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alex41 [277]
3 years ago
5

The number of grams of H2 in 1470 mL of H2 gas. ​

Chemistry
1 answer:
Lorico [155]3 years ago
5 0

Answer:

0.1313 g.

Explanation:

  • It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.
  • Suppose that hydrogen behaves ideally and at STP conditions.

<u><em>Using cross multiplication:</em></u>

1.0 mol of hydrogen occupies → 22.4 L.

??? mol of hydrogen occupies → 1.47 L.

∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.

  • Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:

<em>The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen</em> = (6.563 x 10⁻² mol)(2.0 g/mol) = <em>0.1313 g.</em>

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A teacher moves a box of books weighting 25lbs a distance of 40ft. How much work has she done?
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Answer:

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3 years ago
How many calories would be needed to heat 5.0 lbs of copper from 22 degrees C to 80.0 degrees C? C for copper = 0.092
Pavlova-9 [17]

<u>Answer:</u>

<em>1.21\times 10^4 cal would be needed to heat 5.0 lbs of copper from 22 degrees C to 80.0 degrees C.</em>

<u>Explanation:</u>

Q=m \times c \times \Delta T

where

\Delta T = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for copper 0.092 cal/(g℃))  

m is the mass of water

plugging in the values

$Q=5.01 b s \times 0.092 \frac{c a l}{g^{\circ} \mathrm{c}} \times\left(80.0^{\circ} \mathrm{C}-22^{\circ} \mathrm{C}\right)$

Please Note:

1 lb = 453.592grams  

So,  

5 lbs = 5 × 453.592g  = 2268 g

$\begin{aligned} Q &=2268 g \times 0.092 \frac{c a l}{g^{\circ} \mathrm{C}} \times 58^{\circ} \mathrm{C} \\\\ Q &=12102 \mathrm{cal} \end{aligned}$

=1.21\times10^4 cal (Answer)

4 0
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Darya [45]

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4 years ago
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Answer:

6.94 × 10⁴ J

Explanation:

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W = -P . ΔV

where,

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Knowing that 1 atm.L = 101.325 J, the work of compression is:

W=-P\times \Delta V=-16.7atm\times(80.0L-121.0L)\times\frac{101.325J}{1atm.L} =6.94\times10^{4} J

The positive sign of the work means that the surroundings do work on the system.

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AveGali [126]

Answer:

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