All categories

3 years ago

Why are the wind speeds on gas giant planets so much greater than the wind speeds on earth?

1 answer:

8 0

reason is that there is no land to slow down the wind. Also, wind is caused by differences in air pressure

You might be interested in

Which of the following reactions could be an elementary reaction? 2 NO2(g) + F2(g) → 2NO2F(g) Rate = k[NO2][F2] H2(g) + Br2(g) →

podryga [215]

Answer: The correct answer is $NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2]$

Explanation:

Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction.

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.

For the given reactions:

Equation 1: $2NO_2(g)+F_2(g)\rightarrow 2NO_2F(g);Rate=k[NO_2][F_2]$

Molecularity of the reaction = 2 + 1 = 3

Order of the reaction = 1 + 1 = 2

This is not considered as an elementary reaction.

Equation 2: $H_2(g)+Br_2(g)\rightarrow 2HBr(g);Rate=k[H_2][Br_2]^{1/2}$

Molecularity of the reaction = 1 + 1 = 2

Order of the reaction = $1+\frac{1}{2}=\frac{3}{2}$

This is not considered as an elementary reaction.

Equation 3: $NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2]$

Molecularity of the reaction = 1 + 1 = 2

Order of the reaction = 1 + 1 = 2

This is considered as an elementary reaction.

Equation 4: $NO_2(g)+CO(g)\rightarrow NO(g)+CO_2(g);Rate=k[NO_2]^2$

Molecularity of the reaction = 1 + 1 = 2

Order of the reaction = 2 + 0 = 2

In this equation, the order with respect to each reactant is not equal to its stoichiometric coefficient which is represented in the balanced chemical reaction. Hence, this is not considered as an elementary reaction.

Hence, the correct answer is $NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2]$

3 0

3 years ago

The orbit of a planet is an ellipse with the sunat one of the foci.. This is Kepler's........ second law third law first law

Svetradugi [14.3K]

Answer:

its keplers first law

Explanation:

answer from gauth math

6 0

3 years ago

Choose all the answers that apply. Which of the following will decrease reaction rate? increase pressure decrease temperature in

zavuch27 [327]

The following quantities will effect the reaction rate as follows:

1. On increasing Concentration of the reactant: Rate of the reaction will increases.

2. On increasing pressure : Increases the rate of reaction to the side where there are fewer number of molecules.

3.On increasing temperature of an endothermic reaction: Increases the rate of reaction

4. On decreasing temperature of an endothermic reaction: Increases the rate of reaction.

So the answer is increase pressure, decrease temperature, increase concentration will increases the rate of the reaction.

6 0

3 years ago

Read 2 more answers

A substance with a specific heat of 5 J/g°C absorbs 25 Joules of energy and increases from 75°C to

Ber [7]

Explanation:

0 10 grams that that should be the answer

8 0

3 years ago

A chemist dissolves 0.096 g of CuSO4 · 5 H2O in water and dilutes the solution to the mark in a 500-mL volumetric flask. A 6-mL

scoray [572]

Answer:

(A) 4.616 * 10⁻⁶ M

(B) 0.576 mg CuSO₄·5H₂O

Explanation:

The molar weight of CuSO₄·5H₂O is:

63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol

The molarity of the first solution is:

(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M

The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.

Now we use the dilution factor in order to calculate the molarity in the second solution:

(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M

To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:

4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O

2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O

5 0

3 years ago