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2 years ago

Which section of the ocean floor makes up half of the surface of Earth?

Chemistry

2 answers:

Ostrovityanka [42]2 years ago

7 0

Answer:

Abyssal plain

Explanation:

torisob [31]2 years ago

6 0

The section of the ocean floor that makes up half of the surface of earth IS abyssal plain

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Now we add 0.15 mol of naoh to 1.00 liter of the solution that was given in part 1. what is the ph of the solution after the nao

AnnyKZ [126]

Hello there.

<span>Now we add 0.15 mol of naoh to 1.00 liter of the solution that was given in part 1. what is the ph of the solution after the naoh addition? answer in units of ph.

</span><span>not sodium nitrate</span>

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3 years ago

Which of these would have the largest volume??

DerKrebs [107]

Yea, i think it's a shopping mall

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2 years ago

Besides temperature and precipitation, which other factor does the Köppen climate classification system use to

Morgarella [4.7K]

Answer:

Explanation:

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2 years ago

Read 2 more answers

Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation: H2O(g) + 2NO(g) rig

xxTIMURxx [149]

Answer:

A. $r=k[NO]^2[H_2]$

B. $k=0.121\frac{L^2}{mol^2*s}$

C. Second-order with respect to NO and first-order with respect to H₂

Explanation:

Hello,

In this case, for the reaction:

$H_2O(g) + 2NO(g) \rightarrow N_2O + H_2O(g)$

The rate law is determined by writing the following hypothetical rate laws:

$3.822x10^{-3}=k[0.3]^m[0.35]^n\\\\1.529x10^{-2}=k[0.6]^m[0.35]^n\\\\3.058x10^{-2}=k[0.6]^m[0.7]^n$

Whereas we can compute m as follows:

$\frac{3.822x10^{-3}}{1.529x10^{-2}} =\frac{[0.3]^m[0.35]^n}{[0.6]^m[0.35]^n} \\\\0.25=(0.5)^m\\\\m=\frac{log(0.25)}{log(0.5)} \\\\m=2$

Therefore, the reaction is second-order with respect to NO. Thus, for hydrogen, we find n:

$\frac{1.529x10^{-2}}{3.058x10^{-2}} =\frac{[0.6]^2[0.35]^n}{[0.6]^2[0.7]^n} \\\\0.5=(0.5)^n\\\\n=\frac{log(0.5)}{log(0.5)}\\ \\n=1$

A) Therefore, the reaction is first-order with respect to H₂. In such a way, we conclude that that the rate law is:

$r=k[NO]^2[H_2]$

B) Rate constant is computed from one kinetic data:

$k=\frac{1.529x10^{-2}\frac{mol}{L*s} }{(0.6\frac{mol}{L} )^2(0.35\frac{mol}{L})}\\\\k=0.121\frac{L^2}{mol^2*s}$

C. As mentioned before, reaction is second-order with respect to NO and first-order with respect to H₂.

Best regards.

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2 years ago

Write the formula for the conjugate base acid of hexanoic acid

valentina_108 [34]

Explanation:

Chemical formula of hexanoic acid is $CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}COOH$ . According to Bronsted-Lowry, acid upon dissociation gives conjugate base and base upon dissociation gives conjugate acid.

Hexanoic acid will dissociates as follows.

$CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}COOH \rightarrow CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}COO^{-} + H^{+}$

Therefore, we can conclude that conjugate acid of hexanoic acid is $CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}COO^{-}$

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3 years ago