The following precipitation reaction can be used to determine the amount of copper ions dissolved in solution. A chemist added 5.00 × 10−2 L of a solution containing 0.173 mol L−1 Na3PO4(aq) to a 5.00 × 10−2 L sample containing CuCl2(aq). This resulted in a precipitate. The chemist filtered, dried, and weighed the precipitate. If 1.21 g of Cu3(PO4)2(s) were obtained, and assuming no copper ions remained in solution, calculate the following: the concentration of Cu2+(aq) ions in the sample solution. the concentrations of Na+(aq), Cl−(aq), and PO43−(aq) in the reaction solution (supernatant) after the precipitate was removed.

The color of the object appears as yellow/orange.

The visible light spectrum spans a wavelength range of 400 to 700 nanometers (nm), with shorter wavelengths appearing as blue or violet and longer wavelengths as red.

This object is absorbing light with wavelengths between 430 and 400 nm, indicating that it is absorbing light that is visible in the blue and violet spectrum.

We know that this color that we see is actually a complementary color to blue/violet from the color wheel.

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Answer: The answer is incus

The mass of Fluorine in grams in the second sample is 2066 g. The concept law of definite proportions is used here to obtain the mass of Fluorine in the second sample.

This law was stated by a French chemist, Louis Proust. It states that 'a chemical compound always contains the same elements combined together in the same proportion by mass'.

The mass of Mg in sample 1 is 1.66 kg and that of F is 2.56 kg. The mass of Mg in sample 2 is 1.34 kg and we have to find out the mass of F in second sample.

The mass of Fluorine in sample 2 can be obtained by using the mass ratio of the elements in sample 1.

In sample 1:

Mass of magnesium ÷ Mass of Fluorine = 1.66 kg ÷ 2.56 kg

In sample 2:

By using dimensional analysis,

Mass of Fluorine = 2.56 kg Fluorine ÷ 1.66 kg Magnesium × 1.34 kg Magnesium

Therefore the mass of Fluorine = 2.066 kg

The mass in kilogram can be converted into grams as:

Mass of Fluorine = 2.066 kg × 1000 g ÷ 1 kg = 2066 g

Thus the mass of Fluorine in the second sample is 2066 g.

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The pair of compounds forms a homogeneous solution when combined are : A. homogeneous solution, B. non homogeneous solution, C. homogeneous solution, D. non homogeneous solution.

Part A) :- (a) CH₃CH₂CH₂CH₂CH₃ and CH₃CH₂CH₂CH₂CH₂CH₃ form a homogeneous solution because they are both nonpolar molecules and therefore have similar intermolecular forces.

(b) CBr₄ and H₂O do not form a homogeneous solution because CBr₄ is a nonpolar molecule and H₂O is a polar molecule, so they have different intermolecular forces and are not miscible.

(c) LiNO₃ and H₂O form a homogeneous solution because LiNO₃ is an ionic compound that dissolves in water to form a solution.

(d) CH₃OH and CH₃CH₂CH₂CH₂CH₃ do not form a homogeneous solution because CH3OH is a polar molecule and CH₃CH₂CH₂CH₂CH₃ is a nonpolar molecule, so they have different intermolecular forces and are not miscible.

Part B)

For the homogeneous solutions, the type of forces involved are:

(a) CH₃CH₂CH₂CH₂CH₃ and CH₃CH₂CH₂CH₂CH₂CH₃: London dispersion forces (induced dipole-induced dipole)

(c) LiNO₃ and H₂O: Ion-dipole forces.

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The liquid state of water is more dense than the solid state; option C.

Density is the mass per unit volume of a substance.

Density = mass/volume

The density of a given mass of water in the solid, liquid and gaseous state can be described thus:

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Answer: c- the liquid state is more dense than the solid state.

Explanation:

Answer:

Ca3(PO4)2

Explanation:

Ca3(PO4)2 or calcium phosphate is insoluble in water.

Answer:yes

Explanation:

Explanation:

The brain is like a computer that controls the body's functions, and the nervous system is like a network that relays messages to parts of the body. Click through this slideshow to learn more about the brain and nervous system.

The percentage abundance of the isotope Z-20 is 10.8 % and that of second isotope Z-22 will be 100 -10.8 = 89.2 %.

