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Elanso [62]
2 years ago
11

A real gas behaves more like an ideal gas when the gas molecules are

Chemistry
1 answer:
Firlakuza [10]2 years ago
5 0
The answer is D, far apart and have weak attractive forces between them. The ideal gas means that the volume of molecule and the forces between them can be ignored.
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Which season will the northern hemisphere experience if the axis of Earth is tilted away from the sun?
pashok25 [27]

the answer would be winter because the north would be facing away from the son therefor making the northern states cold.

8 0
3 years ago
Read 2 more answers
The rate of effusion of oxygen to an unknown gas is 0.935. what is the other gas?
Kamila [148]

Answer:

N2

Explanation:

Rate of effusion is defined by Graham's Law:  

(Rate 1/Rate 2) = (sqrt (M2)/ sqrt (M1))

(Where M is the molar mass of each substance. )

Molar Mass of oxygen, O2, is 32 (M1).

Rate of effusion of O2 to an unknown gas is .935(Rate 1).  

Rate 2 is unknown so put 1.

Solve for x (M2).

.935/1 = sqrt x/ sqrt32

.935 x sqrt 32 = sqrt x

5.29 = sq rt x

5.29^2 = 27.975 = 28  

N2 has a molar mass of 28 so it is the correct gas.

4 0
3 years ago
What is the most important monosaccharide on Earth?
Lelechka [254]

Answer:

the answer is A Glucose

hope it helps!

6 0
3 years ago
The total volume required to reach the endpoint of a titration required more than the 50 mL total volume of the buret. An initia
Margaret [11]

Answer:

The endpoint volume is 50.52 ±  0.14 mL

Explanation:

In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:

V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)

V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL

V = 50.52 ±  0.14 mL

It is necessary to consider the sum of the errors too.

7 0
3 years ago
Calculate the work (kJ) done during a reaction in which the internal volume expands from 20 L to 43 L against an outside pressur
Alenkinab [10]

Answer:

\large \boxed{\text{-10.0 kJ}}

Explanation:

1. Calculate the work

w = - pΔV = -4.3 atm × (43 L - 20 L) = -4.3 × 23 L·atm = -98.9 L·atm

2. Convert litre-atmospheres to joules

w = \text{-98.9 L\cdot$atm } \times \dfrac{\text{101.3 J}}{\text{1 L$\cdot$atm }} = \text{-10000 J} = \textbf{-10.0 kJ}\\\\\text{The work done is $\large \boxed{\textbf{-10.0 kJ}}$}

The negative sign indicates that the work was done against the surroundings.

4 0
3 years ago
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