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3 years ago

A cylinder is filled with 2.00 moles of nitrogen, 3.00 moles of argon, and 5.00 moles of helium.

Chemistry

2 answers:

Leona [35]3 years ago

7 0

Answer:

228 torr

Explanation:

Aleks [24]3 years ago

5 0

PRACTICE MULTIPLE CHOICE QUESTIONS FROM UNIT 10

1.) A cylinder is filled with 2.00 moles of nitrogen, 3.00 moles of argon, and 5.00 moles of helium. If

the gas mixture is at STP, what is the partial pressure of the argon?

(A) 152 torr (B) 228 torr (C) 380. torr (D) 760. torr

2.) Compared to the average kinetic energy of 1 mole of water at 0 oC, the average kinetic energy

of 1 mole of water at 298 K is

(A) the same, and the number of molecules is the same

(B) the same, but the number of molecules is greater

(D) greater, but the number of molecules is the same

3.) If the pressure on a given mass of gas in a closed system is increased and the temperature

remains constant, the volume of the gas will

(A) decrease (B) increase (C) remain the same

4.) Which gas has approximately the same density as C2H6 at STP?

(A) NO (B) NH3 (C) H2S (D) SO2

5.) At a temperature of 273 K, a 400. milliliter gas sample has a pressure of 760. millimeters of

mercury. If the pressure is changed to 380. millimeters of mercury, at which temperature will

this gas sample have a volume of 551 milliliters?

(A) 100 K (B) 188 K (C) 273 K (D) 546 K

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3 years ago

In a classical long run supply model the economy is always doing what

ch4aika [34]

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3 years ago

A waste treatment pond is 50 m long and 25 m wide, and has an average depth of 2 m. The density of the waste is 75.3lbm/ft^3 .Ca

umka21 [38]

Answer:

$W = 6.65 \cdot 10^{6} lbf$

Explanation:

To find the weight (W) of the pond contents first we need to use the following equation:

$W = m\cdot g$ (1)

Where m the mass and g is the gravity

Also, we have that the mass is:

$m = \rho*V$ (2)

Where ρ is the density and V the volume

We cand calculate the volume as follows:

$V = L*w*d$ (3)

Where L is the length, w is the wide and d is the depth

By entering equation (2) and (3) into (1) we have:

$W = \rho*L*w*d*g$

$W = 75.3 lbm/ft^{3}*50 m*25 m*2 m*9.81 m/s^{2}$

$W = 75.3 lbm/ft^{3}*\frac{(1 ft)^{3}}{(0.3048 m)^{3}}*\frac{0.454 kg}{1 lbm}*50 m*25 m*2 m*9.81 m/s^{2} = 2.96 \cdot 10^{7} N}*\frac{0.2248 lbf}{1 N} = 6.65\cdot 10^{6} lbf$

Therefore, the weight of the pond is 6.65x10⁶ lbf.

I hope it helps you!

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3 years ago

Aqueous solutions of isopropyl alcohol are commonly sold as rubbing alcohol. The boiling point of isopropyl alcohol is 82.4 °C.

Monica [59]

Answer:

This is due to more hydrogen bonding in ethylene glycol than it is in isopropyl alcohol

Explanation:

The boiling point of isopropyl alcohol is 82.4 °C it contains only a single OH group, hence intermolecular hydrogen bonding is solely responsible for it's boiling point, whereas Ethylene glycol (CH2OHCH2OH) contains 2-OH group and both intermolecular and intramolecular hydrogen bonding are responsible for the higher boiling point of ethylene glycol at 198 °C.

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3 years ago

A scientist measures the standard enthalpy change for the following reaction to be -17.2 kJ : Ca(OH)2(aq) 2 HCl(aq)CaCl2(s)

elixir [45]

Answer: $\Delta H^{0}=-173.72$ kJ/mol

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The standard enthalpy of formation of HCl is calculated as:

$\Delta ^{0}=\Sigma H_{products}-\Sigma H_{reactants}$