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4 years ago

What does it mean for a strong base to be in equilibrium

Chemistry

1 answer:

bonufazy [111]4 years ago

8 0

The position of equilibrium lies far to the right, with products being favored.

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What is the correct name for Mn(NO3),

kupik [55]

Answer:

Manganese trinitrate or manganese(III) nitrate

Explanation:

6 0

3 years ago

Please write balanced equations for these reactions! WILL GIVE BRAINLIEST! 50 POINTS

Romashka [77]

Answer:

(NH4)2S(aq) + Pb(NO3)2(aq) --> 2NH4NO3 (aq) + PbS (s)

3 0

3 years ago

The equilibrium constant for the formation of ammonia from nitrogen and hydrogen is 1.6 × 102. what is the form of the equilibri

Nimfa-mama [501]

Answer: The expression for equilibrium constant is $\frac{[NH_3]^2}{[H_2]^3[N_2]}$

Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as $k_c$

For a general reaction:

$aA+bB\rightleftharpoons cC+dD$

The equilibrium constant is written as:

$k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

Chemical reaction for the formation of ammonia is:

$N_2+3H_3\rightleftharpoons 2NH_3$

$k_c=1.6\times 10^2$

Expression for $k_c$ is:

$k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}$

$1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}$

8 0

3 years ago

Select all that identify a covalent bond.

saw5 [17]

Heres the best help i can give you There is a couple different ways to determine if a bond is ionic or covalent. By definition, an ionic bond is between a metal and a nonmetal, and a covalent bond is between 2 nonmetals. So you usually just look at the periodic table and determine whether your compound is made of a metal/nonmetal or is just 2 nonmetals

8 0

3 years ago

Draw all four products obtained when 2-ethyl-3-methyl-1,3-cyclohexadiene is treated with HBr at room temperature and show the me

LenKa [72]

Answer:

See explanation below

Explanation:

In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.

At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.

Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.

In the picture below, you have the mechanism for all the 4 products.

Hope this helps

5 0

3 years ago