Classify each of the following as a compound, a homogeneous mixture, or a heterogeneous mixture: (a) orange juice; (b) vegetable soup; (c) cement; (d) calcium sulfate; (e) teas

Answers

Answer 1

Answer:

Explanation:

A compound is a substance that contains more than two different atoms in a fixed ratio by mass. Elements of a compound are chemically combined together.

A homogeneous mixture is defined as a mixture in which solute particles are uniformly distributed into the solvent.

Whereas a heterogeneous mixture is defined as a mixture in which solute particles are non-uniformly distributed in a solvent.

In both homogeneous and heterogeneous mixture there is no chemical combination between solute and solvent particles.

Therefore, each given substance is classified as follows.

(a) Orange juice : As it contains pulp of orange along with the liquid which are not present in uniform composition. Hence, orange juice is a heterogeneous mixture.

(b) vegetable soup : This soup also contains vegetable pieces and spices which are non-uniformly distributed. Hence, it is also a heterogeneous mixture.

(c) cement : It contains several different compound present in definite composition. Hence, cement is a homogeneous mixture.

(d) calcium sulfate : Its chemical formula is $CaSO_(4)$ and in this compound both calcium and sulfate atoms are chemically combined together. Therefore, calcium sulfate is a compound.

(e) Tea : It is a homogeneous mixture as there is uniform composition of tea water in milk.

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What two processes are occurring in this picture A.erosion and deposition
B.weathering and deposition
C.weathering and erosion

Answers

Answer:

C. weathering and erosion

Explanation:

What is the daughter nucleus produced when 63 Zn undergoes electron capture? Replace each question mark with the appropriate integer or symbol.

Answers

Answer: The daughter nuclei is $_(29)^(63)\textrm{Cu}$

Explanation:

Electron capture is defined as the process in which an electron is drawn to the nucleus where it combines with a proton to form a neutron and a neutrino particle.

$_Z^A\textrm{X}+e^-\rightarrow _(Z-1)^A\textrm{Y}+\gamma e$

The chemical equation for the reaction of electron capture of Zinc-63 nucleus follows:

$_(30)^(63)\textrm{Zn}+e^-\rightarrow _(29)^(63)\textrm{Cu}+\gamma e$

The parent nuclei in the above reaction is Zinc-63 and the daughter nuclei produced in the above reaction is copper-63 nucleus.

Hence, the daughter nuclei is $_(29)^(63)\textrm{Cu}$

Final answer:

When Zinc-63 undergoes electron capture, it results in the creation of a Copper-63 daughter nucleus. This is due to the atomic number decreasing by one (from 30 to 29) during electron capture, but the mass number remaining unchanged.

Explanation:

Electron capture is a process where a proton-rich nucleus absorbs an inner shell electron, which results in a conversion of a proton into a neutron, and the emission of an electron neutrino. In doing so, the atomic number decreases by one, while the mass number stays the same. Therefore, in the case of 63 Zn (zinc-63), the atomic number is 30 prior to electron capture. After electron capture, the atomic number will decrease by one to become 29, leading to the production of 63 Cu (copper-63).

Remember that the atomic number (bottom number), also known as the proton number, determines the element. Therefore, in our example, Zn changes to Cu. The fact that the mass number (top number) remains the same is due to the total number of protons and neutrons (nucleons) being conserved.

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The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when [NOBr]0 5 0.072 M and [NOBr]0 5 0.054 M.

