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goblinko [34]
3 years ago
5

What does creativity have to do with Science?​

Chemistry
1 answer:
Lady_Fox [76]3 years ago
5 0

creativity is the very element in science that helps us understand things. if we didn't have creative minds, we would never conduct experiments that help us discover things.

hope this helps, good luck!:)

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I need help with question 2 I will brainliest you if I’m able to
erik [133]

Answer:

Blank 1: Microscope

Blank 2: Cells

4 0
3 years ago
Read 2 more answers
A buffer solution is composed of 7.50 mol of acid and 4.75 mol of the conjugate base. If the pKa of the acid is 4.90 , what is t
Mamont248 [21]

Answer:

The correct answer is: pH= 4.70

Explanation:

We use the <em>Henderson-Hasselbach equation</em> in order to calculate the pH of a buffer solution:

pH= pKa + log   \frac{ [conjugate base]}{[acid]}

Given:

pKa= 4.90

[conjugate base]= 4.75 mol

[acid]= 7.50 mol

We calculate pH as follows:

pH = 4.90 + log (4.75 mol/7.50 mol) = 4.90 + (-0.20) = 4.70

7 0
3 years ago
For a half-reaction in an acidic solution, which substances should you add to balance the equation?
solmaris [256]
Hello!

I believe the correct answer to this question is H+ and H2O.

I hope you found this helpful! :)
8 0
4 years ago
A chemist fills a reaction vessel with mercurous chloride solid, mercury (I) aqueous solution, and chloride aqueous solution at
makvit [3.9K]

Answer:

ΔG° = -533.64 kJ

Explanation:

Let's consider the following reaction.

Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)

The standard Gibbs free energy (ΔG°) can be calculated using the following expression:

ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)

where,

ni are the moles of reactants and products

ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products

ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)

ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)

ΔG° = -533.64 kJ

3 0
3 years ago
How many grams are there in 2.3*10^24 atoms of silver nitrate ?​
Paul [167]
410g Ag

2.3*10^24 atoms

1 molcule Ag- 6.02g*10^3
6 0
3 years ago
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