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Lelechka [254]
1 year ago
12

You prepare a standard by weighing 10.751 mg of compound X into a 100 mL volumetric flask and making to volume. You further dilu

te this solution 5 mL to 25 mL. This standard gives an area of 4,374. Your sample is prepared by adding 5 mL of sample solution into a 50 mL flask and making to volume. This gives an area count of 2,582. Calculate the concentration of compound X in the sample - prior to dilution.
Chemistry
1 answer:
ArbitrLikvidat [17]1 year ago
4 0

Answer:

0.12693 mg/L

Explanation:

First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:

  • 10.751 mg / 100 mL = 0.10751 mg/mL

Then we <u>calculate the concentration of compound X in the standard after dilution</u>:

  • 0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L

Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:

  • 2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L

Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:

  • 0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L
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Balance the following oxidation-reduction reaction: Fe(s)+Na+(aq)→Fe2+(aq)+Na(s) Express the coefficients as integers separated
aliina [53]

Answer:

The answer to your question is: 1, 2, 1, 2

Explanation:

                       1 Fe(s)  + 2 Na⁺(aq)  → 1 Fe²⁺(aq)  + 2 Na(s)

                             Fe⁰   -   2e⁻       ⇒           Fe⁺²        Oxidases

                             Na⁺   +  1 e⁻       ⇒           Na⁰         Reduces

                      1 x ( 1 Fe⁰      ⇒         1 Fe⁺²)      Interchange number of

                      2 x ( 2Na⁺       ⇒       2 Na⁰ )      electrons

6 0
2 years ago
Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate ∙ x-hydrate (CuSO4 ∙ xH2O), where x is an integer. Part of t
Gnom [1K]

Answer:

  • <em>The unknown integer X in the formula is </em><u>5</u><em>.</em>

Explanation:

<u>1) Data:</u>

a) Mass of CuSO₄ ∙ XH₂O = 1.50 g

b) Mass of CuSO₄ = 0.96

c) X = ?

<u>2) Additional needed data:</u>

a) Molar Mass of CuSO₄ = 159,609 g/mol

b) Molar mass of H₂O = 18,01528 g/mol

<u>3) Chemical principles and formulae used:</u>

a) Law of conservation of mass

b) Molar mass = mass in grams / number of moles = m / n

<u>4) Solution:</u>

a) Law of conservation of mass:

  • Mass of CuSO₄ ∙ XH₂O = mass of CuSO₄ + mass of H₂O

  • 1.50g = 0.96g + mass of H₂O ⇒ mass of H₂O = 1.50g - 0.96g = 0.54g

b) Moles

  • n = m / molar mass

  • CuSO₄: n = 0.96g / 159.609 g/mol = 0.0060 mol

  • H₂O: 0.54g / 18.01528 g/mol = 0.030 mol

c) Proportion:

Divide both mole amounts by the least of the two numbers, i.e. 0.0060

  • CuSO₄:  0.0060 / 0.0060 = 1

  • H₂O: 0.030 mol / 0.0060 = 5

Then, the ratio of CuSO₄ to H₂O is 1 : 5 and the chemical formula is:

  • <u>CuSO₄ . 5H₂O.</u>

Hence, the value of X is 5.

5 0
2 years ago
Read 2 more answers
What volume (mL) of a 0.3428 M HCl(aq) solution is required to completely neutralize 23.55 mL of a 0.2350 M Ba(OH)2(aq) solution
EastWind [94]

Answer:

The volume required for complete neutralize is 32.29 mL

Explanation:

The computation of the volume required for complete neutralize is shown below:

As we know that, the balanced equation is

Ba(OH)_2 + 2Hcl\rightarrow Bacl_2 + 2H_2O

Now

The number of moles of Ba(OH)_2 = n_1 = 1

And, the number of moles of Hcl = n_2 = 2

Therefore

The equation i.e. to be used to find out the volume is given below:

\frac{M_1V_1}{n_1} =  \frac{M_2V_2}{n_2}

V_2 = \frac{M_1V_1}{n_1} \times \frac{n_2}{M_2} \\\\ = \frac{0.2350 \times 23.55}{1} \times \frac{2}{0.3428} \\\\ = \frac{11.0685}{0.3428}

= 32.29 mL

Hence, the volume is 32.29mL

6 0
2 years ago
What is the origin of first (mass of 157.836 amu) peak (of what isotopes does each consist)? express your answers as isotopes se
Klio2033 [76]

Answer and explanation;

-Bromine molecule (Br2) consists of two bromine atoms (Br-Br). These two atoms may be originated from the same type of isotopes Br2(11) and Br2(22) or from two types of isotopes, Br2(12).

The intensity of the peak depends on the abundance of the isotope. The larger the intensity of the peak, the greater the abundance of the isotope. For Br, the relative size of the peak for Br 2 molecule consisting of two different isotopes will be larger than the Br molecule consisting of same isotopes, i.e relative size of the peak for Br molecules consisting of different isotopes is twice as that of Br molecule consisting of same isotopes.

-Hence, from the data in the table we could say that the peak of mass 157.836 represents 79Br - Br peak, 159.834 represents Br - Br peak and peak of mass 161.832 represents Br - 81 81 Br

-The first peak will represent the lighter Br2 molecule, the third peak will represent the heavier Br2 molecules and the middle peak will represent the intermediate Br2 molecule which is Br2(12) .


3 0
2 years ago
Select all the correct answers.
nevsk [136]

Answer:

The energy gained by the peas is lost by the water;

Energy is transferred from the fire to the pot, then to the water, and then to the peas

Explanation:

According to the fundamental laws of thermodynamics, heat flows from hotter objects to colder ones.

Let's analyze the heat flow in each of the systems we have:

  • A pot of water is heated over a fire. We have a system of water/fire. Since water is heated, it gains heat. This means fire loses heat to water.
  • Frozen peas are added to the hot water. We have a system of peas/water. In this scenario, water is hot. Thus, heat flows from water to peas (heat is lost by water and gained by peas).

Analyzing the answer choices, firstly we notice that the energy gained by the peas is lost by the water, as water is hotter than the peas. Hence, this is true.

Secondly, the peas gain energy, as energy is a synonym to heat. This means the fact that peas lose energy is false.

Thirdly, it's false to claim that energy is transferred from the peas to the water and the pot, as we actually have a reverse process: both pot and water are at a higher temperature and heat flows from them towards the peas at a lower temperature.

Fourthly, it's also false to claim that the water receives energy both from the ire and from the frozen peas: the water only gains energy from the fire that is at a higher temperature, but it loses energy to the frozen peas, as the water is hotter than the peas.

Finally, it's true that energy is transferred from the fire to the pot, then to the water, and then to the peas. This is simply understood knowing that the fire here is at the highest temperature and it directly interacts with the pot which transfers energy to the water. The water is now at a higher temperature relatively to the peas, so it transfers energy to the peas afterwards.

3 0
2 years ago
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