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Marina CMI [18]
3 years ago
11

Fill in the table below based on common properties of acids and bases:

Chemistry
2 answers:
icang [17]3 years ago
8 0
Sour tasting= acid
Bitter= base
8-14= base
1-6= acid
Wewaii [24]3 years ago
4 0
Bsbsbsbsbs s a a sbbsbsbsbsbsbsb z snnzjz
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An(A)=25,n(B)=40,n(AuB)=50<br><br>AnB​
puteri [66]

Answer:

5879

Explanation:

3896_+746864uyrhufhj

5 0
3 years ago
6. Which group of compounds is described as insoluble?
Sedaia [141]

Answer:phosphates are insoluble in water

Explanation:

3 0
3 years ago
of the following warnings, which one refers to a chemical property of a substance? fragile flammable handle with care shake well
coldgirl [10]
Flammable is ur answer
6 0
3 years ago
Linolenic acid (C18H30O2 - M.W. = 278.42 g/mol) reacts with hydrogen gas according to the equation: C18H30O2 + 3H2 (g) → C18H36O
Ludmilka [50]

Answer:

2.53 L is the volume of H₂ needed

Explanation:

The reaction is: C₁₈H₃₀O₂ + 3H₂ → C₁₈H₃₆O₂

By the way we can say, that 1 mol of linolenic acid reacts with 3 moles of oxygen in order to produce, 1 mol of stearic acid.

By stoichiometry, ratio is 1:3

Let's convert the mass of the linolenic acid to moles:

10.5 g . 1 mol / 278.42 g  = 0.0377 moles

We apply a rule of three:

1 mol of linolenic acid needs 3 moles of H₂ to react

Then, 0.0377 moles will react with (0.0377 . 3 )/1 = 0.113 moles of hydrogen

We apply the Ideal Gases Law to find out the volume (condition of measure are STP) → P . V = n . R . T → V = ( n . R .T ) / P

V = (0.113 mol . 0.082 L.atm/mol.K . 273.15K) 1 atm = 2.53 L

4 0
3 years ago
How many grams of aluminum chloride are produced when 5.96 grams of aluminum are reacted with excess chlorine gas? Start with a
Vadim26 [7]

Answer:

29.47 g of AlCl₃.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2Al + 3Cl₂ —> 2AlCl₃

Next, we shall determine the mass of Al that reacted and the mass of AlCl₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 2 × 27 = 54 g

Molar mass of AlCl₃ = 27 + (35.5× 3)

= 27 + 106.5

= 133.5 g/mol

Mass of AlCl₃ from the balanced equation = 2 × 133.5 = 267 g

SUMMARY:

From the balanced equation above,

54 g of Al reacted to produce 267 g of AlCl₃.

Finally, we shall determine the mass of AlCl₃ produced by the reaction of 5.96 g of Al. This can be obtained as follow:

From the balanced equation above,

54 g of Al reacted to produce 267 g of AlCl₃.

Therefore, 5.96 g of Al will react to produce = (5.96 × 267)/54 = 29.47 g of AlCl₃.

Thus, 29.47 g of AlCl₃ were obtained from the reaction.

3 0
2 years ago
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