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Schach [20]
3 years ago
15

Who knows how to do this please help

Chemistry
1 answer:
Fofino [41]3 years ago
5 0

Answer: You forgot to zero the balance

Explanation:

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Explain ionization suppressants as applied to atomic spectrometry
levacccp [35]

An ionization suppressor is an alkali metal capable of preventing ionization, which can be used in atomic spectroscopy to determine matter composition.

<h3>What is ionization?</h3>

Ionization refers to the phenomena capable of converting neutral atoms/molecules to electrically charged atoms/ions.

Ionization is a process by which radiation (e.g., alpha, beta, gamma rays) can pass energy to inert matter.

Some examples of ionization suppressors include salts of alkali metals (for example, potassium), which can be used in atomic spectroscopy to determine matter composition.

Learn more about ionization here:

brainly.com/question/1445179

8 0
2 years ago
A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field
maksim [4K]

Answer:

the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  })

Substituting the values into the equation, it becomes

E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  }) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }) = 7.12 × 10⁵(1 - \frac{0.12}{0.1216}) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

7 0
3 years ago
Select all the correct statements, there is more than one.
alexgriva [62]

Answer:

Experiments to determine mechanisms involve looking at indirect evidence to help support or disprove a proposed mechanism.

Most intermediates are not typically isolated to determine reaction mechanisms.

Carbocations are very reactive and are typically not isolated for analysis.

Scientists can prove that a specific mechanism exists.

Evidence of intermediates sometimes can be seen using techniques such as nuclear magnetic resonance spectroscopy

Explanation:

The study of reaction mechanism and chemical kinetics often form the main thrust of study in organic, inorganic and physical chemistry.

We often want to know the actual processes involved in the conversion of one specie to another. Unfortunately, this information may have to be obtained indirectly by certain chemical reactions or by the use of new instrumental methods such as nuclear magnetic resonance spectroscopy.

Many organic reactions have carbocation intermediates. These carbocations are relatively short-lived and are transient intermediates which are rarely isolated unless they are isolated in a molecular cage using a macromolecule or in superacids.

By intensive study, scientists can proof or disprove the authenticity of any proposed mechanism.

We must know that a transition state has partial bonds. It is often an extremely short-lived specie which cannot be isolated.

4 0
3 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
The simplest type of stream cipher, one in which one letter or character is exchanged for another, is known as what?
QveST [7]

Answer: This is known as substitution cipher.

Explanation:

5 0
3 years ago
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