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3 years ago

A storm is

Chemistry

2 answers:

IceJOKER [234]3 years ago

8 0

Answer:

I think it is B I'm not Sure

STALIN [3.7K]3 years ago

4 0

The answer is D. because A is a storm surge, B. Is air mass, and C is an occluded front (i think). i hope i’m correct i’m pretty sure i am.

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6 moles of H2O is equal to how many molecules?

alexandr1967 [171]

6 miles of H2O is equal to 12 Hydrogen molecules and 6 oxygen molecules. Equaling 18 in total.

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3 years ago

What is the pH of pure water at 40.0Â°C if the Kw at this temperature is 2.92 - 10-147 OA) 6.767 O B) 0 465 O c) 7.000 OD) 7 233

Sergeeva-Olga [200]

Answer:

PH= 6.767 (answer is the A option)

Explanation:

first we need to correct the value in Kw at this temperature is 2.92*10^-14

so, in this case we have that:

Kw=2.92*10^-14 M²

[ H3O^+] [ H3O^+]

$[H_{3}O^{+} ] [OH^{-} ] = Kw = 2.92*10^{-14} M^{2} \\\\$

at 40ºC

$[H_{3}O^{+} ] = [OH^{-} ]$

$[H_{3}O^{+} ]^{2} = 2.92*10^{-14} M^{2}$

$[H_{3}O^{+} ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M$

$PH= -log10[H_{3}O^{+} ] = -log10(1.71*10^{-7} ) = 6.767$

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2 years ago

Name the elements present of carbon monoxide

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3 years ago

The ionization constant for water (kw) is 9.311 × 10−14 at 60 °c. calculate [h3o+], [oh−], ph, and poh for pure water at 60 °c.

Sonbull [250]

As,

Kw = [H+] [OH-]

For water, [H+] = [OH-]

Therefore we can write

Kw = [H+]²

9.311 × 10-14 = [H+]²

[H+] = 3.04 × 10-7 = [OH-]

Ph = - log [H+]

= - log ( 3.04 × 10-7)

= 6.52

Thus, Ph = PoH = 6.52

5 0

3 years ago

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$

$molH_{2} O = 0,783 mol$

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴ $MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}$

$MassOfH_{2}O = 14,09g$

5 0

3 years ago