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romanna [79]
2 years ago
11

Explain the difference between a solar and a lunar eclipse.

Chemistry
1 answer:
Zinaida [17]2 years ago
7 0

Answer:

A lunar eclipse occurs at night and a solar eclipse occurs during the day. There are only certain times when either of them can occur. A lunar eclipse can only occur when the moon is directly opposite the Sun in the sky — a full moon. Even though there is a full moon each month, obviously a lunar eclipse does not occur on a monthly basis because the Sun isn't exactly in line with the Earth and the moon. The moon's orbit is actually tilted 5 degrees more than that of the Earth; otherwise, we would see a lunar eclipse each month.

Explanation:

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I need to know what occupied vs full means for chemistry
Anastaziya [24]
You can get the answer on quizlet or google
7 0
3 years ago
Part 1: Use complete sentences to explain why solar storms occur.
Virty [35]
A geomagnetic storm (commonly referred to as a solar storm) is a temporary disturbance of the Earth's magnetosphere caused by a solar wind shock wave and/or cloud of magnetic field that interacts with the Earth's magnetic field.


Solar flares, coronal mass ejections, high-speed solar wind, and solar energetic particles are all forms of solar activity.
4 0
3 years ago
What are the characteristics of the actinides?
iren [92.7K]
The Actinide series contains elements with atomic #'s 89 to 103 and is the sixth group in the periodic table. This series is below the Lanthanide series, which is located under the main body of the periodic table. Both Actinide and Lanthanide series are refered to as Rare Earth Metals. Hope this helped
6 0
3 years ago
Calculate the liters of a 4.40 M KCl solution to obtain 0.200 mole of KCI
Yuki888 [10]

Answer:

\boxed {\boxed {\sf 0.0455 \ L}}

Explanation:

We are asked to calculate the liters given the molarity and moles in a solution.

Molarity is a measure of concentration in moles per liter. It is calculated with the following solution:

molarity= \frac{moles \ of \ solute}{liters \ of \ solution}

The molarity of the solution is 4.40 molar of potassium chloride (KCl). 1 molar is equal to 1 mole per liter, so the molarity is also 4.40 moles of potassium chloride per liter. There are 0.200 moles of potassium chloride or solute. The liters of solution is unknown, so we can use the variable x.

  • molarity= 4.40 moles KCl / L
  • moles of solute = 0.200 mol KCl
  • liters of solution = x

4.40 \ mol \ KCl/ L = \frac{ 0.200 \ mol \ KCl}{x}

Since we are solving for x, we must isolate the variable. First, cross multiply. Multiply the first numerator by the second denominator, then the first denominator and the second numerator.

\frac {4.40 \ mol \ KCl/ L}{1} = \frac{ 0.200 \ mol \ KCl}{x}

4.40 \ mol \ KCl / L * x= 1 * 0.200 \ mol \ KCl

4.40 \ mol \ KCl / L * x=  0.200 \ mol \ KCl

x is being multiplied by 4.40 moles of potassium chloride per liter. The inverse operation of multiplication is division. Divide both sides by 4.40 mol KCl/L

\frac {4.40 \ mol \ KCl / L * x}{4.40 \ mol \ KCl / L}=  \frac {0.200 \ mol \ KCl}{{4.40 \ mol \ KCl / L}}

x=  \frac {0.200 \ mol \ KCl}{{4.40 \ mol \ KCl / L}}

The units of moles of potassium chloride cancel.

x=  \frac {0.200}{{4.40  L}}

x= 0.0454545454545 \ L

The original measurements of liters and moles have 3 significant figures, so our answer must have the same. For the number we calculated, that is the ten-thousandth place (0.0454545454545). The 5 to the right of this place (0.0454545454545) tells us to round the 4 up to a 5.

x \approx 0.455 \ L

There are approximately <u>0.455 liters</u> in a 4.40 molar solution with 0.200 moles of solute.

6 0
3 years ago
Sodium benzoate (C6H5COONa), the sodium salt of the weak acid benzoic acid, is used as a food preservative. A solution is prepar
Aleks [24]

Answer:

8.6

Explanation:

Using the expression :

K_a\times K_b=K_w

Where, K_w is the dissociation constant of water.

At 25\ ^0C, K_w=10^{-14}

Thus, for benzoic acid , pKa = 4.20

Thus, K_a=10^{-4.20}=6.31\times 10^{-5}

K_b for Sodium benzoate can be calculated as:

K_a\times K_b=K_w

6.31\times 10^{-5}\times K_b=10^{-14}

K_b=1.58\times 10^{-10}

The benzoate ion will dissociate as:-

C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}

K_b expression is:-

K_{b}=\frac {\left [ C_6H_5COOH^{+} \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}

Given that:-

Moles = 0.100 moles

Volume = 1.00 L

Thus, Concentration = 0.100/ 1.00 M = 0.1 M

Considering the ICE table in the image below.

So,

1.58\times 10^{-10}=\frac{x^2}{0.1-x}

1.58\left(0.1-x\right)=10^{10}x^2

Solving for x, we get that, x=3.97\times 10^{-6}

Thus, [OH^-]=3.97\times 10^{-6}

Also,

pOH=-log[OH^-]=-log(3.97\times 10^{-6})=5.4

Also, pH + pOH = 14  

<u>So, pH = 14 - 5.4 = 8.6</u>

4 0
3 years ago
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