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forsale [732]
2 years ago
10

How many grams are contained in a 0.893 mol sample of methane, ch4?

Chemistry
1 answer:
irina [24]2 years ago
6 0

We are given with a compound, Methane (CH4), with a molar mass of 0.893 mol sample. We are tasked to solve for it's corresponding mass in g. We need to solve first the molecular weight of Methane, that is

C=12 g/mol

H=1g/mol

 

CH4= 12 g/mol +1(4) g/mol = 16 g/mol

With 0.893 mol sample, its corresponding mass is

g CH4= 0.893 mol x 16g/mol =14.288 g

Therefore, the mass of methane is 14.288 g

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Cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture. The initial rea
Dima020 [189]

Answer : The correct option is, (C) ^{58}\textrm{Fe}

Explanation :

Neutron capture : In this decay process, an atomic nucleus and one or more number of neutrons collide and combine to form a heavier nucleus. The mass number changes in this process.

The neutron capture equation is represented as,

_Z^A\textrm{X}+_{0}^1\textrm{n}\rightarrow _{Z}^{A+1}\textrm{X}+\gamma

(A is the atomic mass number and Z is the atomic number)

Beta emission or beta minus decay : It is a type of decay process, in which a neutrons gets converted to proton, an electron and anti-neutrino. In this the atomic mass number remains same.

The beta minus decay equation is represented as,

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0e

(A is the atomic mass number and Z is the atomic number)

As per question, the cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture.

Process 1 : Neutron capture.

_{26}^{58}\textrm{Fe}+_{0}^1\textrm{n}\rightarrow _{26}^{59}\textrm{Fe}+\gamma

Process 2 : Beta emission.

_{26}^{59}\textrm{Fe}\rightarrow _{27}^{59}\textrm{Co}+_{-1}^0e

Process 3 : Neutron capture.

_{27}^{59}\textrm{Co}+_{0}^1\textrm{n}\rightarrow _{26}^{60}\textrm{Co}+\gamma

From this we conclude that, the initial reactant in the production of cobalt-60 is _{26}^{58}\textrm{Fe}

Hence, the correct option is, (C) ^{58}\textrm{Fe}

7 0
2 years ago
93.2 mL of a 2.03 M potassium fluoride (KF) solution
Marrrta [24]

Answer:

1.98 M

Explanation:

Given data

  • Initial volume (V₁): 93.2 mL
  • Initial concentration (C₁): 2.03 M
  • Volume of water added: 3.92 L

Step 1: Convert V₁ to liters

We will use the relationship 1 L = 1000 mL.

93.2mL \times \frac{1L}{1000mL} = 0.0932 L

Step 2: Calculate the final volume (V₂)

The final volume is the sum of the initial volume and the volume of water.

V_2 = 0.0932L + 3.92 L = 4.01L

Step 3: Calculate the final concentration (C₂)

We will use the dilution rule.

C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.03 M \times 3.92L}{4.01L} = 1.98 M

3 0
2 years ago
Read 2 more answers
2.1. The lithium ion in a 250,00 mL sample of mineral water was
juin [17]

Answer:

11482 ppt of Li

Explanation:

The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:

<em>Moles Mg²⁺:</em>

0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA

<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>

0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium

That means mass of lithium is (Molar mass Li=6.941g/mol):

4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:

0.00287g * (1000000μg / g) = 2870μg of Li

As ppt is μg of solute / Liter of solution, ppt of the solution is:

2870μg of Li / 0.250L =

<h3>11482 ppt of Li</h3>

4 0
2 years ago
What is the total number of sp2-hybridized carbon atoms present in the fluorophore used in the experiments
Mandarinka [93]

Answer:

9

Explanation:

The structure of fluorophore used in the experiments has been drawn in the attachment. And from the drawing counting we can say that there are 9 sp2-hybridized carbon atoms present. Fiuorophores are a fluorescent chemical compound that can re-emit light upon light excitation. Normally used to produce absorbance and emission spectra.

3 0
2 years ago
Compound X has the same molecular formula as butane but has a different boiling point and melting point. What can be concluded a
Vsevolod [243]
Compound X is an isomer of butane with different chain types. It is either straight chain or bend chain.
3 0
2 years ago
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