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2 years ago

Why did mendeleev leave blank spaces in his version of the periodic table?.

Chemistry

2 answers:

irga5000 [103]2 years ago

6 0

Answer:

Mendeleev left blank spaces in his version of the periodic table because he left the gaps to place elements not known at the time

Leviafan [203]2 years ago

3 0

<h2>Answer: He left blank spaces for <em>undiscovered</em> elements to be added to the table.</h2>

Explanation:

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How many valence electrons does carbon have

Vinvika [58]

Carbon has 4 valence electrons, oxygen has 6

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3 years ago

Read 2 more answers

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

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Give two examples of activities performed by the veterinarian

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Answer:

exp:

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How can I balance this equation? ____ KClO3 ---> ____ KCl + ____ O2

tensa zangetsu [6.8K]

__ KClO₃ → __ KCl + __ O₂

Left Side:
1 K
1 Cl
3 O

Right Side:
1 K
1 Cl
2 O

Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.

This gives us 2KClO₃ → __ KCl + 3O₂.

However, this equation is still not balanced.

Left Side:
2 K
2 Cl
6 O

Right Side:
1 K
1 Cl
6 O

In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.

2KClO₃ → 2KCl + 3O₂

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