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Vladimir [108]
2 years ago
8

Write the salient features of long form of Periodic Table .​

Chemistry
2 answers:
WITCHER [35]2 years ago
5 0

Main Features of the Long Form of the Periodic Table :

<u>(1) It shows arrangement of elements based on modern periodic law</u>.

<u> (2) There are 18 vertical columns known </u><u> </u><u> </u><u>a</u><u>s</u><u> </u><u>groups. </u>

<u>(3) There are 7 horizontal rows known as periods.</u>

frez [133]2 years ago
3 0

Answer:

(1) It shows arrangement of elements based on modern periodic law.

(2) There are 18 vertical columns known as groups.

(3) There are 7 horizontal rows known as periods.

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Que diferencia hay entre drogas estimuladoras depresoras y perturbadoras del sistema nervioso
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3 years ago
What is the frequency of this radiation
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6 0
3 years ago
Consider the following specific heats: copper, 0.384 J/g· ◦C; lead, 0.159 J/g· ◦C; water, 4.18 J/g· ◦C; glass, 0.502 J/g· ◦C. Wh
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Answer:

The correct answer is

2. Water

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7 0
3 years ago
A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound
Karolina [17]

Answer:

The empirical formula of the compound is C_{0.504}HO_{1.008}.

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen (H) is scaled up to a mole, since it has the molar mass, and both carbon (C) and oxygen (O) are also scaled up in the same magnitude. The empirical formula is of the form:

C_{x}HO_{y}

Where x, y are the number of moles of the carbon and oxygen, respectively.

The scale factor (r), no unit, is calculated by the following formula:

r = \frac{M_{H}}{m_{H}} (1)

Where:

m_{H} - Mass of hydrogen, in grams.

M_{H} - Molar mass of hydrogen, in grams per mole.

If we know that  M_{H} = 1.008\,\frac{g}{mol} and m_{H} = 0.2\,g, then the scale factor is:

r = \frac{1.008}{0.2}

r = 5.04

The molar masses of carbon (M_{C}) and oxygen (M_{O}) are 12.011\,\frac{g}{mol} and 15.999\,\frac{g}{mol}, then, the respective numbers of moles are: (r = 5.04, m_{C} = 1.2\,g, m_{O} = 3.2\,g)

Carbon

n_{C} = \frac{r\cdot m_{C}}{M_{C}} (2)

n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }

n_{C} = 0.504\,moles

Oxygen

n_{O} = \frac{r\cdot m_{O}}{M_{O}} (3)

n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }

n_{O} = 1.008\,moles

Hence, the empirical formula of the compound is C_{0.504}HO_{1.008}.

3 0
2 years ago
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