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maria [59]
3 years ago
10

A container with a fixed volume, filled with hydrogen gas at -104°C and 71.8 K PA is heated until the pressure reaches 225.9 K P

A. What is the temperature of the hydrogen gas in degrees Celsius
Chemistry
1 answer:
Daniel [21]3 years ago
4 0

Answer:

The correct answer is 532 K

Explanation:

The Gay-Lussac law describes the behavior of a gas at constant volume, by changing the pressure or temperature. When is heated, the change in pressure of the gas is directly proportional to it absolute temperature (in Kelvin or K).

We have the following initial conditions:

P1= 71.8 kPa

T1= -104ºC +273 = 169 K

If the pressure increases until reaching 225.9 kPa (P2), we can calculate the final temperature of the gas (T2) by using the Gay-Lussac derived expression:

P1 x T2 = P2 x T1

⇒T2= (P2 x T1)/P1 = (225.9 kPa x 169 K)/71.8 kPa= 531.7 K ≅ 532 K

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Which of the following pieces of laboratory glassware should be used to most accurately measure out a 25.00 ml sample of a solut
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Answer:

The correct answer is 25 mL graduated cylinder (it should be used in all the cases)

Explanation:

In order to measure 25.00 ml sample of a solution it should be used a 25 mL graduated cylinder, as it is previously and properly calibrated. The other laboratory glassware, beaker and erlenmeyer, have graduations which are approximate, so they are used when exact volumes are not needed.

ii) graduated cylinder has the least uncertainly. It is more accurate than a beaker or erlenmeyer (to within 1%)

iii) A 25 mL graduated cylinder should be used because it is the most accurate lab glassware (between those were mentioned: beaker, erlenmeyer).

6 0
3 years ago
How many grams of solid nickel will be plated if an aqueous nickel(II) sulfate solution is electroplated over 10.5 min with a co
Anna71 [15]

Answer:

613 mg

Explanation:

$Ni^{2+} + 2e \rightarrow Ni(s)$

Number of fargday's  $=\frac{It}{96500}$

Here, I = 9.20 A

        t = 10.5 min

          = 10.5 x 60 seconds

So, $\frac{It}{96500}$

  $=\frac{3.20 \times 10.5 \times 60}{96500}$

 = 0.0208 F

Here, 2e, 2F

2F = 1 mol of Ni

$0.0208 \ F = \frac{0.0208}{2} = 0.0104 \ mol \ Ni$

1 mol = 59 gm of Ni

0.0104 mol = 59 x0.0104 gm Ni

                    = 0.613 gm Ni

                    = (0.613 x 1000 ) mg of Ni

                    = 613 mg of Ni

3 0
3 years ago
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