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4 years ago

How many molecules are in 3.8 mols of copper?

Chemistry

1 answer:

ioda4 years ago

5 0

Answer:

Hi there!

The correct answer is 2.29 x $10^{24}$ molecules

Explanation:

To convert from the moles to molecules:

1 mole = 6.022 x 10²³ molecules

so basically you multiply 3.8 by 6.022 x 10²³

and you get: 2.29 x $10^{24}$

You might be interested in

Calculate the ph after 0.020 mol hcl is added to 1.00 l of each of the four solutions below.

melisa1 [442]

A) according to this reaction:
by using ICE table:
NH2OH(aq) + H2O(l) → HONH3+(aq) + OH-
initial 0.4 M 0 0
change -X +X +X
Equ (0.4-X) X X

when Kb = [OH-][HONH3+]/[NH2OH]
when we have Kb = 1.1x10^-8 so,
by substitution:
1.1x10^-8 = X^2/(0.4-X) by solving this equation for X

∴X = 6.6x10^-5 M
∴[OH] = 6.6x10^-5 M
when POH = - ㏒[OH]
∴POH = -㏒(6.6x10^-5)= 4.18
∴PH = 14 - POH = 14 - 4.18
= 9.82
when PH = -㏒[H+]
∴[H+] = 10^9.82 = 1.5x10^-10 M+0.02molHcl
= 0.02
∴ the new value of PH = -㏒(0.02)
∴PH = 1.7

B) according to this reaction:
by using ICE table:
HONH3+(aq) → H+(aq) + HONH2(aq)
intial 0.4 0 0
change -X +X +X
Equ (0.4-X) X X

when Ka HONH3Cl = 9.09x10^-7
and Ka = [H+][HONH2] / [HONH3+]

So by substitution and we can assume [HONH3+] = 0.4 as the value of Ka is so small so,
9.09x10^-7 = X^2 / 0.4 by solving for X
∴ X = 6 x 10 ^-4
∴[H+] = 6x10^-4
PH = -㏒[H+]
= -㏒ (6x10^-4) = 3.22
when [H+] = 6x10^-4 + 0.02 m HCl
∴new value of PH = -㏒(6x10^-4+0.02)
= 1.69
C) when we have pure H2O and PH of water = 7
So we can get [H+] when PH = -㏒[H+]
∴[H+] = 10^-7 + 0.02MHCl
= 0.02
∴new value of PH = -㏒0.02
PH = 1.7
d) when HONH2 & HONH3Cl have the same concentration and Hcl added to them so we can assume that PH=Pka
and when we have Ka for HONH3Cl = 9.09x10^-7
So we can get the Pka:

Pka = -㏒Ka
= -㏒9.09x10^-7
= 6.04
∴PH = 6.04
and because of the concentration of the buffer components, HONH2 & HONH3Cl have 0.4 M and the adding of HCl = 0.02 M So PH will remain very near to 6

4 0

3 years ago

You are working in a laboratory, and you are given the task of converting cyclopentene into 1, 5-pentanediol. Your first thought

Helen [10]

Answer:

Reagent A = $H_3O^+$

Reagent B= $Na_2Cr_2O_7~H_2SO_4~H_2O$

Intermediate C= δ-Valerolactone

Explanation:

In the reaction from the alkene to the alcohol, we can use the alkene hydration in which the hydronium ion is added to the double bond followed by the attack of water to produce the alcohol.

Then in the conversion from alcohol to ketone can be produced if an oxidant reactive is used. In this case the Jones reagent ( $Na_2Cr_2O_7~H_2SO_4~H_2O$ ).

The intermediate is a structure produced by a peroxyacid. This reaction would introduce an ester group in the cycle generating the δ-Valerolactone (Figure 1).

7 0

3 years ago

Rank these transition metal ions in order of decreasing number of unpaired electrons.

lesya692 [45]

Answer: The given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}$

Explanation:

In atomic orbitals, the distribution of electrons of an atom is called electronic configuration.

The electronic configuration in terms of noble gases for the given elements are as follows.

Atomic number of Fe is 26.

$Fe^{3+} - [Ar] 3d^{5}$

So, there is only 1 unpaired electron present in $Fe^{3+}$ .

Atomic number of Mn is 25.

$Mn^{4+} - [Ar]3d^{3}$

So, there are only 3 unpaired electrons present in $Mn^{4+}$ .

Atomic number of V is 23.

$V^{3+} - [Ar] 3d^{2}$

So, there are only 2 unpaired electrons present in $V^{3+}$ .

Atomic number of Ni is 28.

$Ni^{2+} - [Ar] 3d^{8}$

So, there will be 2 unpaired electrons present in $Ni^{2+}$ .

Atomic number of Cu is 29.

$Cu^{+} - [Ar] 3d^{10}$

So, there is no unpaired electron present in $Cu^{+}$ .

Therefore, given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}$

Thus, we can conclude that given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}$

7 0

3 years ago

What is the density of lead (in g/cm^3 3 ) if a rectangular bar measuring 0.500 cm in height, 1.55 cm in width, and 25.00 cm in

Viktor [21]

Answer:

Density = 11.4 g/cm³

Explanation:

Given data:

Density of lead = ?

Height of lead bar = 0.500 cm

Width of lead bar = 1.55 cm

Length of lead bar = 25.00 cm

Mass of lead bar = 220.9 g

Solution:

Density = mass/ volume

Volume of bar = length × width × height

Volume of bar = 25.00 cm × 1.55 cm × 0.500 cm

Volume of bar = 19.4 cm³

Density of bar:

Density = 220.9 g/ 19.4 cm³

Density = 11.4 g/cm³

3 0

3 years ago

What is a negatively charged ion

dem82 [27]

an electrically charged atom or group of atoms formed by the loss or gain of one or more electrons, as a cation (positive ion) , which is created by electron loss and is attracted to the cathode in electrolysis, or as an anion (negative ion) , which is created by an electron gain and is attracted to the anode.

7 0

3 years ago