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alukav5142 [94]
2 years ago
10

Sodium hydroxide reacts with sulfuric acid according to the equation:

Chemistry
1 answer:
Lelechka [254]2 years ago
6 0

Answer:

27) Double replacement

28) 2NaOH + H2SO4 --> Na2SO4 + 2H2O

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What are the four protein structures?
daser333 [38]
Hello there.

<span>What are the four protein structures?
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<span>Primary structure 

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4 0
3 years ago
Use standard enthalpies of formation to calculate ΔH∘rxn for the following reaction______.2H2S(g)+3O2(g)→2H2O(l)+2SO2(g) ΔH∘rxn
Ahat [919]

Answer:

-1,103.39KJ/mol

Explanation:

We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.

In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.

The standard enthalpies of the molecules above are as follows:

H2S = -20.63KJ/mol

H2O = -285.8KJ/mol

SO2 = -296.84KJ/mol

O2 = 0KJ/mol

ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3

 ΔfH⦵(O2)]

ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]

-[ 3 × -20.63)]

= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol

6 0
3 years ago
Consider an exceptionally weak acid, HA, with a Ka = 1x10 -20 . You make a 0.1M solution of the salt Na
Vedmedyk [2.9K]

Answer:

pH=13

Explanation:

Hello,

In this case, given the acid, we can suppose a simple dissociation as:

HA\rightleftharpoons H^+ + A^-

Which occurs in aqueous phase, therefore, the law of mass action is written by:

Ka=\frac{[H^+][A^-]}{[HA]}

That in terms of the change x due to the reaction's extent we can write:

1x10^{-20}=\frac{x*x}{0.1M-x}

But we prefer to compute the Kb due to its exceptional weakness:

Kb=\frac{Kw}{Ka}=\frac{1x10^{-14}}{1x10^{-20}}  =1x10^{-6}

Next, the acid dissociation in the presence of the base we have:

Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}

Whose solution is x=0.0999M which equals the concentration of hydroxyl in the solution, thus we compute the pOH:

pOH=-log([OH^-])=-log(0.0999)=1

Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:

pH+pOH=14\\\\pH=14-pOH=14-1\\\\pH=13

Regards.

5 0
3 years ago
Read 2 more answers
at room temperature table salt is a solid and mercury is a liquid. what conclusion can you draw about the melting points of thes
Tju [1.3M]
Table slat has a low melting pointy while mercury has a high one
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5 0
3 years ago
Just write a C.E.R (claim, evidence, reasoning)​
Charra [1.4K]

Answer:

I have attached pictures below. Maybe those will help?

Explanation:

5 0
3 years ago
Read 2 more answers
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