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Ray Of Light [21]
3 years ago
14

Question 1 of 15 2 Points What is another name for a pure substance? Answer here

Chemistry
1 answer:
maksim [4K]3 years ago
4 0

Answer:

element

Explanation:

elements consist of pure substances

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This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): an analytical chemist has det
zhannawk [14.2K]
<h3>Answer:</h3>

                    3.2 moles of Oxygen

<h3>Explanation:</h3>

                        Acetic Acid having chemical formula C₂H₄O₂ and structural formula attached below is the second member of carboxylic family in organic compounds. It is commonly used as vinegar<em> i.e</em>. a mixture containing 5 % Acetic acid and 95 % water.

As shown in structure 1 mole of Acetic acid contains 2 moles of Carbon atoms, 4 moles of Hydrogen atoms and 2 moles of Oxygen atoms respectively. Hence, the number of moles of Oxygen atoms contained by acetic acid containing 3.2 moles of Carbon is calculated as,

                      2 moles of C accompany  =  2 moles of O

So,

               3.2 moles of C will accompany  =  X moles of O

Solving for X,

                     X  =  (3.2 mol C × 2 mol O) ÷ 2 mol C

                      X =  3.2 mol of Oxygen

6 0
3 years ago
How many liters of a 0.5M sodium hydroxide solution would contain 2 mols of solute
Airida [17]
Molarity = mol/L
(0.5M) = (2mol)/L
(2mol)/(0.5M) = L
4 liters
4 0
3 years ago
Read 2 more answers
Propane gas (C3H8) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide gas (CO2) and water vapor (H2O).
Sindrei [870]
Oxen beacon is great for your self halllway
7 0
3 years ago
which of the following will stay constant, no matter if the substance is in the solid, liquid, or gas state?
pogonyaev
Liquid I think it might be
6 0
3 years ago
As an FDA physiologist, you need 0.625 L of phosphoric acid acid / dihydrogen phosphate (H3PO4 (aq) / H2PO4 - (aq) ) buffer with
aksik [14]

Answer:

0.4058L of 1.0M H3PO4

0.2192L of 1.5M NaOH

Explanation:

The pKa of the H3PO4 / H2PO4- buffer is 2.12

To solve this question we must use H-H equation for this system:

pH = pKa + log [H2PO4-] / [H3PO4]

2.75 = 2.12 +  log [H2PO4-] / [H3PO4]

0.63 = log [H2PO4-] / [H3PO4]

<em>4.2658 = [H2PO4-] / [H3PO4] (1)</em>

Where [] could be taken as the moles of each reactant

As you have H3PO4 solution, the reaction with NaOH is:

H3PO4 + NaOH → H2PO4- + Na+ + H2O

As you can see, both H3PO4 and H2PO4- comes from the same 1.0M H3PO4 solution

The moles of H3PO4 are:

[H3PO4] = Moles H3PO4 - Moles NaOH

And for H2PO4-:

[H2PO4-] = Moles NaOH added

Replacing in (1):

4.2658 = [Moles NaOH] / [Moles H3PO4 - Moles NaOH]

4.2658 Moles H3PO4 - 4.2658 moles NaOH = Moles NaOH

4.2658 Moles H3PO4 = 5.2658 moles NaOH <em>(1)</em>

<em></em>

In volume:

0.625L = Moles H3PO4 / 1.0M + Moles NaOH / 1.5M

0.625 = Mol H3PO4 + 0.6667 Moles NaOH <em>(2)</em>

Replacing (2) in (1):

4.2658 Moles H3PO4 = 5.2658 (0.625 - Mol H3PO4 / 0.6667)

4.2658 Moles H3PO4 = 5.2658 (0.625 - Mol H3PO4) / 0.6667

4.2658 Moles H3PO4 = 5.2658*(0.9375 - 1.5 mol H3PO4)

4.2658 Moles H3PO4 = 4.9367 -7.8983 mol H3PO4

12.1641 mol H3PO4 = 4.9367

Mol H3PO4 = 0.4058moles * (1L / 1.0moles) =

0.4058L of 1.0M H3PO4

And:

0.625L - 0.4058L =

0.2192L of 1.5M NaOH

5 0
3 years ago
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