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avanturin [10]
3 years ago
7

To determine the concentration of X in an unknown solution, 1.00 mL of 8.48 mM S was added to 3.00 mL of the unknown X solution

and the mixture was diluted to 10.0 mL. After chromatographic separation, this solution gave peak areas of 5473 and 4851 for X and S, respectively. Determine the concentration of S in the 10.0 mL solution.
Chemistry
1 answer:
kogti [31]3 years ago
6 0

Answer:

positif

Explanation:

3.87169.843826 = y = x = .ion \: in \: cells = y = x. >  \\  \geqslant  {8}^{2}  \times \frac{4}{3}  | \geqslant |  \times \frac{68.1 < }{3 = 8}

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Which equation is used to help form the combined gas law?<br> eP, V, P, V, т.
alexdok [17]

Answer:

The combined gas law is formulated from PV/T =K.

Explanation:

The combined gas law comprises of Boyle's law, Charles's law and Gay lusaac's law. This laws were not discovered but simply put together considering other cases of ideal gas law. It states that if the amount of gas is left unchanged, the ratio between the pressure, volume, and temperature is constant.

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3 years ago
What would be the IUPAC name of PoTs2
Vikentia [17]

Answer:

6-methylhept-1-en-3-one

Explanation:

6 0
3 years ago
What is the IUPAC name for this product?
mihalych1998 [28]

The IUPAC name for the given product is 2 chloro Butane.

<h3>What is IUPAC nomenclature?</h3>

IUPAC stands for 'International Union of Pure and Applied Chemistry', which givers some rule for designing the name of compounds of chemistry.

  • In the given product total four carbon atoms are present and between all of them single bonds are present.
  • In the second carbon atom, chlorine group is present.
  • During the nomenclature process, first we write down the name of the attached group which is followed by the alkane chain.

Hence name of the product is 2 chloro Butane.

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7 0
2 years ago
In a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes. Determine the average relative atomic
Burka [1]

The average relative atomic mass of a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes is 9.6g.

<h3>What is Atomic mass? </h3>

Atomic mass is defined as the whole mass of an atom.

It is also defined as the sum of atomic number and number of neutrons.

Atomic mass = Atomic number + neutrons

<h3>What is Isotopes?</h3>

Isotopes are the those element which have same atomic number but have different mass number and number of neutrons.

The average relative atomic mass can be calculated as

mass of isotopes/ mass of sample

mass of all isotopes = (70 × 9X) + (30 × 11X)

=(630 + 330) X

= 960X

Average relative atomic mass = 960X/ 100 X

= 9.6 g

Thus, we concluded that the average relative atomic mass of a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes is 9.6g.

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5 0
1 year ago
A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.50 grams of aluminum foil in a solut
Irina-Kira [14]

Answer:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

Explanation:

  • It is a stichiometry problem.
  • We should write the balance equation of the mentioned chemical reaction:

<em>2Al + 3CuCl₂ → 3Cu + 2AlCl₃.</em>

  • It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.
  • Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:

<em>n = mass / molar mass</em>

  • The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.
  • The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³  mol.
  • <em>From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.</em>

∴ 3.85 x 10⁻³  mol of Al foil reacts completely with 5.578 x 10⁻³  mol of CuCl₂ with <em>(2:3)</em> ratio and CuCl₂ is the limiting reactant while Al foil is in excess.

  • From the stichiometry 3.0 moles of  CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
  • So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
  • Finally, we can calculate the mass of copper produced using:

mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³  mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.

  • <u><em>So, the answer is:</em></u>

<em>Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.</em>

5 0
3 years ago
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