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2 years ago

what property of the noble gases most likely prevented the gases from being readily/easily discovered?

Chemistry

1 answer:

Vlad [161]2 years ago

7 0

These gases very rarely react, with others and also noble gases are odourless and colourless.

Explanation:

Noble gases will not react with anything so that is the reason why they are known as an inert gas.

Noble gases are present in group 18 on the periodic table and following the rule of the octet which is they completed their orbital by s2p6 which is the highest energy level.

Most elements are discovering through their reactivity with the other elements, commonly with oxygen. In the case of a noble gas, it is difficult for a scientist to work with the gases which have very less or no chemical property in terms of their reactivity.

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6. What is the molarity of 1.35 mol H2So4 in 245<br><br> mL of solution?

vovikov84 [41]

Answer:

5.51mol/L

Explanation:

Number of moles = 1.35moles

Volume of the solution = 245mL = 245*10^-3L = 0.245L

Molarity of a solution is the defined as the number of moles of a solute dissolved in 1L of the solution.

1.35 moles = 0.245L

X moles = 1L

X = (1.35 * 1) / 0.245

X = 5.51mol/L

The molarity of the solution is 5.51mol/L

7 0

2 years ago

Automobile battery acid is 38% H2SO4 and has a destiny of 1.29g/ml. Calculate the molality and the molarity of this solution.

baherus [9]

Answer:

$M=5.0M\\\\m=6.2m$

Explanation:

Hello,

In this case, 38 % is commonly a by mass concentration, meaning that we have 38 grams of solute (sulfuric acid) per 100 grams of solution (water+sulfuric acid):

$38\%=\frac{m_{H_2SO_4}}{m_{H_2SO_4}+m_{H_2O}}$

Hence, we compute the moles of sulfuric acid in 38 grams by using its molar mass (98 g/mol):

$n_{H_2SO_4}=38g*\frac{1mol}{98g}=0.39mol H_2SO_4$

Next, the volume of the solution in litres by using the density of the solution:

$V_{solution}=100g*\frac{1mL}{1.29g}*\frac{1L}{1000mL} =0.0775L$

This is done since the molarity is defined as the ratio of the moles of the solute to the volume of the solution in litres, thus we have:

$M=\frac{n}{V}=\frac{0.38mol}{0.0775L}=5.0M$

On the other hand, the molality is defined as the ratio of the moles of the solute to the mass of the solvent in kilograms, thus, we compute the mass of water (solvent) as shown below:

$m_{H_2O}=100g-38g=62g*\frac{1kg}{1000g}=0.062kg$

So compute the molality:

$m=\frac{n_{solute}}{m_{solvent}}=\frac{0.39mol}{0.062kg}=6.2m$

Regards.

7 0

2 years ago

Read 2 more answers

THE SPENT ACID FROM A NITRATION PROCESS CONTAINS 33% H2SO4, 36% HNO3 AND 31% H2O. THIS IS TO BE STRENGTHENED BY THE ADDITION OF

Nady [450]

Answer:

a) 217.15 kg of spent acid

b) 556.14 kg of concentrate H₂SO₄

c) 726.70 kg of concentrate HNO₃

Explanation:

First, let's do a sketch for the problem (image below) and let's choose a basis of 1500 kg of production of the desired acid. Calling A the flow of spent acid, B the amount of concentrated H₂SO₄, C the flow of concentrated HNO₃, and P the flow of the product.

a) The global mass balance is:

input = output

A + B + C = P

A + B + C = 1500

The mass balance for H₂SO₄:

0.33A + 0.95B = 0.4x1500

0.33A + 0.95B = 600

0.95B = 600 - 0.33A

B = (600 - 0.33A)/0.95

The mass balance for HNO₃:

0.36A + 0.78C = 0.43x1500

0.36A + 0.78C = 645

0.78C = 645 - 0.36A

C = (645 - 0.36A)/0.78

Replacing B and C in the global mass balance:

A + (600 - 0.33A)/0.95 + (645 - 0.36A)/0.78 = 1500

0.1911A + 1458.5 = 1500

0.1911A = 41.5

A = 217.15 kg of spent acid

b) B = (600 - 0.33A)/0.95

B = (600 - 0.33x217.15)/0.95

B = 556.14 kg of concentrate H₂SO₄

c) C = (645 - 0.36A)/0.78

C = (645 - 0.36x217.15)/0.78

C = 726.70 kg of concentrate HNO₃

5 0

2 years ago

A __________ bond is a chemical bond between atoms with similar electronegativities; a sharing bond; valence electrons are share

11Alexandr11 [23.1K]

Covelent bonds wich is a bond between atoms

5 0

2 years ago

Read 2 more answers

You work for a cutlery manufacturer who wants to electrolytically precipitate 0.500 g of silver onto each piece of a batch of 25

ICE Princess25 [194]

Answer:

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

Explanation:

Mass of silver to be precipitated on ecah spoon = 0.500 g

Number of silver spoons = 250

Total mass of silver = 250 × 0.500 g = 125 g

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of AgCN = n = $\frac{125 g}{134 g/mol}=0.9328 mol$

Volume of AgCN solution =V

Molarity of the AgCN = 2.50 M

$V=\frac{0.9328 mol}{2.50 M}=0.3731 L=373.1 mL$

(1 L = 1000 mL)

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

6 0

2 years ago