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natulia [17]
3 years ago
6

Plz help 40 pts i need itthank you

Chemistry
1 answer:
Scorpion4ik [409]3 years ago
7 0

Answer:

answer is in the link because it says my answer contains inappropriate words

Explanation:

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Which statement is true for most autotrophic organisms?
lara [203]

omg yees

eExplanation:

5 0
2 years ago
Consider the following reaction: NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2 How many mol of CO2 would be produced from the complete
Assoli18 [71]

Answer:

0.208mole of CO2

Explanation:

First, let us calculate the number of mole of HC3H3O2 present.

Molarity of HC3H3O2 = 0.833 mol/L

Volume = 25 mL = 25/100 = 0.25L

Mole =?

Mole = Molarity x Volume

Mole = 0.833 x 0.25

Mole of HC3H3O2 = 0.208mole

Now, we can easily find the number of mole of CO2 produce by doing the following:

NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2

From the equation,

1mole of HC2H3O2 produced 1 mole of CO2.

Therefore, 0.208mole of HC2H3O2 will also produce 0.208mole of CO2

6 0
3 years ago
Read 2 more answers
Find the ph of of 100 ml of an aqueous 0.43m baoh2 solution
denpristay [2]
Answer is: pH of barium hydroxide is 13.935.
Chemical dissociation of barium hydroxide in water:
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).
c(Ba(OH)₂) = 0.43 M.
V(Ba(OH)₂) = 100 mL ÷ 1000 mL/L = 0.1 L.
n(Ba(OH)₂) = 0.43 mol/L · 0.1 L.
n(Ba(OH)₂) = 0.043 mol.
From chemical reaction: n(Ba(OH)₂) : n(OH⁻) = 1 : 2.
n(OH⁻) = 0.086 mol.
c(OH⁻) = 0.86 mol/L.
pOH = -logc(OH⁻).
pOH = 0.065.
pH = 14 - 0.065 = 13.935.
4 0
2 years ago
Thsof<br> is which acid H2S04
Crank

Explanation:

H2So4=sulphuric acid , strong acid

6 0
1 year ago
Which of these reactions involves an acid and a covalent base?
Juli2301 [7.4K]
I think the correct answer would be D. The reaction that involves an acid and a covalent base would be the reaction of sulfuric acid and water or H2SO4 + 2H2O → 2H3O+ + SO42– . The acid would be H2SO4 and the covalent base would be H2O since it is being held by covalent bonds and when in solution it will have equal amounts of OH- and H+ ions.
5 0
3 years ago
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