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2 years ago

What can you do to see what season another planet experiences?

Chemistry

2 answers:

Katen [24]2 years ago

6 0

Answer:

Every planet in the solar system has seasons. Extraterrestrial seasons are hardly noticeable on some planets (Venus), mindbogglingly extreme on others (Uranus) and in some cases simply impossible to define (Mercury).

julia-pushkina [17]2 years ago

3 0

Answer:

Observe the other planet throughout the year. Look at the environment when the season changes to see what changes in the environment.

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76. Which of the following choices is not an example of an adaptation?

Fiesta28 [93]

Answer:

Explanation:

The wolf with shorter legs would not be an adaptation but more of a birth defect as a wolf with short legs would be at a disadvantage.

7 0

2 years ago

What is the volume of 0.5 M solution of NaOH that contains 3.6 mol of solute?

nasty-shy [4]

Answer:

The volume of NaOH solution is 0.532 L.

Step 1. Calculate the moles of NaOH.

Moles = 11.5 g NaOH * (1 mol NaOH/40.00 g NaOH) = 0.2875 mol NaOH

Step 2. Calculate the volume of NaOH

Volume = 0.2875 mol NaOH * (1 L NaOH/0.540 mol NaOH) = 0.532 L

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6 0

2 years ago

what temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution (specific heat capacity = 3

ivolga24 [154]

Answer:

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

Explanation:

Mass of ethylene glycol = m = 100 g

Specific heat capacity of ethylene glycol = c = 3.5 J/g°C

Change in temperature of ethylene glycol = ΔT

Heat loss by the ethylene glycol = Q = 350 J

$Q=mc\Delta T$

$\Delta T=\frac{Q}{mc}=\frac{350 J}{100 g\times 3.5 J/g^oC}$

ΔT = 1°C

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

3 0

3 years ago

A container of gas is initially at 0.25 atm and 0 ˚C. What will the pressure be at 125 ˚C?

Pani-rosa [81]

Answer:

0.37atm

Explanation:

Given parameters:

Initial pressure = 0.25atm

Initial temperature = 0°C = 273K

Final temperature = 125°C = 125 + 273 = 398K

Unknown:

Final pressure = ?

Solution:

To solve this problem, we use a derivative of the combined gas law;

$\frac{P1}{T1}$ = $\frac{P2}{T2}$

P and T are pressure and temperature

1 and 2 are initial and final values

$\frac{0.25}{273}$ = $\frac{P2}{398}$

P2 = 0.37atm

3 0

2 years ago

Mazyrski [523]

$1) {ca}^{2 + } = cation \\ {br}^{2 - } = anion$

7 0

2 years ago

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