Activity 4: Crystallization Materials: Glass rod, crystallization (trough) pan, copper sulfate, distilled water, beaker(250mL). Procedure: In a beaker containing 100 mL distilled water, carefully add copper sulfate crystals and stir with the glass rod in order to dissolve the maximum amount of copper sulfate crystals. - Pour the contents of the beaker into a crystallization trough and allow the solution to stand until there is complete evaporation of the water. Analysis: a- What do you observe at the bottom of the trough? b- What are some benefits of crystallization? Figure 2
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Answer:

cooking an egg is a chemical reaction because you can change the cooked egg into it original form

Explanation:

pls pls give brainliest pls

Answer:

Yes,Cooking an egg is a chemical reaction.

Explanation:

There are two types of reactions

Answer:

D Cobalt

Explanation:

The volume of the sphere is  40 -25 = 15 cm^3

Density = mass/volume = 133 gm / 15 cm^3 = 8.87 gm/cm^3

  which corresponds to Cobalt from the chart

Answer:

It really depends on the situation. If you have 1000 elk and 2 wolves, they really would play a better role in killing off elk so that the elk population doesn't get over populated and diseased, causing a major population collapse. But if there were only 10 elk and 50 wolves, the wolves would play a very negative role on the population.

Hope this helps!

Explanation:

A. The properties of matter and how matter changes.

Answer: A. the properties of matter and how matter changes

Explanation:

just got it right

Answer:

The answer is true.

Explanation:

The deeper you go the more water there's on top of you pushing down.

Answer:

True

Explanation:

The moles of Ba3(PO4)2 precipitated is 0.00332 moles. To calculate the moles of Ba3(PO4)2 precipitated, we need to first calculate the mass of Ba3(PO4)2 formed.

Mass of Ba3(PO4)2 = Mass of filter paper and Ba3(PO4)2 – Mass of filter paper

= 5 g – 3 g

= 2 g

Now, we need to calculate the moles of Ba3(PO4)2 formed using its molar mass.

Molar mass of Ba3(PO4)2 = 601.9 g/mol

Moles of Ba3(PO4)2 = Mass of Ba3(PO4)2 / Molar mass of Ba3(PO4)2

= 2 g / 601.9 g/mol

= 0.00332 moles

Therefore, the moles of Ba3(PO4)2 precipitated is 0.00332 moles.

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Answer: The answer is a. limiting, excess. I took the test.

Explanation:

Temperature of the sample helium with an root-mean-square velocity of 450.0 m/s is 32.47K

The value of the square root of the sum of the squares of the stacking velocity values divided by the number of values is the root-mean square (rms) velocity.

As we know, from Maxwell's speed distribution, the rms speed of velocity of gas is given by,

\(V_{rms}\) = \(\sqrt{3RT/M}\)

450 = \(\sqrt{(3*8.3145*T)/4*10^{-3} }\)       [ Mass of helium = 4.00 u]

T = 32.47 K

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Answer:

D

Explanation:

The ratio of C6H12O6 (which will be referred to as "the carb") to oxygen is 1 to 6, so if we find an answer which has the same ratio, it should be chosen. A is 1:3
B is even worse with a ratio of the carb to oxygen of 1:2
C is the same as B, 1:2
D has a ratio of the carb to oxygen of 1:6, which is what we are looking for.

When ammonium chloride(NH₄Cl) is dissolved in water (H₂O), the temperature of the solution falls because the ammonium chloride is pulling heat in from its surroundings. The given statement is true.

When ammonium chloride(NH₄Cl) is dissolved in water (H₂O), the temperature of the solution falls down because ammonium chloride absorbs heat from the water and due to this the temperature of the solution reduces significantly.

This type of energy change is considered as endothermic.

Generally, the reaction in which absorption of heat takes place is known an endothermic reaction.

Hence, the given statement When ammonium chloride dissolves in water, the temperature of the solution reduces is correct.

