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3 years ago

Please explain to me how to answer no.2 questions

Chemistry

1 answer:

bulgar [2K]3 years ago

4 0

Answer: -

IE 1 for X = 801

Here X is told to be in the third period.

So n = 3 for X.

For 1st ionization energy the expression is

IE1 = 13.6 x Z ^2 / n^2

Where Z =atomic number.

Thus Z =( n^2 x IE 1 / 13.6)^(1/2)

Z = ( 3^2 x 801 / 13.6 )^ (1/2)

= 23

Number of electrons = Z = 23

Nearest noble gas = Argon

Argon atomic number = 18

Number of extra electrons = 23 – 18 = 5

a) Electronic Configuration= [Ar] 3d34s2

We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.

So more the first ionization energy, less is the atomic radius.

X has more IE1 than Y.

b) So the atomic radius of X is lesser than that of Y.

c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom.

Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.

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Answer:

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Explanation:

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3 years ago

How many grams of HF are needed to react with 3.0 moles of Sn?

Flauer [41]

Answer:

120g

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction.

Sn + 2HF —> SnF2 + H2

Step 2:

Determination of the number of mole HF needed to react with 3 moles of Sn.

From the balanced equation above,

1 mole of Sn and reacted with 2 moles of HF.

Therefore, 3 moles Sn will react with = 3 x 2 = 6 moles of HF.

Step 3:

Conversion of 6 moles of HF to grams.

Number of mole HF = 6 moles

Molar Mass of HF = 1 + 19 = 20g/mol

Mass of HF =..?

Mass = number of mole x molar Mass

Mass of HF = 6 x 20

Mass of HF = 120g

Therefore, 120g of HF is needed to react with 3 moles of Sn.

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3 years ago

Which of the outer planets is the biggest<br> A.jupiter<br> B.saturn<br> C.your anus<br> D.neptune

Alina [70]

A. Jupiter. is the correct answer. Mark as brainliest please.

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3 years ago

Read 2 more answers

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

If the Dry bulb reads 25 degrees Celsius and the wet bulb reads 22

Nitella [24]

Answer:

Relative humidity is low .

Explanation:

The wet bulb reads low temperature because due to low humidity of atmosphere , evaporation of water takes place from the wet bulb which makes the bulb cool and therefore it reads lower temperature . In the process of evaporation , heat equal to latent heat of vaporization is taken from the bulb and it loses temperature.

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3 years ago