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RoseWind [281]
3 years ago
15

Extremely lost... the world is cold. Plz help

Chemistry
1 answer:
blagie [28]3 years ago
7 0

Answer:

um to small

Explanation:

to small! TO small1

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17) A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with what volume? HI
galina1969 [7]

Answer: A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

Explanation:

Given: M_{1} = 0.20 M,      V_{1} = 15.0 mL

M_{2} = 0.10 M,            V_{2} = ?

Formula used is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula s follows.

M_{1}V_{1} = M_{2}V_{2}\\0.20 M ]times 15.0 mL = 0.10 M ]times V_{2}\\V_{2} = 30 mL

Thus, we can conclude that a 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

4 0
3 years ago
Coffee is<br><br> homogeneous.<br><br> heterogeneous.
Basile [38]
The answer is homogeneous
3 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
I need help with this and quickyly if anyone is good in Chemistry??
Dominik [7]
The correct answer is d) chrima
6 0
3 years ago
Br2(g) cl2(g)⇌2brcl(g) δh∘f for brcl(g) is 14. 6 kj/mol. Δs∘f for brcl(g) is 240. 0 j/mol
max2010maxim [7]

The Change in Gibb's free energy, ΔG for the reaction at 298K is; -56.92KJ.

<h3>Gibb's free energy of reactions</h3>

It follows from the Gibb's free energy formula as expressed in terms of Enthalpy and Entropy that;

  • ΔG = ΔH - TΔS

On this note, it follows that;

  • ΔG = 14.6 - (298× 0.24)

Hence, the Gibb's free energy for the reaction is;

  • ΔG = 14.6 - 71.52
  • ΔG = -56.92KJ

Remarks: The question requires that we determine the Gibb's free energy for the reaction at 298K.

Read more on Gibb's free energy;

brainly.com/question/13765848

5 0
2 years ago
Read 2 more answers
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