How many km is 250 m?

1.Oxygen (O) is a gas found in the 16th column of the periodic table. What statement is true about oxygen and the other elements

Lady_Fox [76]

1. A. All the elements in the column have similar chemical properties.

2. Substances on the periodic table cannot be broken down into other substances and are therefore elements.

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Why are all of the substances on the periodic table classified as elements?

Readme [11.4K]

All elements are pure substances.

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Which pair shares the same empirical formula?

ZanzabumX [31]

Answer:- 3. $CH_3$ and $C_2H_6$

Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.

For example, the molecular formula of benzene is $C_6H_6$ . The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.

In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and the empirical formula is $CH_2$ . In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.

In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is $CH_2$ . The C to H ratio for second molecule is 1:4, so the empirical formula is $CH_4$ . Here also, the empirical formulas are not same and hence it is also not the right choice.

In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is $CH_3$ . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also $CH_3$ . Hence. this is the correct choice.

In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is $CH_2$ . As the empirical formulas are different, it is not the right choice.

So, the only and only correct pair is the third one. 3. $CH_3$ and $C_2H_6$

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Name three physical properties that differentiate oxides from hydroxide group of minerals.

erik [133]

Answer:

2k20

Explanation:

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g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig

kumpel [21]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

$v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s$

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

$K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J$

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

$\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K$

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, $\Delta S_{env} = -1.18 J/K$

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: brainly.com/question/22655760

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