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3 years ago

In dry climates, rates of evaporation exceed what?

Chemistry

1 answer:

Dennis_Churaev [7]3 years ago

3 0

<h2>Answer:</h2>

<u>In dry climates, rates of evaporation exceed precipitation</u>

<h2>Explanation:</h2>

Evaporation is an important process in the global water cycle. Sunrays hits the surface of water or land and causes water to change state from a liquid to a gas. This is what we call evaporation. When evaporation occurs, at the same time precipitation occurs in the atmosphere which is the water falling from the clouds as rain, freezing rain, sleet, snow, or hail. As the rate of evaporation increases, the rate of precipitations increases too because both are connected to each other.

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A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ = 0.00375 mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

$HN_3 + OH- ---> N^-_{3} + H_2O$

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L

Concentration of $HN_3$ = $\dfrac{0.001755}{0.0383}$ = 0.0458 M

Concentration of $N^{-}_3$ = $\dfrac{ 0.001995 }{0.0383}$ = 0.0521 M

GIven that :

Ka = $1.9 x 10^{-5}$

Thus; it's pKa = 4.72

$pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})$

$pH =4.72 + log(1.1376)$

$pH =4.72 + 0.05598$

$pH =4.77598$

pH ≅ 4.80

3 0

3 years ago

How many moles of glucose are in 19.1g of glucose?

Crank

Answer:

0.106 mol (3s.f.)

Explanation:

To find the number of moles, divide the mass of glucose (in grams) by its Mr. Glucose has a chemical formula of C6H12O6. To find the Mr, add all the Ar of all the atoms in C6H12O6.

Ar of C= 12, Ar of H= 1, Mr of O= 16

These Ar values can be found on the periodic table.

Mr of glucose= 6(12)+ 12(1) + 6(16)= 180

Moles of glucose

= mass ÷ mr

= 19.1 ÷ 180

= 0.106 mol (3 s.f.)

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3 years ago

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xz_007 [3.2K]

Energy of reactants is higher than the products , I am not sure through

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4 years ago

Estimate the pH of the resulting solution if you mixed equal amounts of hydrochloric acid (HCI) and sodium hydroxide (NaOH).

ioda

It obtains a neutral pH (7). The combination of a strong acid and a strong base results in a neutral pH. The pH of HCl is about 1, and the pH of NaOH is 14.

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4 years ago

What is the percent composition of Phosphorus in Li3PO3

Vilka [71]

Answer:

you will get 17.983 g of lithium for every 100 g of lithium phosphate.

Explanation:

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3 years ago