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Varvara68 [4.7K]
2 years ago
4

One mole of iron A. is heavier than one mole of lead (Pb). B. is 77.0 g of iron. C. is 26.0 g of iron. D. weighs the same as one

mole of lead. E. None of the above.
Chemistry
1 answer:
zalisa [80]2 years ago
3 0

Answer:

E. None of the above.

Explanation:

Given 1 mole of iron. We need to find the mass of the iron. And we will compare mass of the iron with the given options.

The molar mass of the iron is 55.845 g/mol.

So, for 1 mole the mass of the iron would be 55.845\times 1=55.845\ g

Also, the molar mass of the lead is 207.2 g/mol.

So, for 1 mole the mass of the lead would be 207.2\times 1=207.2\ g

We can see that one mole of lead is heavier than one mole of iron. So, the only option is correct is  E. None of the above.

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Give two examples (i.e. list 2 elements that are examples) of: a. an atom with a half-filled subshell b. an atom with a complete
elena55 [62]

Answer:

an atom with a half-filled subshell - hydrogen

an atom with a completely filled outer shell - argon

an atom with its outer electrons occupying a half-filled subshell and a filled subshell- copper

Explanation:

The outermost shell or the valence shell of the atom is the last shell in the atom. Chemical reactions occur at this outer most shell. The number of electrons on the outermost shell of an atom determines the group to which it belongs in the periodic table as well as its chemical properties.

Hydrogen has a half filled 1s sublevel. Only one electron is present in this sublevel.

Let us consider argon

1s2 2s2 2p6 3s2 3p6

The outermost ns and np levels are completely filled. Thus the outermost shell is completely filled.

In the last case; let us look at the electronic configuration of nitrogen;

1s2 2s2 2p3

The outermost 2p subshell is exactly half filled while the 2s sublevel is fully filled. The outermost shell of nitrogen is made up of 2s2 and 2p3 sublevels.

4 0
2 years ago
A 1.00 l solution contains 3.50×10-4 m cu(no3)2 and 1.75×10-3 m ethylenediamine (en). the kf for cu(en)22+ is 1.00×1020. what is
Alik [6]
<span>Answer: A 1.00 L solution containing 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en). contains 0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 moles of Cu(en)2^2+ Kf for Cu(en)2^2+ is 1x10^20. so 1 Cu+2 & 2 en --> Cu(en)2^2+ Kf = [Cu(en)2^2+] / [Cu+2] [en]^2 1x10^20. = [0.000300] / [Cu+2] [0.00180 ]^2 [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6) Cu+2 = 9.26 e-19 Molar since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2</span>
4 0
2 years ago
A sample of an ideal gas occupies 2.78 x 10^3 mL at 25°C and 760 mm Hg.
iris [78.8K]

Answer: It will occupy 4.45\times 10^3ml at the same temperature and 475 mm Hg.

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

P_1V_1=P_2V_2    (At constant temperature and number of moles)

where,

P_1 = initial pressure of gas = 760 mm Hg

P_2 = final pressure of gas = 475 mm Hg

V_1 = initial volume of gas = 2.78\times 10^3ml

V_2  = final volume of gas = ?

Putting in the values:

760mm Hg\times 2.78\times 10^3ml=475 mm Hg\times V_2

V_2=4.45\times 10^3ml

Thus it will occupy 4.45\times 10^3ml at the same temperature and 475 mm Hg

5 0
2 years ago
Read 2 more answers
22. How many atoms are there in 344.75 g of gold nugget? a. 1.05 x 10 to the power of 24 atoms b. 1.05 x 10 to the power of 23 a
mixas84 [53]

Answer:

1.053×10²⁴ atoms of gold

Explanation:

Hello,

Gold nugget are usually the natural occurring gold and they contain 85% - 90% weight of pure gold.

In this question, we're required to find the number of atoms in 344.75g of a gold nugget.

We can use mole concept relationship between Avogadro's number and molar mass.

1 mole = molar mass

Molar mass of gold = 197 g/mol

1 mole = Avogadro's number = 6.022 × 10²³ atoms

Number of mole = mass / molar mass

Mass = number of mole × molar mass

Mass = 1 × 197

Mass = 197g

197g is present in 6.022×10²³ atoms

344.75g will contain x atoms

x = (344.75 × 6.022×10²³) / 197

X = 1.053×10²⁴ atoms

Therefore 344.75g of gold nugget will contain 1.053×10²⁴ atoms of gold

5 0
2 years ago
When Z-4,5-dimethyloct-4-ene is treated with hydrogen chloride, HCl, the result is:________.
Vika [28.1K]

Answer:

The  correct option is  c

Explanation:

The chemical equation for the reaction of  Z-4,5-dimethyloct-4-ene and HCl is shown on the first uploaded image

Now looking at the product we see that there are two who has four different groups attached to them this carbon are known as chiral carbons hence the product formed is a pair of diastereomers

6 0
2 years ago
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