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Marta_Voda [28]
3 years ago
14

What do we call compounds that will not dissolve in water

Chemistry
2 answers:
denis-greek [22]3 years ago
8 0

Answer:

covalent compounds

Explanation:

sasho [114]3 years ago
7 0

Answer:

oil duh.................

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Can an intermediate appear as a reactant in the first step of a reaction mechanism?.
saveliy_v [14]

Answer:

No

Explanation:

An intermetiate in a chemical reaction is something that is formed and reused in a reaction.

8 0
2 years ago
How many moles of tin atoms are in a pure tin cup with a mass of 37.6 g ?
Georgia [21]
Tin is an element called Stannum and has the symbol Sn. Molar mass is the mass of 1 mol of a compound, 1 mol of any substance is made of 6.022 x 10²³ units, these units could be atoms making up an element or molecules making up a compound. 
While the number of atoms making up 1 mol is the same for any element, the weight of 1 mol of substance varies from one another.
In tin(Sn) molar mass - 118.71 g/mol
In 118.71 g - there's 1 mol of tin
therefore in 37.6 g of tin - 1 x 37.6 / 118.71 = 0.31 mol 
In 37.6 g of tin, there's 0.31 mol 
3 0
3 years ago
Can radiation carry thermal energy from object to object?
Viktor [21]

Answer:

Radiation is the transfer of thermal energy by waves that can travel through empty space. When the waves reach objects, they transfer thermal energy to the objects.

3 0
2 years ago
Read 2 more answers
Draw the conjugate base for the Brønsted-Lowry acid-base reaction that occurs when the following acid reacts with water. Show al
DanielleElmas [232]

Answer:

The structures are shown below.

Explanation:

When the acid reacts with water, it loses one proton (H⁺) and forms a base, which is the conjugate base of its acid.

The formal charge of an atom can be calculated by:

FC = X - (Y + Z/2)

Where X is the valence electrons of the neutral atom, Y is the unshared electrons, and Z is the shared electrons in the molecule.

a) When HCl deprotonates, it forms Cl⁻ as the conjugate base. The neutral atom Cl has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1 The structure is shown below in figure a.

b) When Hbr deprotonates it forms Br- as the conjugate base. The neutral atom has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1. The structure is shown below in figure b.

c) When CH3COOH loses a proton, it forms the conjugate base CH3COO⁻. The carbon as 4 valence electrons, hydrogen has 1 valence electron and oxygen has 6 valence electrons. The first carbon make simple bonds with each hydrogen and with the second carbon, and so, all the electrons are shared, and it has FC = 4 - (0 + 8/2) = 0, as so the hydrogens have FC = 1 - (0 + 2/2) = 0.

The second carbon does 1 simple bond with the first carbon, a double bond with one oxygen, and a simple bond with the other oxygen, and so doesn't have unshared electrons, and FC = 4 - (0 + 8/2) = 0.

The first oxygen does a double bond with the carbon, and so it has 4 unshared electrons, so FC = 6 - (4 + 4/2) = 0. The second oxygen does a simple bond with the carbon, and so has 5 unshared electrons, so FC = 6 - (5 + 2/2) = 0.

The structure is shown in figure c.

7 0
3 years ago
Read 2 more answers
Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv
Triss [41]

Answer:

1. 3.57\times 10^{-4}was the K_a value calculated by the student.

2. 5.93\times 10^{-4}was the K_b of ethylamine value calculated by the student.

Explanation:

1.

The pH value of Aspirin solution = 2.62

pH=-\log[H^+]

[H^+]=10^{-2.62}=0.00240 M

Moles of s asprin = \frac{2.00 g}{180 g/mol}=0.01111 mol

Volume of the solution = 0.600 L

The initial concentration of Aspirin  = c = \frac{0.01111 mol}{0.600 L}=0.0185 M

HAs\rightleftharpoons As^-+H^+

initially

c       0    0

At equilibrium

(c-x)      x   x

The expression of dissociation constant :

K_a=\frac{[As^-][H^+]}{[HAs]}:

K_a=\frac{x\times x }{(c-x)}

=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}

K_a=3.57\times 10^{-4}

3.57\times 10^{-4}was the K_a value calculated by the student.

2.

The pH value of ethylamine = 11.87

pH+pOH=14

pOH=14-11.87=2.13

pOH=-\log[OH^-]

[OH^-]=10^{-2.13}=0.00741 M

The initial concentration of ethylamine = c = 0.100 M

C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-

initially

c                    0    0

At equilibrium

(c-x)                x   x

The expression of dissociation constant :

K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}:

K_b=\frac{x\times x}{(c-x)}

=\frac{0.00741\times 0.00741}{(0.100-0.00741)}

K_b=5.93\times 10^{-4}

5.93\times 10^{-4}was the K_b of ethylamine value calculated by the student.

3 0
3 years ago
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