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sleet_krkn [62]
3 years ago
8

Combined gas law, PLEASE HELP!

Chemistry
1 answer:
katrin2010 [14]3 years ago
5 0
You correctly said that you need the combined gas law. 

combine gas law----> P1V1/T1= P2V2/T2

P1= ?
V1= 2.58 L
T1= 368K

P2= 777 torr
V2= 1.53 L
T2= 495K

Now we plug in the values. 

(P1 X 2.58)/ 368= (777 X 1.53)/ 495 K

P1= 343 torr
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Properties of metals
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Answer:

Physical Properties of Metals :

Metals are lustrous, malleable, ductile, good conductors of heat and electricity. Other properties include: State: Metals are solids at room temperature with the exception of mercury, which is liquid at room temperature (Gallium is liquid on hot days).

Hope this helps! :D

3 0
4 years ago
3. Round the number 10.4656 to the tenths place *<br><br> 10 <br><br> 10.4 10.5 or 10.6
motikmotik
Answer: 10.5 hope this helps!
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3 years ago
Bromocresol green, Ka=2.0×10−5, is yellow in its protonated form (HX) and blue in its ionized form (X−). At what pH will bromocr
nydimaria [60]

Answer:

pH = 4,70

Explanation:

For bromocresol, as HX is yellow and X⁻ is blue, the indicator will be a perfect green color when HX = X⁻.

Using Henderson-Hasselbalch formula:

pH = pka + log [X⁻]/[HX] <em>(1)</em>

Where pka is -log ka = -log2,0x10⁻⁵ = 4,70

As HX = X⁻, [HX] / [X⁻] = 1

Replacing in (1):

pH = 4,70 + log 1

pH = 4,70 + 0

<em>pH = 4,70</em>

<em></em>

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4 0
4 years ago
(02.01 LC)
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4 0
3 years ago
A 5.41 g sample of carbon reacts with hydrogen to form 6.32 g of an organic compound. What is the empirical formula of the organ
zhannawk [14.2K]

Answer:

CH₂

Explanation:

From the question given above, the following data were obtained:

Mass of Carbon (C) = 5.41 g

Mass of compound = 6.32 g

Empirical formula =?

Next, we shall determine the mass of Hydrogen (H) in the compound. This can be obtained as follow:

Mass of Carbon (C) = 5.41 g

Mass of compound = 6.32 g

Mass of Hydrogen (H) =?

Mass of H = mass of compound – mass of C

Mass of H = 6.32 – 5.41

Mass of H = 0.91 g

Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:

Mass of C = 5.41 g

Mass of H = 0.91 g

Divide by their molar mass

C = 5.41 / 12 = 0.451

H = 0.91 / 1 = 0.91

Divide by the smallest

C = 0.451 / 0.451 = 1

H = 0.91 / 0.451 = 2

Thus, the empirical formula for the compound is CH₂

3 0
3 years ago
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