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Marizza181 [45]
3 years ago
13

On a cold morning, the smoke coming out of a chimney is observed to be in blue color. What could be the reason? Name the effect

and the reason for the effect
Chemistry
1 answer:
nirvana33 [79]3 years ago
6 0

Stack effect? I'm not totally sure about this...

Explanation:

Large amount of tiny particles of water droplets, dust and smoke are present on a misty day. These tiny particles in the air scatter blue colour of white light passing through it. When this scattered light reaches our eyes, the smoke appears blue.

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Explain why vanadium (radius=134 pm) and copper (radius=128 pm) have nearly identical atomic radii, even though the atomic numbe
Viktor [21]
<span>There are few main factors affecting the atomic radii, the outermost electrons and the protons in the nucleus and also the shielding of the internal electrons. I would speculate that the difference in radii is given by the electron clouds since the electrons difference in these two elements is in the d orbital and both has at least 1 electron in the 4s (this 4s electron is the outermost electron in all the transition metals of this period). The atomic radio will be mostly dependent of these 4s electrons than in the d electrons. Besides that, you can see that increasing the atomic number will increase the number of protons in the nucleus decreasing the ratio of the atoms along a period. The Cu is an exception and will accommodate one of the 4s electrons in the p orbital.

</span><span>Regarding the density you can find the density of Cu = 8.96g/cm3 and vanadium = 6.0g/cm3. This also correlates with the idea that if these two atoms have similar volume and one has more mass (more protons;  density is the relationship between m/V), then a bigger mass for a similar volume will result in a bigger density.</span>
7 0
3 years ago
Whhat are the maximum number of molecules present in 10g of O2 gas at STP?
Romashka [77]

Of course, at STP, dioxygen is a gas, but 10.0 g is still 10.0 g. We could calculate its volume at STP, which is 22.4 L × its molar quantity, approx. 8⋅L . There are 1.51×1023molecules O2 in 10.0 g O2 .

6 0
3 years ago
1. Using the Slater rule, determine the effective nuclear charge of platinum.
AleksandrR [38]

Answer:

Z* = 3.55

Explanation:

Slater rule says that:

Z*= Z - S

Z* be the nuclear effective charge

Z is the nuclear charge

S is the shielding constant

First we write the electronic configuration of platinum:1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 4f^{14} 5d^{9} 6s^{1}

The first Slater rule says that we need to group:

(1s^{2}) (2s, 2p)^{8} (3s, 3p)^{8} (3d^{10}) (4s, 4p)^{8} (4d^{10}) (5s, 5p)^{8} (4f^{14}) (5d^{9}) (6s^{1})

The second rule says that the electrons to the right are not shielding, but we are going to solve the exercise for the last level (6s), so we don't have electrons to the right.

For the third rule we have two considerations, if is ns or np and if is nd or nf:

For our case, we have an electro that is in ns, so the rule says that

-electrons within same group shield 0.35, except the 1s which shield 0.30

-electrons within the n-1 group shield 0.85

-electrons within the n-2 or lower groups shield 1.00

Now we can proceed with the calculation:

The first consideration in the third rule does not apply as we only have one electron on this level.

The second consideration will be as follow for the level 5, where we have 17 electrons.

Finally the third consideration will be for levels 1, 2, 3 and 4, where we have 14 for 4f, 10 for 4d, 8 for 4s and 4p, 10 for 3d, 8 for 3s and 3p, 8 for 2s and 2p and finally 2 for 1s, which gives 60 electrons.

So the result for S=(60*1.00 + 17*0.85) = 74.45

And the equation is: Z* = 78 - 74.45

So Z* = 3.55

3 0
3 years ago
How many moles of Na2SO4 can be made with 7.28 mol of H2SO4?
Slav-nsk [51]

Answer:

7.28 mol Na2SO4

Explanation:

Since it is already in moles, all we have to do is use a molar ratio

A molar ratio is the proportions of reactants and products using the balanced equation. When writing a mole ratio, the given information must cross out with the right thing.

7.28 mol H2SO4 * 1 mol Na2SO4/1 H2SO4 = 7.28 mol Na2SO4

*notice how the H2SO4 crosses out

5 0
3 years ago
If iodine-131 has a half-life of 8 days, how much of a 64.0 g sample of iodine-131 will remain after 32 day?
Mariulka [41]

Answer:

4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.

Explanation:

Half life (t1/2) = 8 days

Original mass (No) = 64 g

Elapsed time (t) = 32 days

Mass remaining (Nt) = ?

Using the half life equation we can obtain the mass remaining (Nt)

Nt = No (1/2) ^t/t1/2

Substituting the values, we have;

Nt = 64 * ( 1/2 ) ^32/8

Nt = 64 * (1/2) ^4

Nt = 64 * 0.0625

Nt = 4 g

So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.

8 0
3 years ago
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