Answers

Explanation:

2NOBr(g) --> 2NO(g) 1 Br2(g)

Rate constant, k = 0.80

a) Initial concentration, Ao = 0.086 M

Final Concentration, A = ?

time = 22s

These parameters are connected with the equation given below;

1 / [A] = kt + 1 / [A]o

1 / [A] = 1 / 0.086 + (0.8 * 22)

1 / [A] = 11.628 + 17.6

1 / [A] = 29.228

[A] = 0.0342M

b) t1/2 = 1 / ([A]o * k)

when [NOBr]0 5 0.072 M

t1/2 = 1 / (0.072 * 0.80)

t1/2 = 1 / 0.0576 = 17.36 s

when [NOBr]0 5 0.054 M

t1/2 = 1 / (0.054 * 0.80)

t1/2 = 1 / 0.0432 = 23.15 s

Answer:

(a)

$0.0342M$

(b)

$t_(1/2)=17.36s\nt_(1/2)=23.15s$

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

$(1)/([NOBr])=kt +(1)/([NOBr]_0)\n(1)/([NOBr])=(0.8)/(M*s)*22s+(1)/(0.086M)=(29.3)/(M)\n$

$[NOBr]=(1)/(29.2/M)=0.0342M$

(b) Now, for a second-order reaction, the half-life is computed as shown below:

$t_(1/2)=(1)/(k[NOBr]_0)$

Therefore, for the given initial concentrations one obtains:

$t_(1/2)=(1)/((0.80)/(M*s)*0.072M)=17.36s\nt_(1/2)=(1)/((0.80)/(M*s)*0.054M)=23.15s$

Best regards.

How many sodium ions are in 1.4 kg of sodium chloride, NaCl?

Answers

Answer:

1.44 x 10²⁵ ions of Na⁺

Explanation:

Given parameters:

Mass of NaCl = 1.4kg = 1400g

Unknown:

Number of ions of sodium = ?

Solution:

The compound NaCl in ionic form can be written as;

NaCl → Na⁺ + Cl⁻

In 1 mole of NaCl we have 1 mole of sodium ions

Now, let us find the number of moles in NaCl;

Number of moles = $(mass)/(molar mass)$

Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

Number of moles = $(1400)/(58.5)$ = 23.93mol

So;

Since 1 mole of NaCl gives 1 mole of Na⁺

In 23.93 mole of NaCl will give 23.93 mole of Na⁺

1 mole of a substance = 6.02 x 10²³ ions of a substance

23.93 mole of a substance = 6.02 x 10²³ x 23.93

= 1.44 x 10²⁵ ions of Na⁺

VSEPR theory predicts that an atom with one lone pair and three bonding pairs (such as the N-atom in aniline) will have a tetrahedral electron geometry and a trigonal pyramidal molecular geometry due to steric repulsions between H-atoms and the N-atom lone pair. However, in question 5 you observed that the N-atom in aniline is not perfectly sp3 hybridized (i.e. the hybridization is different from that predicted for a tetrahedral electron geometry). Briefly describe all of the factors that result in the calculated hybridization of the N-atom lone pair

Answers

Answer: The lone pair of electron on nitrogen is accommodated in a 2p orbital hence it interacts with the pi system in aniline.

Explanation:

Aniline is less basic than amines. This is because, the nitrogen atom in aniline is not purely sp3 hybridized. Its actual hybrization state is closer to sp2 because the lone pair on nitrogen is accommodated in a 2p orbital. The nitrogen atim in aniline is planar and its

lonely pair interacts with the pi electron system of aniline. This makes the lone pair unavailable for protonation hence aniline is less basic than amines.

Final answer:

The calculated hybridization of the N-atom lone pair in aniline is affected by electron-electron repulsions, resonance, and steric effects from substituents on the aromatic ring.

Explanation:

The calculated hybridization of the N-atom lone pair in aniline is different from the predicted sp3 hybridization due to a combination of factors:

The presence of a lone pair on the nitrogen atom leads to electron-electron repulsions, causing distortions in the molecule's geometry.
The lone pair on the nitrogen atom can participate in resonance, resulting in delocalization of electrons and a change in hybridization.
The presence of the substituents on the aromatic ring can affect the hybridization of the N-atom lone pair by exerting steric effects.

Overall, these factors contribute to the observed hybridization of the N-atom lone pair in aniline, deviating from the predicted tetrahedral electron geometry.

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A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 170 C, and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to 500. mL. Then he filters it and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 3.96 g.Using only the information above, can you calculate yes the solubility of X in water at 17.0 C? If you said yes, calculate it.