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3 years ago

Which instrument is most often used to measure acid volume before a titration begins?

2 answers:

4 0

Good Morning!

To measure the volume of an acid we use in the laboratory an instrument called burette (a). This instrument is used to calculate the volumes of chemical solutions. At first glance, these tools may be confused with syringes, but have different functions.

Hugs!

Rufina [12.5K]3 years ago

4 0

its actually a volumetric pipette

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Why is it appropriate to say the constitution is designed to "bend like a willow?

olga nikolaevna [1]

The Constitutions was designed to be flexible so new laws can be added as needed and old laws can be reinterpreted

3 0

3 years ago

Aqueous solutions of lithium sulfate and calcium nitrate are mixed, forming the precipitate

Mashcka [7]

Answer:

(aq) Li2SO4 + (aq) Ca(NO3)2 --> (s) CaSO4 + (aq) 2LiNO3

Explanation:

Aqueous Lithium Sulfate is Li2SO4

Aqueous Calcium Nitrate is Ca(NO3)2

When they're mixed, we get the double displacement solution Calcium Sulfate and Lithium Nitrate.

Precipitate is solid.

We got solid Calcium Sulfate so the other solution must be aqueous.

CaSO4 (s) --> Calcium Sulfate product

LiNO3 (aq) --> Lithium Nitrate product

We balance equation now. 2 Lithiums total and 2 Nitrates total. (Cuz there are 2 on the other side, 1 here)

4 0

2 years ago

Consider a generic redox reaction?

vesna_86 [32]

The Nernst equation allows us to predict the cell potential for voltaic cells under conditions other than the standard conditions of 1M, 1 atm, 25°C. The effects of different temperatures and concentrations may be tracked in terms of the Gibbs energy change ΔG. This free energy change depends upon the temperature & concentrations according to ΔG = ΔG° + RTInQ where ΔG° is the free energy change under conditions and Q is the thermodynamic reaction quotient. The free energy change is related to the cell potential Ecell by ΔG= nFEcell

so for non-standard conditions
-nFEcell = -nFE°cell + RT InQ

or

Ecell = E°cell - RT/nF (InQ)

which is called Nernst equation.

5 0

3 years ago

50.0 g of NaNO3 are dissolved into enough water to make 250 mL of solution. What is the molarity of this solution?

d1i1m1o1n [39]

Answer:

2.35 M

Explanation:

Molarity is mol/L of solution. We have to convert the g to mol and the mL to L. G to mol uses the molar mass of the compound. The molar mass of NaNO₃ is 85.00g/mol.

$50.0gNaNO3*\frac{1molNaNO3}{85.00gNaNO3} = 0.588molNaNO3$

Then you have to convert mL to L.

$250mL*\frac{1L}{1000mL} = 0.250L$

Now divide the mol by the L.

$\frac{0.588mol NaNO3}{0.250L} = 2.352 M$

Round to the smallest number of significant figures = 2.35M

7 0

3 years ago

The arsenic in a 1.223 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and

Vladimir79 [104]

Answer:

5.471% As₂O₃ in the sample.

Explanation:

...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).

First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:

Moles KSCN = Moles Ag⁺ in the filtrate:

0.01127L * (0.100mol / L)= 0.001127moles Ag⁺

Total moles Ag⁺:

0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺

Moles of Ag⁺ in the precipitate:

0.0031564 - 0.001127 = 0.0020294 moles Ag⁺

Moles AsO₄ = Moles As:

0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄

Moles As₂O₃:

6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =

3.382x10⁻⁴ moles As₂O₃

Mass As₂O₃:

3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃

Percent is:

0.0669g As₂O₃ / 1.223g sample * 100 =

<h3>5.471% As₂O₃ in the sample</h3>

7 0

3 years ago