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denpristay [2]
3 years ago
8

What are alternative periodic tables

Chemistry
1 answer:
aksik [14]3 years ago
5 0
Tabulations of chemical elements differing in their organization from the traditional seen periodic system
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Question 2 (1 point)
Alexxandr [17]

Answer:

a mole have have wrong becouse i understand

4 0
2 years ago
A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the
Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, C_6H_4N_2O_4

Explanation :

First we have to calculate the mass of benzene.

\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}

\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g  = 0.04395 kg

Formula used :  

\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}

where,

\Delta T_f = change in freezing point  = 5.53-1.37=4.16^oC

\Delta T_s = freezing point of solution

\Delta T^o = freezing point of benzene

Molal-freezing-point-depression constant (K_f) for benzene = 5.12^oC/m

m = molality

Now put all the given values in this formula, we get

4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}

\text{Molar mass of unknown compound}=180.6g/mol

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles

Moles of N = \frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{3.575}{1.193}=2.99\approx 3

For H = \frac{2.4}{1.193}=2.01\approx 2

For N = \frac{1.193}{1.193}=1

For O = \frac{2.381}{1.193}=1.99\approx 2

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_3H_2N_1O_2

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{180.6}{84}=2

Molecular formula = (C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4

Therefore, the molecular of the compound is, C_6H_4N_2O_4

3 0
3 years ago
Please give me the reasons of the solution!
irinina [24]

1. The answer is option E, that is None of the above is correct.

As a polymer becomes more crystalline,

its melting point doesn't decreases, its density doesn't decreases, its stiffness doesn't decreases and its yield stress doesn't decreases.

2. The answer is option B, that is the molecules are arranged in sheets, with their long axes parallel and their ends aligned as well.

In the smectic A liquid-crystalline phase, molecules are arranged in sheets, with their long axes parallel and their ends aligned as well.

3.  For a substitutional alloy to form, the two metals combined must have similar atomic radii and chemical bonding properties.

6 0
2 years ago
Which of these is an acid?
julsineya [31]
My best guess would be c
8 0
3 years ago
Calculate the mass of each element in the
Lelechka [254]

Answer:

For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively

Explanation:

The computation of the mass of each element is given below:

As we know that

A1 mole of ammonium nitrate i.e. 2 mol N, 4 mol H, 3 mol

Now we multiply each of above  by the molar masses

For N

= 14.0 g/mol × 2

= 28.0 gN

For H

= 1.0 g/mol × 4

= 4.0 gN

ANd, for O

= 16.0 g/mol × 3

= 48.0 gN

Hence, For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively

7 0
2 years ago
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