2O \rightarrow HCN + OH^-$

$CN^-$ is Bronsted lowry base and HCN is its conjugate acid.

$H_2O$ is Bronsted lowry acid and $OH^-$ is its conjugate base.

(c)

$H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-$

$H_2SO_4$ is Bronsted lowry acid and $HSO_4^-$ is its conjugate base.

Cl^- is Bronsted lowry base and HCl is its conjugate acid.

(d)

$HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O$

$HSO_4^-$ is Bronsted lowry acid and $SO_4^{2-}$ is its conjugate base.

OH^- is Bronsted lowry base and $H_2O$ is its conjugate acid.

(e)

$O_{2-}+H_2O \rightarrow 2OH^-$

$O_{2-}$ is Bronsted lowry base and OH- is its conjugate acid.

$H_2O$ is Bronsted lowry acid and OH- is its conjugate base.

6 0

3 years ago

The following reaction shows sodium carbonate reacting with calcium hydroxide.

velikii [3]

Ans: 15.1 grams

Given reaction:

Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3

Mass of Na2CO3 = 20.0 g

Molar mass of Na2CO3 = 105.985 g/mol

# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol

Mass of NaOH produced = 0.3774*39.996 = 15.09 grams

5 0

2 years ago

29.47 mL of a solution of the acid HBr is titrated, and 72.90 mL of 0.2500-M NaOH is required to reach the equivalence point. Ca

Klio2033 [76]

The original concentration of the acid solution is 6.175 $\times$ 10^-4 mol / L.

<u>Explanation:</u>

Concentration is the ratio of solute in a solution to either solvent or total solution. It is expressed in terms of mass per unit volume

HBr + NaOH -----> NaBr + H2O

There is a 1:1 equivalence with acid and base.

Moles of NaOH = 72.90 $\times$ 10^-3 $\times$ 0.25

= 0.0182 mol.

[ HBr ] = moles of base / volume of a solution

= 0.0182 / 29.47

= 6.175 $\times$ 10^-4 mol / L.

4 0

2 years ago

If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio

bulgar [2K]

Answer:

$m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

$n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2$

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

$m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Best regards.

4 0

2 years ago

Write and balance the half-reaction for the oxidation of white phosphorous P4 to the phosphate ion PO3^−4 in a basic solution.

aalyn [17]

Since the half-reaction is occurring in a basic solution, add 32OH− to each side of the equation to eliminate the H+ ions.

P₄ +16H₂O + 32OH⁻ ⟶ 4PO₃⁻⁴ + 32H⁺ +32OH⁻

Final reaction :

P₄ + 32OH⁻ ⟶ 4PO₃⁻⁴ + 16H₂O + 20e⁻

A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.

The concept of half-reactions is used to describe what occurs in an electrochemical cell, such as a Galvanic cell battery. Half-reactions can be written to describe both the metal undergoing oxidation (known as the anode) and the metal undergoing reduction (known as the cathode).

Half-reactions are often used as a method of balancing redox reactions. For oxidation-reduction reactions in acidic conditions, after balancing the atoms and oxidation numbers, one will need to add H+ ions to balance the hydrogen ions in the half reaction.

For oxidation-reduction reactions in basic conditions, after balancing the atoms and oxidation numbers, first treat it as an acidic solution and then add OH- ions to balance the H+ ions in the half reactions (which would give H2O).

Learn more about Half reactions here : brainly.com/question/2491738

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3 0

1 year ago