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The rate expression is k [BF₃]²[NH₃], overall order is 3, rate constant k is 1.13 × 10⁻³ mol⁻² dm⁶ s⁻¹.

The rate expression for this reaction can be written as:

Rate = k [BF₃]^m[NH₃]^n

The method of initial rates can be used to calculate the values of m and n. When we compare experiments 1 and 2, notice that halving the concentration of [NH3] reduces the initial rate of reaction. This suggests that the reaction is first order in terms of [NH3], implying that n = 1.

When comparing experiments 1 and 3, notice that increasing the concentration of [BF3] by a factor of 2.5 increases the initial rate of reaction by a factor of 9.39 (i.e., 2.13/0.227). This shows that the reaction is about second order in relation to [BF3], i.e., m 2.

Therefore, the rate expression for the reaction is:

Rate = k [BF₃]²[NH₃]

The overall order of the reaction is m + n = 3.

Using the data from experiment 4, substitute the values and solve for k:

Rate = k [BF₃]²[NH₃]

1.02 × 10-¹ = k (3.00 × 10⁻¹)² (1.00 x 10⁻¹)

k = 1.02 × 10-¹ / (3.00 × 10⁻¹)² (1.00 x 10⁻¹)

k = 1.13 × 10⁻³ mol⁻² dm⁶ s⁻¹

Therefore, the rate constant k is 1.13 × 10⁻³ mol⁻² dm⁶ s⁻¹.

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Answer:

Explanation:

Consider the given reaction,

A+B +C

\(rate=k[A]^m[B]^m\)

Reaction is first order with respect to reactant B.

The pseudo order rate is

\(rate=k_1 [B]^m,\\\\where k_1 =k[A]^m\)

Pseudo order reaction is the reaction in which the reactant concentration remains constant; its concentration is very high so it does not involve in the rate of the reaction. Here, the concentration of  A is sufficiently high so this reaction becomes the pseudo first order reaction.

The reaction is first order with respect to [B] with rate constant k1 .

Using the given table, concentration of  B is decreasing the rate also decreasing

Therefore, order of the reaction with respect to A  is 1.

The order of the reaction in reactant A is 1, means the value of m for given rate should be 1.

Pseudo order reactions are those reactions in which two reactants are present but the concentration of one of the reactant is present in excess quantity and it seems like first order reaction.

Given chemical reaction is: A + B +C

According to the question, given reaction is first order with respect to the reactant B and rate law will be represented as:

Rate = k[A]m[B]m

Also given that reactant A is present in high concentration in the reaction, so the rate of pseudo order reaction is:

Rate = k1[B]n, where

k1 = k[A]m

According to the given table with respect to concentration of reactant B and rate k1, it is clear that rate of the reaction decreases as the concentration of reactant B decreses. So, to get the required result order of the reaction or value of m with respect to the reactant A should be 1.

Hence option (B) is correct i.e. 1.

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The 132 L  volume would it be under these conditions.

What is volume ?

Space is occupied by every three-dimensional object. Its volume serves as a gauge for this area. The area contained by an object's limits in three-dimensional space is referred to as its volume. Another name for it is an object's capacity.

What is mole ?

A mole is the atom's elementary particle, an ion. The mole of the substance is always related to the Avogadro number. The mole is always associated with the weight or mass of the element or substance. The standard unit of a mole is mol. The mole is a significant factor of the reactant and products to form an equation. A mole calculates the atom, ion, and substance weighs.

Therefore, 132 L  volume would it be under these conditions.

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Answer:

60 kilometres per hour is the slowest rate of travel

Explanation:

we convert

20 kilometer per minute into kilometer per second

it comes out 1/3 kilometer per sec

Answer:

c

Explanation:

a- 5/1km per second

b-1/3km per second

c. 1/60km per second

the larger the denominator the smaller the fraction

Ammonia (\(NH_3\)) and ammonium chloride (\(NH_4Cl\)) mixture could be a useful buffer in a solution. Option C

A buffer is a solution that can resist changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base or a weak base and its conjugate acid. The buffer system works by the principle of Le Chatelier's principle, where the equilibrium is shifted to counteract the changes caused by the addition of an acid or a base.

