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ss7ja [257]
3 years ago
13

The term used to describe the Moon phases as the light spreads across the surface of the moon or grows.

Chemistry
1 answer:
katrin [286]3 years ago
5 0

Answer:

Explanation:

New moon

Waxing crescent moon

First quarter moon

Waxing gibbous moon

Full moon

Waning gibbous moon

Last quarter moon

Waning crescent moon

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You may feel cold when you touch certain kinds of matter because some matter gives off cold air. I’m NOT SURE
8 0
3 years ago
What is the pressure, in atm, exerted by 41.6 g oxygen in a 22.0L container at 30.0C ?
Reika [66]

Answer:

1.47 atm

Explanation:

Step 1: Calculate the moles corresponding to 41.6 g of oxygen

The molar mass of oxygen is 32.00 g/mol.

41.6 g × 1 mol/32.00 g = 1.30 mol

Step 2: Convert 30.0 °C to Kelvin

We will use the following expression.

K = °C + 273.15 = 30.0 + 273.15 = 303.2 K

Step 3: Calculate the pressure exerted by the oxygen

We will use the ideal gas equation.

P × V = n × R × T

P = n × R × T / V

P = 1.30 mol × (0.0821 atm.L/mol.L) × 303.2 K / 22.0 L = 1.47 atm

3 0
3 years ago
Is fire a solid liquid or gas?
Doss [256]
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4 0
3 years ago
Read 2 more answers
) Do you think the pH of 1,0 M tri-methyl ammonium (CH3)3NH+, pKa = 9.80, will be higher or lower than that of 1.0 M phenol, C6H
Elanso [62]

Answer:

1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).

2. The difference in pH values is 4.95.

Explanation:

1. The pH of a compound can be found using the following equation:

pH = -log([H_{3}O^{+}])

First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.

<u>Trimethyl ammonium</u>:

We can calculate [H₃O⁺] using the Ka as follows:

(CH₃)₃NH⁺ + H₂O  →  (CH₃)₃N + H₃O⁺    

1.0 - x                               x           x  

Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}

10^{-pKa} = \frac{x*x}{1.0 - x}

10^{-9.80}(1.0 - x) - x^{2} = 0    

By solving the above equation for x we have:  

x = 0.097 = [H₃O⁺]

pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01                                      

<u>Phenol</u>:

C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺

1.0 - x                        x             x

Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}

10^{-10} = \frac{x^{2}}{1.0 - x}

1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0

Solving the above equation for x we have:

x = 9.96x10⁻⁶ = [H₃O⁺]

pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00

Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.

2. The difference in pH values for the two acids is:

\Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95

Therefore, the difference in pH values is 4.95.

I hope it helps you!

7 0
3 years ago
What type of solid is this? This solid has high melting point, does not conduct electricity, and does not dissolve in water. Met
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The awnser It is iconic
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