Isotopes are atoms of same atomic number and different mass number. An element can have two or more isotopes.  The percentage abundance of each isotopes in the nature will be different.

Given that the average atomic mass of Z is 21.5 amu. It is  the sum of products of the mass and abundance of each isotopes. Let x be the % abundance of Z-20 and 100-x for Z-22, now it can be written as:

20 x +  22 (100 - x ) = 21.5

20 x - 2200 -22x = 21.5

 2x = 2200-21.5

 x/100 = 10.8 %.

Therefore percentage abundance of Z-20 is 10.8 % and that of Z-22 is 100 - 10.8 = 89.2 %.

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Answer:

6 different frequencies

Explanation:

From energy level 1 to 2 is one frequency, from energy level 1 to 3 is one frequency and From energy level 1 to 4 is one frequency. So, we have a total of 3 frequencies for transition from energy level 1.

From energy level 2 to 3 is one frequency and from energy level 2 to 4 is one frequency. So, we have a total of 2 frequencies for transition from energy level 2.

From energy level 3 to 4 is one frequency.

So we have a total of 3 + 2 + 1 different frequencies = 6 different frequencies.

Note that the reverse process for each step produces the same frequency as the step in consideration.

Answer:

Bathypelagic

54% of the ocean lies in the Bathypelagic (aphotic) zone into which no light penetrates. This is also called the midnight zone and the deep ocean. Due to the complete lack of sunlight, photosynthesis cannot occur and the only light source is bioluminescence.

Explanation:

The small surface zone that has light is the photic zone. The entire rest of the ocean does not have light and is the aphotic zone.

Permanent darkness below 1000 meters with bioluminescent bacteria, bottom feeders, and angler fish is where no light reaches.

This is referred to the state of being dark as a result of absence of light in the area.

The light ray penetration decreases with increase in depth thereby making areas below 1000 meters dark with bioluminescent bacteria, bottom feeders, and angler fish which is why option C was chosen.

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Answer:

I am almost done

Explanation:

21.0 L

↓

2H₂(g) + O₂(g) → 2H₂O(l)

↑

4.67 × 10²⁴ moles

[This is done so the equation is balanced and the reaction can occur]

[subscripts are the number of atoms, and the coefficients are the number of molecules]

______________

Since there are about 6.02 × 10²³ moles in one mol of a substance.

4.67 × 10²⁴ / 6.02 × 10²³ =

7.7547174437 mol.

Next, we must determine the grams per mole of each substance, this can be done with the periodic table.

Since hydrogen, and oxygen gas are diatomic, they will consist of two atoms each. So the atomic mass will be multiplied by 2 to get the molecular mass. Therefore H₂ will have a molecular mass of 1.00794 × 2 = 2.01588 g/mol.

Answer:

I can not see the file

Explanation:

Answer: Stibine could be a poisonous gas that has impacts similar to those of arsine, but stibine isn't experienced as often as possible as arsine. The foremost likely course of introduction to arsine and stibine is inward breath. Arsine poisonous quality is diverse from arsenic harmfulness. Arsine causes gigantic hemolysis and comes about in frailty, jaundice, and hemoglobin uric renal disappointment.

Explanation:

The given white substances can be divided into two groups based on their solubility.

Ionic compounds potassium iodide, sodium chloride, and calcium carbonate are all strong electrolytes that conduct electricity well when dissolved. They all dissolve readily in water.

Citric acid, sucrose, and phenyl salicylate are weak or non-electrolytes. Sucrose is more soluble in water than the other two substances are in ethanol.

The chemicals could also be divided into groups according to whether they dissolve in water producing hydrogen ions, hydroxide ions, or neither, according to whether they have high or low melting temperatures, according to whether they are optically active. Chemical classifications are extremely useful; they are applied as necessary.

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Answer:

as in shapes

Explanation:

Considering the phase diagram, the state of matter that exists in area B is gas.

The correct option is A.

A phase diagram is a graphical representation that shows the conditions of temperature and pressure at which different phases or states of a substance exist.

The axes of a phase diagram typically represent temperature (usually on the horizontal axis) and pressure (usually on the vertical axis). The diagram is divided into regions that correspond to different phases, and the lines separating these regions represent phase boundaries.

The point where three phase boundaries meet is known as the triple point, which represents the temperature and pressure at which all three phases can coexist in equilibrium.

