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djyliett [7]
2 years ago
9

2. What is the temperature (°C) of 1.50 moles of gas stored in a 10.0 L container at 1558 mm

Chemistry
1 answer:
nalin [4]2 years ago
6 0

Answer:

-106°C

Explanation:

1558 mmHg / 1 × 1 atm / 760. mmHg = 2.05 atm

PV = nRT

2.05 atm (10.0 L) = 1.50 mol (0.082 atmL/molK) T

20.5 atmL = (0.123 atmL/K) T

K/0.123 atm L × 20.5 atmL = (0.123 atmL/K) T × K/0.123 atmL

166.666 K = T

167 K = T (sig figs)

K = °C + 273

167 - 273 = °C - 273

-106 = °C

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Mashutka [201]

Answer:

The density of the metal is 0.561 g/mL

Explanation:

The computation of the density of the metal is shown below;

As we know that

The Density of the metal is

= \frac{mass}{volume}

where,

Mass = 4.9g

Change in volume = 6.9 mL

Now place these values to the above formula

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3 years ago
2Al + 6HCl --> 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L
dem82 [27]

Answer:

8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

<em>Moles H2:</em>

PV = nRT; PV/RT = n

<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>

Replacing:

1atm*11L/0.082atmL/molK*273.15K = n

n = 0.491 moles of H2 must be produced

<em />

<em>Moles Al:</em>

0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required

<em />

<em>Mass Al -Molar mass: 26.98g/mol-:</em>

0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries

7 0
3 years ago
A certain gas occupies a volume of 26L when the applied pressure is 5.7atm. find the pressure in the system when the gas is comp
sergiy2304 [10]
P1 = 5.7atm V1 = 26L
P2 = ? V2 = 6.5 L

By Boyles Law,
P1V1 = P2V2
5.7 × 26 = P2 × 6.5
By solving,
P2 = 22.8atm.
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Explanation:

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3 years ago
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