In option A, acetic acid (\(CH_3CO_2H\)) is a weak acid, but hydrochloric acid (HCl) is a strong acid. This combination does not form a buffer because HCl is completely dissociated in water and cannot provide a significant concentration of its conjugate base.

Option B consists of sodium hydroxide (NaOH), which is a strong base, and elemental sodium (Na), which is a metal. This combination does not form a buffer as there is no weak acid-base pair involved.

Option D contains acetic acid (\(CH_3CO_2H\)), a weak acid, and ammonia (\(NH_3\)), a weak base. Although they are weak acid and base, they do not form a buffer system together as they are both weak acids or bases and lack the required conjugate acid-base pair.

Option C, ammonia (\(NH_3\)), is a weak base, and ammonium chloride (\(NH_4Cl\)) is its conjugate acid. This combination can form a buffer system. When ammonia reacts with water, it forms ammonium ions (NH4+) and hydroxide ions (OH-).

The ammonium ions act as the weak acid, while the ammonia acts as the weak base. The addition of a small amount of acid will be counteracted by the ammonium ions, and the addition of a small amount of base will be counteracted by the ammonia, thus maintaining the pH of the solution relatively stable.

Therefore, option C, consisting of ammonia (\(NH_3\)) and ammonium chloride (\(NH_4Cl\)), is the suitable mixture that could be a useful buffer in a solution.

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a. The de Broglie wavelength of the 50.0 g mass travelling at 0.2 m/s is \(6.626 * 10^{-32} m\).

b. The de Broglie wavelength of the 50.0 g mass travelling at 45 km/s is \(2.949 * 10^{-37} m.\)

c. The de Broglie wavelength of an He atom travelling at 1000 m/s is \(9.961 * 10^{-11} m.\)

The de Broglie wavelength (λ) of a particle can be calculated using the following formula: λ = h / p, where h is Planck's constant (\(6.626 * 10^{-34} J s\)) and p is the momentum of the particle, given by: p = m * v, where m is the mass of the particle and v is its velocity.

(a) For a mass of 50.0 g travelling at 0.2 m/s:

\(p = m * v = 50.0 g * 0.2 m/s = 10 g m/s\)

Converting the mass to kilograms:

m = 50.0 g = 0.0500 kg

Therefore,

\(p = 0.0500 kg * 0.2 m/s = 0.010 kg m/s\)

Using the de Broglie wavelength formula:

λ = \(h / p = 6.626 * 10^{-34} J s / 0.010 kg m/s = 6.626 * 10^{-32} m\)

Therefore, the de Broglie wavelength of the 50.0 g mass travelling at 0.2 m/s is \(6.626 * 10^{-32} m\).

(b) For the same mass travelling at 45 km/s:

Converting the velocity to m/s:

v = 45 km/s = 45,000 m/s

p = \(m * v = 0.0500 kg * 45,000 m/s = 2,250 kg m/s\)

Using the de Broglie wavelength formula:

λ = \(h / p = 6.626 * 10^{-34} J s / 2,250 kg m/s = 2.949 * 10^{-37} m\)

Therefore, the de Broglie wavelength of the 50.0 g mass travelling at 45 km/s is \(2.949 * 10^{-37} m.\)

(c) For an He atom travelling at 1000 m/s:

The mass of an He atom is approximately \(6.646 * 10^{-27} kg\).

