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Bas_tet [7]
3 years ago
8

if you add sugar to lemonade but notice that the sugar is sitting un dissolved at the bottom is the lemonade saturated, pure sat

urated or unsaturated
Chemistry
1 answer:
devlian [24]3 years ago
8 0

Answer:

Saturated

Explanation:

If the sugar sits undissolved at the bottom even after vigorous stirring, the lemonade has dissolved all the sugar it can hold. The lemonade is saturated.

If the lemonade were <em>unsaturated</em>, it could hold more sugar and that at the bottom would continue to dissolve.

The lemonade cannot be <em>supersaturated </em>because, if it were, the solid at the bottom would serve as nuclei on which the excess sugar in the solution could form more crystals.

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Lubov Fominskaja [6]

Answer: Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. It is the foundation of all quantum physics including quantum chemistry, quantum field theory, quantum technology, and quantum information science.

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7 0
3 years ago
Read 2 more answers
The spectrochemical series is I &lt; Br&lt; &lt; Cl^- &lt; F
vfiekz [6]

Answer:

b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

Explanation:

The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.

Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.

Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.

Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

6 0
3 years ago
Reacting 35.4 ml of 0.220 m agno3 with 52.0 ml of 0.420 m k2cro4 results in what mass of solid formed
laila [671]
Answer is: 1.29 grams <span>of solid formed.
</span>Chemical reaction: 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq).
n(AgNO₃) = c(AgNO₃) · V(AgNO₃).
n(AgNO₃) = 0.220 M · 0.0351 L.
n(AgNO₃) = 0.0078 mol; limiting reactant.
n(K₂CrO₄) = 0.420 M · 0.052 L.
n(K₂CrO₄) = 0.022 mol.
From chemical reaction: n(AgNO₃) : n(Ag₂CrO₄) = 2 : 1.
n(Ag₂CrO₄) = 0.0078 mol ÷ 2.
n(Ag₂CrO₄) = 0.0039 mol.
m(Ag₂CrO₄) = 0.0039 mol · 331.73 g/mol.
m(Ag₂CrO₄) = 1.29 g.

7 0
3 years ago
The half-life of sr-90 is 28 years. after 56 years of decay only 0. 40 g of a sample remains. what was the mass of the original
Svetradugi [14.3K]

Half-life is the time taken for the concentration of the substance to reduce by 50%. The original sample of strontium had a mass of 1.6 gms. Thus, option d is correct.

<h3>What is half-life?</h3>

The half-life of any radioactive substance is the time period at which the concentration will get reduced to half the initial amount. The initial mass of Sr-90 is calculated as,

N(t) = N_{0} (\dfrac{1}{2})^{ \frac{t }{t 1/2}}

Given,

Quantity of the remaining substance N (t) = 0.40 gm

Initial radioactive substance quantity N_{0} =?

Time duration (t) = 56 years

Half-life = 28 years

Substituting values above:

\begin{aligned} 0.40 &= N_{0} (\dfrac{1}{2}) ^{{\frac{56}{28}}\\\\0.40 &= N_{0} (\dfrac{1}{2})^{2}\end{aligned}

= 1.6 gm

Therefore, option d. the initial mass of Sr is 1.6 gm.

Learn more about half-life here:

brainly.com/question/16145921

#SPJ4

7 0
2 years ago
The
goldfiish [28.3K]

Answer: …

Explanation:

5 0
2 years ago
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