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Anika [276]
2 years ago
11

An a transition metal with 91 protons and electrons?

Chemistry
2 answers:
ser-zykov [4K]2 years ago
6 0
Protactinium

Hshsbsbwhwhhsnsnanananananababsbsbsbsbsbsbababjaajajajnsnsnsnana
grin007 [14]2 years ago
3 0

Answer:

Protactinium

Explanation:

the atomic number lines up with the number of protons and electrons an element has.

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Separate this redox reaction into its balanced component half-reactions. What is the oxidation and reduction half reactions
erma4kov [3.2K]
<span>Separate this redox reaction into its component half-reactions. 
Cl2 + 2Na ----> 2NaCl 

reduction: Cl2 + 2 e- ----> 2Cl-1 
oxidation: 2Na ----> 2Na+ & 2 e- 


2) Write a balanced overall reaction from these unbalanced half-reactions: 

oxidation: Sn ----> Sn^2+ & 2 e- 
reduction: 2Ag^+ & 2e- ----> 2Ag 

giving us 
2Ag^+ & Sn ----> Sn^2+ & 2Ag </span>Steve O <span>· 5 years ago </span><span>
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3 years ago
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in a typical person, the level of glucose is about 85 mg/ 100 mL of blood. if an average body contains about 11 pt of blood, how
Vanyuwa [196]

Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.

We make a proportion out of the word problem

(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose

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3 years ago
How to describe a ufo??​
erastova [34]

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2 years ago
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The standard reduction potentials of lithium metal and chlorine gas are as follows:Reaction Reduction potential(V)Li+(aq)+e−→Li(
meriva

Answer:

A) E° = 4.40 V

B) ΔG° = -8.49 × 10⁵ J

Explanation:

Let's consider the following redox reaction.

2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)

We can write the corresponding half-reactions.

Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq)      E°red = 1.36 V

Anode (oxidation):  2 Li(s) → 2 Li⁺(aq) + 2 e⁻         E°red = -3.04

<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>

The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.

E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V

<em>B) Calculate the free energy ΔG° of the reaction.</em>

We can calculate Gibbs free energy (ΔG°) using the following expression.

ΔG° = -n.F.E°

where,

n are the moles of electrons transferred

F is Faraday's constant

ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J

8 0
3 years ago
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3 years ago
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