\(p = m * v = 6.646 * 10^{-27} kg * 1000 m/s = 6.646 * 10^{-24} kg m/s\)

Using the de Broglie wavelength formula:

λ = \(h / p = 6.626 * 10^{-34} J s / 6.646 * 10^{-24} kg m/s = 9.961 * 10^{-11} m\)

Therefore, the de Broglie wavelength of an He atom travelling at 1000 m/s is \(9.961 * 10^{-11} m.\)

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Answer:

Explanation:

i) in keeping industrial machine cool

ii) in textile inustries for dying clothes

Answer:

B. 2-butene and

D. 3-hexene can show the phenomenon of geometric isomerism.

Explanation:

The compounds that can exhibit geometrical isomerism has been 2-butene and 3-hexene. Thus, option B and D is correct.

Isomers can be defined as the compounds with same unit formula and different structures.

The difference in structure has been resulted in various types of isomers, have been geometrical , structural, functional.

The geometrical isomers has been defined as the compounds with same chemical formula and difference in the spatial arrangement of the atom across the bonds. The geometrical isomerism results in cis and trans isomers.

The geometrical isomerism has been found in the compounds with the compounds exhibiting carbon-carbon double bond. The double bond has been present at the middle of the compound to exhibit geometrical isomerism.

The compounds that can exhibit geometrical isomerism has been 2-butene and 3-hexene. Thus, option B and D is correct.

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Answer:

A(n)___ is the smallest neutral unit of two or more atoms held together by a covalent bond? ​

=>MOLECULE

The amount of excess reactants that will remain unchanged is 0.39 mol, option D.

First, we need to determine which reactant is limiting and which is in excess. The balanced chemical equation shows that 1 mole of N₂H₄ reacts with 2 moles of H₂O₂. Therefore, if we have 0.67 moles of N₂H₄, it will react with 2 × 0.67 = 1.34 moles of H₂O₂.

Since there are only 0.57 moles of H₂O₂, it is the limiting reactant. The amount of N₂H₄ needed to react with 0.57 moles of H₂O₂ is (0.57 mol H₂O₂) / (2 mol H₂O₂/1 mol N₂H₄) = 0.285 mol N₂H₄. Therefore, 0.67 mol - 0.285 mol = 0.385 mol.

Therefore, 0.39 mol of N₂H₄ is in excess and will remain unchanged.

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Taking into account the definition of calorimetry, the specific heat of the unknown solid is 0.285 \(\frac{J}{gC}\).

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

Specific heat of the unknown solid

In this case, you know:

Replacing in the expression to calculate heat exchanges:

For unknown solid: Qsolid= csolid × 35 g× (26.5 - 100)°C

For water: Qwater= 4.186 \(\frac{J}{gC}\)× 50 g× (26.5 - 23)°C

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the solid gives up will be equal to the heat that the water receives. Therefore:

- Qsolid = + Qwater

Replacing the corresponding expressions:

- csolid × 35 g× (26.5 - 100)°C= 4.186 \(\frac{J}{gC}\)× 50 g× (26.5 - 23)°C

Solving:

csolid × 2,572.5 g×°C= 732.55 J

csolid =732.55 J÷ 2,572.5 g×°C

csolid=0.285 \(\frac{J}{gC}\)

Finally, the specific heat of the unknown solid is 0.285 \(\frac{J}{gC}\).

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The molar extinction coefficient is specific to the substance being measured and the wavelength of light used. Accurate and precise values for absorbance, concentration, and path length are necessary for an accurate calculation.

To calculate the molar extinction coefficient (ε) using wavelength (λ) and absorbance (A):

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Answer:

Define/describe/identify the following terms:

a.Acid

b.Acid-base reaction

c.Adhesion

d.Aqueous solution

e.Arrhenius base

f.Base

g.Bent

h.Boiling point

Here are some Revision Links below:

Answer:c

Explanation:c

Explanation:

U can use molar ratio of

Li : Li2SO3

2 : 1

so 3.33÷2=1.665 so 1.67 moles

then u can use the

moles = mass ÷ mr

= 12.33÷ 100.08

= 0.123 moles of CuSO3 then use the molar ratio

CuSO3 : Cu

1:1

so 0.123 moles of cu

hope this make sense:)