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Sophie [7]
3 years ago
12

I need help with #4 please

Chemistry
1 answer:
Sholpan [36]3 years ago
3 0

Answer:

1. Land 2. open 3. broken 4. Lava matter 5. Force of the plates

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rite stepwise equations for protonation or deprotonation of this polyprotic species in water. What are the products of the first
adoni [48]

<u>Answer:</u> The products of the given chemical equation are HCO_3^-\text{ and }OH^-

<u>Explanation:</u>

Protonation equation is defined as the equation in which protons get added in the substance.

The chemical equation for the protonation of carbonate ion in the presence of water follows:

CO_3^{2-}+H_2O\rightarrow HCO_3^-+OH^-

By Stoichiometry of the reaction:

1 mole of carbonate ion reacts with 1 mole of water to produce 1 mole of hydrogen carbonate ion and 1 mole of hydroxide ion

Hence, the products of the given chemical equation are HCO_3^-\text{ and }OH^-

3 0
4 years ago
Plants produce a variety of substances:food,flavoring,drugs,poisons.this is a result of
tatuchka [14]

Answer:

Synthesis

Explanation:

They synthesize these chemicals

7 0
3 years ago
If a buffer solution is 0.220 M in a weak acid ( Ka=7.4×10−5) and 0.540 M in its conjugate base, what is the pH?
valkas [14]

Answer: the pH of the solution is 4.52

Explanation:

Consider the weak acid as Ha, it is dissociated as expressed below

HA     H⁺  +  A⁻

the Henderson -Haselbach equation can be expressed as;

pH = pKa + log( [A⁻] / [HA])

the weak acid is dissociated into H⁺ and A⁻ ions in the solution.

now the conjugate base of the weak acid HA is

HA(aq) {weak acid}     H⁺(aq)  +  A⁻(aq) {conjugate base}

so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;

pKₐ = -logKₐ

pKₐ = -log ( 7.4×10⁻⁵ )

pKₐ = 4.13

now thw pH is

pH = pKₐ  + log( [A⁻] / [HA])

pH = 4.13 + log( [0.540] / [0.220])

pH = 4.13 + 0.3899

pH = 4.5199 = 4.52

Therefore the pH of the solution is 4.52

6 0
3 years ago
DDT was banned from use as a pesticide in the United States because
Dmitry_Shevchenko [17]
Too much money and dangerous 
5 0
3 years ago
How do I find the moles of OH- which reacted (mol) in the titration. Table Attached
BaLLatris [955]

Answer:

It is equal to the number of moles of acid that reacted. When Oxalic acid is your limiting reactant it is the # of moles of oxalic acid used. When NaOH is your limiting reactant it is equal to the number of moles of NaOH used.

4 0
3 years ago
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