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CaHeK987 [17]
2 years ago
15

Mendeleev created this table as he noticed that a

Chemistry
1 answer:
Nimfa-mama [501]2 years ago
4 0

Answer:

Mendeleev had left the noble gases out of his periodic table.

Explanation:

Mendeleev's periodic table is pictured in the image attached to the question.

Mendeleev's table obviously lacked the noble gases. The reason for this grave omission is simple; the noble gases were not known as at the time when he formulated his periodic table. There weren't any known elements  whose properties were similar to the  properties of the noble gases. This would have lead him to suspect their existence.

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Advocard [28]
D The Part That Stays The Same
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3 years ago
The dissolution of 0.200 l of sulfur dioxide at 19 °c and 745 mmhg in water yields 500.0 ml of aqueous sulfurous acid. The solut
aivan3 [116]

Answer:

Molarity=1.22\ M

Explanation:

Given:  

Pressure = 745 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 745 / 760 = 0.9803 atm

Temperature = 19 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (19 + 273.15) K = 292.15 K  

Volume = 0.200 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9803 atm × 0.200 L = n × 0.0821 L.atm/K.mol × 292.15 K  

⇒n = 0.008174 moles

From the reaction shown below:-

H_2SO_3+2NaOH\rightarrow Na_2SO_3+2H_2O

1 mole of H_2SO_4 react with 2 moles of NaOH

0.008174 mole of H_2SO_4 react with 2*0.008174 moles of NaOH

Moles of NaOH = 0.016348 moles

Volume = 13.4 mL = 0.0134 L ( 1 mL = 0.001 L)

So,

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.016348}{0.0134}\ M

Molarity=1.22\ M

8 0
3 years ago
In a solution of a carbonated beverage the water is what? Solute, saturated, solvent or precipitate
Scilla [17]
Https://us-static.z-dn.net/files/d49/33d4ec86853ef95e6f6c14242c663be4.png
5 0
3 years ago
Consider the reaction. mc014-1.jpg How many grams of methane should be burned in an excess of oxygen at STP to obtain 5.6 L of c
garri49 [273]
The reaction for the combustion of methane can be expressed as follows.
                           CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
                            moles of CO2 = (5.6 L) / (22.4 L/1 mole)
                             moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
                              moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
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Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane. 
5 0
3 years ago
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anyanavicka [17]

Answer: B

Explanation: The answer is B

6 0
3 years ago
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