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2 years ago

11

Which of the following will ammonium ion form an insoluble salt? 4/16 a. chloride b. sulfate c. sulfate and carbonate d. carbona

te e. none of the above

1 answer:

Darina [25.2K]2 years ago

6 0

Answer:

e

Explanation:

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Three compounds containing potassium and oxygen are compared. Analysis shows that for each 1.00 g of O, the compounds have 1.22

kicyunya [14]

<span>The law of proportion states that elements combine in whole number ratios. The gram readings for K are multiples of each other, both in grams and moles.
Let us compare the ratios:
</span>2.44 grams/1.22 grams = 2
<span>4.89 grams/2.44 grams = 2</span>

<span>Therefore, Potassium always combines with Oxygen in a ratio of 2 is to 1.</span>

4 0

2 years ago

Convert 7.8 liters of carbon tetra fluoride cfa to grams

dolphi86 [110]

To answer the question, we assume that the given compound is an ideal gas that at STP, one mole of the substance will occupy 22.4 L. From the given volume, we determine the number of moles of substance.
7.8 L / (22.4 L /mole) = 0.3482 moles of cfa
Then, we multiply this number of moles by the molar weight of cfa which is equal to 88 g/mol.
Multiplying,
weight = (0.3482 moles of cfa) x (88 g/mol) = <em>30.64 grams</em>

4 0

2 years ago

Gaseous ICl (0.20 mol) was added to a 2.0 L flask and allowed to decompose at a high temperature:

Ne4ueva [31]

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

2 ICl(g) ⇄ I₂(g) + Cl₂(g)

Initially 0.20 - -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React x x/2 x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

So in equilibrium I will have

(0.20 - x) x/2 x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)

Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).

As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

Kc = (0.07/2 . 0.07/2) / (0.060/2)²

Kc = 1.225x10⁻³ / 9x10⁻⁴

Kc = 1.36

8 0

1 year ago

Place each charge form of alanine under the pH condition where it would be the predominant form. The pKa values for the carboxyl

damaskus [11]

Answer:

(A) pH < 1 the predominant form is the cation: H3C-C(H)(NH3+)-COOH

(B) pH = pl the predominant form is the zwitterion H3C-C(H)(NH3+)-COO-

(C) pH > 11 the predominant form is the anion: H3C-C(H)(NH2)-COO-

(D) Does not occurs in any significant pH: H3C-C(H)(NH2)-COOH

Explanation:

Amino acids are bifunctional because they have an amine group and a carboxyl group. The amine group is a weak base and the carboxyl group is a weak acid, but the pKa of both groups will depend on the whole structure of the amino acid. Also, every amino acid has an isoelectric point (pI), which means the pH were the predominant form of the amino acid is the zwitterion. The structure of the alanine (CH3CH2NH2COOH) shows it has the carboxyl group at C1 with a pKa1 of 2.3 and the amino group at C2 whit the pKa2 of 9.7. The isoelectric poin (pI) of Alanine is 6. Consequently, the protonation of the molecule will depend on the pH of the solution. There are three possibilities:

1) If the pH is under the pKa of the carboxyl group (2.3) the predominant form will be with the amino group protonated, forming a cation (CH3CH(NH3+)COOH).

2) If the pH is between pKa1 (2.3) and pKa2 (9.7) the predominant form will be the zwitterion (CH3CH(NH3+)(COO-)).

3) If the pH is upper the pKa2 of the amino group (9.7) the predominant form will be with the carboxyl group deprotonated, forming an anion (CH3CHNH2(COO-)).

8 0

2 years ago

In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n

bulgar [2K]

<span>Molar mass(C)= 12.0 g/mol
Molar mass (O2)=2*16.0=32.0 g/mol
Molar mass (CO2)=44.0 g/mol

18g C*1mol C/12 g C = 1.5 mol C

C + O2 → CO2

from reaction 1 mol 1 mol 1 mol
from problem 1.5 mol 1.5 mol 1.5 mol

1.5 mol O2*32 g O2/1 mol O2 = 48 g O2

In reality this reaction requires only 48 g O2 for 18 g carbon.
And from 18 g carbon you can get only
1.5 mol CO2*44 g CO2/1 mol CO2=66 g CO2
But these problem has 72g CO2. The best that we can think, it is a mix of CO2 and O2.
So to find all amount of O2 that was added for the reaction (probably people who wrote this problem wanted this)
we need (the mix of 72g - mass of carbon 18 g)= 54 g.
So the only answer that is possible is </span><span>2.) 54 g.</span>

3 0

2 years ago