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serious [3.7K]
3 years ago
9

I'm doing a adopt a element project and I need a slogan for bismuth please tell me a good one

Chemistry
1 answer:
trapecia [35]3 years ago
6 0

Bismuth, it’s elemental!

Bismuth, Agent 83

Bismuth, what a catalyst?

I like you more than Bismuth

Bismuth, we love you

Get things done with Bismuth

It’s a Bismuth thing, you wouldn’t understand

Bismuth, it’s a brittle thing

Got Bismuth?

Bi Bro

Bismuth born Brittle

Silvery by default like Bismuth

Bi the one with Bismuth


http://www.bestslogans.com/list-ideas-taglines/periodic-table-bismuth-slogans/


https://sloganshub.org/bismuth-slogans/

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Difference between vegetative and reproductive part​
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Answer:

The vegetative parts includes the roots, stems,leaves and the reproductive parts include flowers. ... The vegetative parts can reproduce new plants by the process of vegetative propagation but the reproductive parts cannot produce new plants directly instead they produce precursor of new plants that are seeds or fruits.

8 0
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4 years ago
Noble gases (except for He) have _________ electron(s) in their valence shell. A. 1 B. 2 C. 4 D. 8
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3 0
3 years ago
Compute the equilibrium constant,K, at 25 ∘C for the reaction between Ni2+(aq) and Zn(s), which form Ni(s) and Zn2+(aq)
ANEK [815]

Answer:

The answer to the question is;

The equilibrium constant,K, at 25 °C for the reaction between Ni²⁺(aq) and Zn(s), which form Ni(s) and Zn²⁺ is 2.04×10¹⁷.

Explanation:

The half reactions are as follows

Ni²⁺ (aq) + 2e⁻ -> Ni (s)

Zn (s) -> Zn²⁺ (aq) + 2e⁻

For the Ni²⁺/Ni system we have the potential given as -0.23V (Reduction)

For the Zn²⁺/Zn sytem, the potential is -0.76. Here however, we should note that the zinc is being oxidized and therefore the potential is positive, that is;

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Therefore the voltage for the sum of the reactions on both sides of the process is

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We then call upon the Nernst equation to calculate the equilibrium constant as follows

E⁰_{cell} = \frac{0.0592}{n} logK

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Therefore we have

0.53 V = \frac{0.0592}{2} logK

Therefore log K = 17.905

and K =   10^{17.9054} = 2.04×10¹⁷.

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<span>A. the sum of the protons and neutrons in one atom of the element</span>
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