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Answer:

B

Explanation:

4moles of Cl2 produces 4moles of HCl therefore 12moles of Cl2 will produce 12moles of HCl

Answer:

Explanation:

The Lewis structure of ammonia, NH3 , would be three hydrogen atoms bonded to a nitrogen atom in the middle, with a lone pair of electrons on top of the atom. This is the reason why ammonia acts as a Lewis base, as it can donate those electrons

The concentration of H3O+ ions in the solution of aluminum chloride is \(3.16×10^-5\) mol/L.

Aluminum chloride is an acidic salt that contains a cation, Al3+, and an anion, Cl-. When aluminum chloride is dissolved in water, it dissociates into its constituent ions, and the Al3+ cations hydrolyze to produce H+ ions.

This reaction leads to the formation of an acidic solution. The pH of a solution of aluminum chloride is \(4.5×10^0\). We need to determine the concentration of H3O+ ions in this solution.

The concentration of H3O+ ions in a solution is given by the equation: pH = -log[H3O+] where pH is the negative logarithm of the concentration of H3O+ ions in the solution. The negative sign indicates that the pH is inversely proportional to the concentration of H3O+ ions. To determine the concentration of H3O+ ions, we need to rearrange the equation:

[H3O+] = \(10^-pH\) Substituting the value of pH =\(4.5×10^0\), we get: [H3O+] = \(10^-4.5\)

The value of \(10^-4.5\) can be calculated using scientific notation: \(10^-4.5\)= \(3.16×10^-5\) mol/L

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Answer:

question pls.............

Answer:

A

Explanation:

Mn(s) metal will spontaneously react with Zn²⁺(aq), but will not

spontaneously react with Mg²⁺ (aq)

The Eo value of an electrochemical cell determines its spontaneity. Positive Eo electrochemical cells are spontaneous, and vice versa.

The relevant Eo of the half-cell in this instance are as follows for Mn(s) metal

Zn2+/Zn = -0.763v for Eo

2.37v for Eo Mg2+/Mg.

Mn2+/Mn = -1.18v for Eo.

Consequently, the equation for an Eo cell (with Zn as one of the half-cells) is: Eo Zn2+/Zn - Eo Mn2+/Mn = -0.763 - (-1.18) = 0.417v.

On the other hand, the equation for an Eo cell (with Mg as one of the half-cells) is: Eo Mg2+/Zn - Eo Mn2+/Mn = -2.37 - (-1.18) = -1.19v.

As a result, Mn(s) metal will spontaneously react with Zn2+(aq), but not with Mg2+ (aq)

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Answer:

448

Explanation:

K =273+°c

K=273+175

K=448

In order to test for ammonia, a moist red litmus paper should be held over the test tube containing the ammonia.  Option 3.

Ammonia is a substance that is alkaline in nature. All alkaline substances are able to turn red litmus paper to blue. This is in opposition to acidic substances that turn blue litmus paper to red.

Thus, the first test that can be used to determine if a substance is ammonia would be to hold a moist red litmus paper over the test tube containing the suspected substance.

Since ammonia is gas at room temperature, the substance suspected to be ammonia will diffuse from the test tube to turn the moist red litmus paper to blue.

Other methods to test for ammonia include passing a test tube containing hydrochloric acid over the test tube containing the substance. A white fume of ammonium chloride will confirm the presence of ammonia.

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Explanation:

To calculate the concentration of the resulting solution, we can use the formula C₁V₁ = C₂V₂, where C₁ is the initial concentration, V₁ is the initial volume, C₂ is the final concentration, and V₂ is the final volume.

Given:

C₁ = 0.2 mg/mL

V₁ = 3.5 mL

V₂ = V₁ + 0.5 mL = 3.5 mL + 0.5 mL = 4 mL (assuming no significant change in volume upon mixing)

Using the formula, we can solve for C₂:

C₁V₁ = C₂V₂

0.2 mg/mL * 3.5 mL = C₂ * 4 mL

0.7 mg = C₂ * 4

C₂ = 0.7 mg / 4 mL

C₂ ≈ 0.175 mg/mL

Therefore, the concentration of the resulting solution is approximately 0.175 mg/mL.

Answer:

The answer is 412

Explanation:

Because we distracted c1 from h2o due to the chemical combination of the two