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3 years ago

Which numbers are shown in the symbol for a radioactive nuclide?

Chemistry

2 answers:

sammy [17]3 years ago

4 0

Answer:atomic and mass number

Explanation:

Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.

General representation of a radioactive nuclide is given as: $_Z^A\textrm{X}$

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

Thus atomic and mass numbers are shown in the symbol for a radioactive nuclide.

xz_007 [3.2K]3 years ago

3 0

It would be 6 because of the elements

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How many moles of ethanol are produced starting with 500.g glucose?

Monica [59]

<h3>Answer:</h3>

5.55 mol C₂H₅OH

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Tables
Moles

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

<u>Step 3: Stoichiometry</u>

[DA] Set up conversion: $\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})$
[DA} Multiply/Divide [Cancel out units]: $\displaystyle 5.55001 \ mol \ C_2H_5OH$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

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3 years ago

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e-lub [12.9K]

Answer: 7

Explanation:

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3 years ago

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LuckyWell [14K]

To get the value of ΔG we need to get first the value of ΔG°:

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so by substitution:

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then, we can now get the value of ΔG when:

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when ΔG° = -19 KJ

and R is constant in KJ = 0.00831

and T is the temperature in Kelvin = 298 K

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so, by substitution:

ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )

= -40

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3 years ago

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Answer:

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Explanation:

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