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8_murik_8 [283]
2 years ago
5

OxYgen and hydrogen particles are kept in a container. Which of the following

Chemistry
1 answer:
Tema [17]2 years ago
7 0

Answer:

Option D

Oxygen particles mix evenly with hydrogen particles.

Explanation:

Diffusion refers to the movement of gasses from a region of higher concentration to a region of lower concentration.

Diffusion occurs in gases because their particles are in a state of constant random motion.

Due to this continuous random motion, once the oxygen and the hydrogen are added to the container, they will begin to mix freely with each other until they fill up the entire volume of the container.

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Which symbol in a chemical equation separates the reactants from the products?​
max2010maxim [7]

Answer:

arrow

Explanation:

ex. xy+xy>xy+xy

- Hope that helped! Let me know if you need a further explanation.

4 0
3 years ago
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vfiekz [6]
It's either genus or species. I'd go with Genus
6 0
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What is the frequency of yellow light, which has a wavelength of 5.56 x 10^-7 m?
Mnenie [13.5K]
The answer is B 5.40 x 10^14 Hz
7 0
2 years ago
Which one of these is the accepted name for the compound n2o5?
Leya [2.2K]
C is the correct answer. For nonmetal-nonmetal compounds, use the Greek prefixes

Mono - 1
Di - 2
Tri - 3
Tetra - 4
Penta - 5
Hexa - 6
Hepta - 7

And so on...

The reason D is not right is because a and o cannot be next to each other in between prefix and element.

Hope this helps!

6 0
3 years ago
A sample of hydrogen gas at a pressure of 977 mm Hg and a temperature of 32 °C, occupies a volume of 11.3 liters. If the gas is
IrinaK [193]

Answer : The volume of gas sample will be, 9.93 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles of gas.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=11.3L\\T_1=32^oC=(32+273)K=305K\\V_2=?\\T_2=-5^oC=(-5+273)K=268K

Now put all the given values in above equation, we get:

\frac{11.3L}{305K}=\frac{V_2}{268K}\\\\V_2=9.93L

Therefore, the volume of gas sample will be, 9.93 L

6 0
3 years ago
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