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snow_tiger [21]
3 years ago
15

Would it be C???...idk HELP

Chemistry
1 answer:
givi [52]3 years ago
8 0

Answer:

C.  Butanal , is the aldehyde

Explanation:

A . It is carboxylic acid : ---COOH group

B. It is Ester : ----COOR group , Here R = CH3

C. It is Aldehyde : -----CHO group

D. It is ketone : ----C=O group

See image :

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How do exothermic and endothermic differ
Ber [7]

Answer:

Explanation:

exothermic reactions involve release of heat whereas endothermic reaction involve absorption of heat.

6 0
2 years ago
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The alkali earth metal beryllium (Be) engages in a chemical reaction and loses all of its valence electrons.
denpristay [2]

The loss of electron from an results in the formation of cation represented by the positive charge on the element whereas gaining of electron results in the formation of anion represented by the negative charge on the element.

The alkali earth metal beryllium (Be) belongs to the second group of the periodic table. The ground state electronic configuration of Be is:1s^{2}2s^{2}

From the electronic configuration it is clear that it has 2 valence electrons in its valence shell (2s^{2}).

After losing all valence electrons that is 2 electrons from 2s orbital. The electronic configuration will be:

1s^{2}2s^{0}

Since, lose of electron is represented by positive charge on the element symbol. So, the beryllium will have +2 charge on its symbol as Be^{2+}.

Hence, beryllium will have 2+ charge on it after losing all its valence electrons in the chemical reaction.

6 0
3 years ago
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What happens on a molecular level when a diatomic molecule is a gas but is then cooled to a solid?
irina1246 [14]

Answer:

As an example of the processes depicted in this figure, consider a sample of water. When gaseous water is cooled sufficiently, the attractions between H2O molecules will be capable of holding them together when they come into contact with each other; the gas condenses, forming liquid H2O. For example, liquid water forms on the outside of a cold glass as the water vapor in the air is cooled by the cold glass.

Explanation:

Hopefully that helps!

4 0
3 years ago
You begin pouring sodium chloride into a glass of water. For a long time, the sodium chloride just dissolves in the water, but s
Gennadij [26K]

Answer:

  • Option D:<u><em> The water is saturated</em></u>

Explanation:

Solubility is the term used to designate the maximum amount of a solute that can be dissolved in a given amount of solvent, at a given temperature and pressure.

At start, you begin with pure water in the glass. This water is a pure substance, not a solution.

When <em>you begin pouring sodium chloride into the glass of water,</em> a solution, i.e. a homogenous mixture of solute and solvent, is formed.

This solution, at first, is diuted, which means that it contanins just few grams (matter) of the solute dissolved.

As, more sodium chloride is dissolved, the solution becomes more concentrated but is unsaturated. At some point, the water cannot dissolve more sodium chloride, because it has reached the maximum amount that can contain at that temperature and pressure. Then, the solution is saturated.

You can tell that the water is saturated, ie it contaiins the maximum amount of sodium chloride that can be dissolved, by that amount of water, at the given temperature, because from that point, you will note that <em>as you pour more sodium chloride, it begins to pile up at the bottom of the glass</em>. Hence, the true statement is the letter D: <u><em>the water is saturated.</em></u>

5 0
3 years ago
Consider the reaction 4PH3(g) → P4(g) + 6H2(g) At a particular point during the reaction, molecular hydrogen is being formed at
Svetllana [295]

Answer:

The rate at which P_4 is being produced is 0.0228 M/s.

The rate at which PH_3 is being consumed is 0.0912 M/s.

Explanation:

4PH_3\rightarrow P_4(g)+6H_2(g)

Rate of the reaction : R

R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}

The rate at which hydrogen is being formed = \frac{d[H_2]}{dt}=0.137 M/s

R=\frac{1}{6}\frac{d[H_2]}{dt}

R=\frac{1}{6}\times 0.137 M/s=0.0228 M/s

The rate at which P_4 is being produced:

R=\frac{1}{1}\frac{d[P_4]}{dt}

0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}

The rate at which PH_3 is being consumed :

R=\frac{-1}{4}\frac{d[PH_3]}{dt}

0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}

\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s

6 0
3 years ago
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