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3 years ago

Please help! Why is cooler, drier air related to High Pressure

1 answer:

azamat3 years ago

7 0

Due to density differences among two air masses. Due to high air pressure in cooler, drier gases, the air is supposed tobe rather dense and it does, mind you, tend to move towards warm or moist areas.

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A solution is prepared by mixing 25.0 g H2O and 25.0 g C2H5OH. Determine the mole fractions of each substance.

Veronika [31]

Answer:

Mole fraction H₂O → 0.72

Mole fraction C₂H₅OH → 0.28

Explanation:

By the mass of the two elements in the solution, we determine the moles of each:

25 g . 1 mol/ 18g = 1.39 moles of water (solute)

25 g . 1 mol / 46 g = 0.543 moles of ethanol (solvent)

Mole fraction solute = Moles of solute / Total moles

Mole fraction solvent = Moles of solvent / Total moles

Total moles = Moles of solute + Moles of solvent

1.39 moles of solute + 0.543 moles of solvent = 1.933 moles → Total moles

Mole fraction H₂O = 1.39 / 1.933 → 0.72

Mole fraction C₂H₅OH= 0.543 / 1.933 → 0.28

Remember that sum of mole fractions = 1

8 0

3 years ago

Read 2 more answers

PLEASE HELP DUE IN 30 MINUTES NO LINKS PLEASE

hammer [34]

Answer:

B is the correct answer

Explanation:

If this answer is correct plzz mark me as brainliest plzzzz

7 0

3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago

Choose the term that best describes the following statement: a system in equilibrium will oppose a change in a way that helps re

Scorpion4ik [409]

Le Chatelier's principle (D)

3 0

3 years ago

Which of the following is an example of convection?

Dovator [93]

Answer:

<h3>C. hot air rising and cooler air failing. </h3>

Explanation:

<h3>Please mark my answer as a brainliest. That is the correct answer. ❤❤❤❤</h3>

3 0

3 years ago