Effects of warm tropical waters meeting with cool air.

Answers

Answer 1

Answer:

The effect of warm tropical waters meeting with cool air is hurricane.

What is a hurricane?

A hurricane is formed when warm tropical water meet with air that is cool and humid. This creates a zone of pressure over the warm water causing the air to rotate.

Hence, the effect of warm tropical waters meeting with cool air is hurricane.

Learn more about hurricane:brainly.com/question/11891941

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Answer 2

Answer: A hurricane.. I believe

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Which of the following describe chemical properties?Sugar burns in air to form water and carbon dioxide
Water freezes
Titanium is less dense than iron
Iodine ( a purple solid) becomes a purple gas

Answers

sugar burns in air.....................................

Compounds are formed when the atoms of two or more elements are chemically combined.What type of bond is formed if atoms donate electrons to other atoms when the elements are combined?

A. ionic bond
B. hydrogen bond
C. covalent bond

Answers

Fully donating an electron results in an ionic bond, so A is the correct answer.

In comparison, B, a hydrogen bond, is a bond that forms between an H in one compound and an O, N, or F in another. It is among the strongest of intermolecular forces, while ionic/covalent bonds are intramolecular.

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Assuming 100% dissociation, calculate the freezing point and boiling point of 2.59 m K3PO4(aq).

Answers

Answer:

$T_f=19.3^oC\nT_b=105.3^oC$

Explanation:

Hello,

In this case, it firstly necessary to understand how the complete (100%) dissociation of the potassium phosphate is done:

$K_3PO_4(aq)\rightarrow 3K^+(aq)+PO_4^(-3)(aq)$

In such a way, since the freezing point depression and the boiling point elevation are computed as:

$(T_f-T_(pure))=i*m*K_f$

$(T_b-T_(pure))=i*m*K_b$

The corresponding van't Hoff factor is four as 3 potassiums and 1 phosphate as ions are obtained during the dissolution. In addition, the freezing point depression constant and boiling point elevation constant of water are 1.86 and 0.512 °C/m respectively. Moreover, the freezing point of water is 0 °C and the boiling point is 100 °C, therefore, the resulting freezing and boiling temperatures for the 2.59 m solution of potassium phosphate are:

$T_f=0^oC+4*2.59m*1.86^oC/m\nT_f=19.3^oC\n\nT_b=100^oC+4*2.59m*0.512^oC/m\nT_b=105.3^oC$

Best regards.

The freezing point can be obtained from the formula Tf= Tf (solvent)- Kf* m* n. Kf of water is 1.85 C/m. n is equal to 4 since there are 4 ions dissociated. Tf (solvent) is 0 C. Hence, Tf is -19.27 C. The boiling point can be obtained from the formula Tb= Tb(solvent)+ Kb* m* n where Kb is 0.512 C/m. Tb (solvent)Tb is 100 C. Tb then is 105.30 C.

When a voltaic cell operates ions move through the

Answers

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