6)

Which wind belt do we live in<br>

vesna_86 [32]

Answer:

the wind belt one

Explanation:

mhm

4 0

3 years ago

Read 2 more answers

The boiling points for the homonuclear diatomic molecules in the halogen family are 85 K, 238 K, 332 K, 457 K and 610 K. Which s

patriot [66]

Answer:

$Cl_{2}$ has boiling point of 238 K

Explanation:

Boiling point depends on different intermolecular force such as molecular wight, dipole-dipole attraction force, hydrogen bonding, ionic attraction force.

Homonuclear diatomic molecules are covalent non-polar molecules and thereby free from dipole-dipole attraction force, hydrogen bonding and ionic interaction forces.

Hence, boiling point of homonuclear diatomic molecules depends solely on molecular weight.

We know, higher the molecular weight of a molecule, higher will be its boiling point. This phenomenon can be realized in terms of increasing london dispersion force with increase in molecular weight.

Decreasing order of molecular weight of halogen molecules :

$I_{2}$ > $Br_{2}$ > $Cl_{2}$ > $F_{2}$

So, decresing order of boiling point of halogen molecules:

$I_{2}$ > $Br_{2}$ > $Cl_{2}$ > $F_{2}$

Hence $Cl_{2}$ has boiling point of 238 K

4 0

3 years ago

How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C

NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0

3 years ago

I have 2 samples of solid chalk (aka calcium carbonate). Sample A has a total mass of 4.12 g and Sample B has a total mass of 19

IRINA_888 [86]

Answer:

A) Sample B has more calcium carbonate molecules

Explanation:

M = Molar mass of calcium carbonate = 100.0869 g/mol

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

For the 4.12 g sample

Moles of a substance is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}$

Number of molecules is given by

$nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}$

For the 19.37 g sample

$n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}$

Number of molecules is given by

$nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}$

$1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}$

So, sample B has more calcium carbonate molecules.

The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.

7 0

3 years ago

When PCl5 solidifies it forms PCl4+ cations and PCl6– anions. According to valence bond theory, what hybrid orbitals are used by

Vlad1618 [11]

Answer:

sp³

Explanation:

Number of hybrid orbitals = ( V + S - C + A ) / 2

Where

H is the number of hybrid orbitals

V is the valence electrons of the central atom = 5

S is the number of single valency atoms = 4

C is the number of cations = 1

A is the number of anions = 0

For PCl₄⁺

Applying the values, we get:

H = ( 5+4-1+0) / 2

= 4

<u>This corresponds to sp³ hybridization.</u>

5 0

3 years ago

Read 2 more answers