Find the density of a cube volume of 150 cm and a mass of 100g

A 0.0000792 M sample of Compound X in a solvent has an absorbance of 0.341 at 528 nm in a 1.000-cm cuvet. The solvent alone has

natima [27]

Answer:

a) a = 3485 M⁻¹cm⁻¹

b) C = 0,000127 M

Explanation:

Lambert-Beer law says that there is a linear relationship between absorbance and concentration of a chemical substance. The formula is:

A = a×b×C

Where A is absorbance, a is molar absorptivity, b is path length and C is concentration.

a) In the problem Concentration is 0.0000792 M, b is 1,000cm and Absorbance is absorbance of sample-absorbance of blank: 0,341-0,065 = 0,276

Replacing:

0,276 = a×1,000cm×0,0000792M

a = 3485 M⁻¹cm⁻¹

b) As the experiment consist in the same compound in the same solvent, the molar absorptivity will be the same, a = 3485 M⁻¹cm⁻¹, path length will be 1,000cm and absorbance: 0,508-0,065 = 0,443

Replacing:

0,443 = 3485 M⁻¹cm⁻¹×1,000cm×C

C = 0,000127 M

I hope it helps!

8 0

2 years ago

Can someone plz help me???? will mark as brainliest!!!!!!!!!!!

Nookie1986 [14]

1=c 2=a 3=b

i hope these answer your questions

7 0

2 years ago

The question is below!

Agata [3.3K]

The anode is the negative electrode and so will be donating electrons to assist in this chemical reaction occuring. All reactions accept electrons as reactants. The key issue is the reduction potential Eo (+1.8V). This is greatest for the reaction:

Co3+ + e -> Co2+

Therefore this reaction has the greatest tendency to occur.

6 0

2 years ago

How many moles of sodium carbonate are contained by 57.3g of sodium carbonate

Lady_Fox [76]

Answer:

$\boxed {\boxed {\sf 0.541 \ mol \ Na_2CO_3}}$

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

Na: 22.9897693 g/mol
C: 12.011 g/mol
O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

Na₂ = 22.9897693 * 2= 45.9795386 g/mol
O₃ = 15.999 * 3= 47.997 g/mol
Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

$\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

We are converting 57.3 grams to moles, so we multiply by this value.

$57.3 \ g \ Na_2CO_3} *\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

Flip the ratio so the units of grams of sodium carbonate cancel.

$57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386 \ g \ Na_2CO_3}$

$57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }$

$\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3$

$0.5406295944 \ mol \ Na_2CO_3$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

$0.541 \ mol \ Na_2CO_3$

There are approximately 0.541 moles of sodium carbonate in 57.3 grams.

6 0

2 years ago

Calculate the hydronium ion concentration in an aqueous solution with a ph of 9.85 at 25Â°c.

UkoKoshka [18]

Answer:

1.41 × 10⁻¹⁰ M

Explanation:

We have a solution with a pH of 9.85 at 25 °C. We can calculate the concentration of H⁺ using the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH

[H⁺] = antilog -9.85

[H⁺] = 1.41 × 10⁻¹⁰ M

H⁺ is usually associated with water molecules forming hydronium ions.

H⁺ + H₂O → H₃O⁺

Then, the concentration of H₃O⁺ ions is 1.41 × 10⁻¹⁰ M.

8 0

2 years ago

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