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4 years ago

How would you classify a solution if [OH-] = 4.2 x 10-8 M?

Chemistry

2 answers:

MA_775_DIABLO [31]4 years ago

7 0

wrong answer I just took the test

Oksana_A [137]4 years ago

3 0

Answer: Slightly acid, almost neutral.

pOH = -log [OH -] = - log [4.2 x 10^-8] = 7.38

pH + pOH = 14 ==> pH = 14 - pOH = 14 - 7.38 = 6.62.

Remember: 7 is neutral, below 7 is acid. This solution is almost neutral.

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To draw a Lewis structure, first add the number of outer (valence) electrons contributed by each atom to obtain the total number

Alex777 [14]

Answer:

12 is the total number of outer electrons for a molecule of $H_2C=CH_2$

Explanation:

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

Valence electron of carbon = 4

Valence electron of hydrogen = 1

Number of carbon atoms in $H_2C=CH_2$ = 2

Number of hydrogen atoms in $H_2C=CH_2$ = 4

Total number of valence electrons :

$2\times 4+4\times 1 =12$

4 0

4 years ago

What happens if an atom loses a neutron?Its mass is lowered, but it is still the same element. Its mass stays the same, and it i

Whitepunk [10]

Answer:

If an atom were to gain or lose neutrons it becomes an isotope. ... If it gains a neutron it become an isotope called deuterium. Since the atomic mass is the total of the number of protons and neutrons, an isotope would have a different atomic mass, but the same atomic number as the original atom

Explanation:

7 0

3 years ago

Read 2 more answers

The ΔHvap of a certain compound is 49.09 kJ·mol–1 and its ΔSvap is 53.69 J·mol–1·K–1. What is the boiling point of this compound

rusak2 [61]

Hey there !

<span>Convert Joule to KJ :
</span>
1 j ---------------- 0.001 kj
53.69 j ----------- Kj

Kj = 53.69 * 0.001

=> 0.05369 Kj

T = ΔH / <span>ΔS

T = 49.09 / 0.05369

T = 914.32ºC</span>

3 0

3 years ago

Need help with this

choli [55]

Answer:

H20is water andN20is nitrogen 4 oxide

6 0

3 years ago

Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4

inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

$M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol$

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

So you can solve for the moles of the solute:

$n_{solute}=M*V_{solution}$

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

$n_{solute}=1mol/L*1.5L=1.5mol$

But this is just a supposition.

Regards.

4 0

3 years ago