Can anyone help me with my homework

Ice forms as water is poured onto dry ice is a physical change .

So , option D is correct one .

The physical change is explained in following points ,

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The empirical formula of the given ions and name of their compounds are given below:

Empirical formula is the simplest formula of a compound where the elements in the compounds are shown in their simplest combining ratios.

The empirical formula of some compound are equal to the molecular formula of the compound whereas in some other, compounds, the empirical formulas are multiples of the molecular formula.

The empirical formula of the given ions and name of their compounds are given below:

Cation: Co³⁺; Anion; PO³⁻ ; Empirical formula : CoPO₄; Name : Cobalt (iii) phosphate

Cation: Fe²⁺  Anion; CrO₄²⁻  Empirical formula : FeCrO₄ ; Name : Iron (ii) tetraoxochromate (vi)

Cation: Cu²⁺  Anion; OH⁻  Empirical formula : Cu(OH)₂ ; Name : Copper (ii) hydroxide

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A metal (M) forms a compound with formula MCl₂. The compound contains 57.69% Cl by mass and the metal is Chromium. Symbol of Chromium metal is Cr.

Given :

57.69 % Cl by mass means that 57.69 grams of chlorine present in 100 grams of compound.

Mass of compound = 100 g

Mass of chlorine = 57.69 g

Mass of metal = 100 - 57.69= 42.31 g

First, we will calculate the moles of chlorine, using formula

Moles = Given mass

              molar mass

Moles = 57.69

              35.45

Moles = 1.627 moles

The given formula of compound is, MCl₂

Therefore, ratio of M : Cl = 1 : 2

We will calculate the moles of metal,

2 moles of Cl combines with 1 mole of metal

1.627 mole of Cl combines with 1.627 = 0.813 mole of metal

                                                        2

calculate the molar mass of metal -

Molar mass of metal =   Given mass  

                                           moles

Molar mass = 42.31  g     = 52.04 g/mol

                       0.813 mol

The molar mass of metal is 52.04 g/mole.

From this, we can conclude that the metal is Chromium (Cr)  

So, the formula will be, CrCl₂

Hence, the identity of the metal is Chromium (Cr)

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The pH of the solution prepared by dissolving sodium acetate and acetic acid is 5.05.

Given -

CH3COONa = 2.05g

So, Number of moles of CH3COONa = 2.05 / 82.03 = 0.024

∴ Molarity = 0.024  / 0.05

                = 0.48 M

and CH3COOH =  0.25 M  

pH of buffer - The equilibrium constant (Ka) of the weak acid and the ratio of weak base (A-) to weak acid (HA) in solution are the two variables that determine the pH of a buffer.

1) Various weak acids have different equilibrium constants (Ka).

2) Ka tells us how much of the HA in the solution will be dissolved into H+ and A-.

According to the question -

CH3COOH ⇄ CH3COO⁻ (aq) + H⁻ (aq)

Ka = [CH3COO⁻] [H⁺] /  {CH3COOH]

 [H⁺] = Ka x [CH3COOH] / [CH3COO⁻]

[H⁺] = 1.75 x 10⁻⁵ [0.25] / [ 0.48]

[H⁺] = 0.43 x 10⁻⁵ / 0.48

      = 0.89 x 10⁻⁵

pH =  -log [H⁺]

∴ pH = - log [0.89 x 10⁻⁵]

pH =  5.05

Hence, the pH of the solution will be 5.05.

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The total number of atoms in 0.160g of P2O5 is 3 x 10^21 atoms.

Atoms are defined as a smallest unit into which a matter can be divided without the release of electrically charged particles.

First calculate the number of mole of P2O5

Mass of P2O5 = 0.160 g

Molar mass of P2O5 = 142g/mole

Number of mole of P2O5 = mass/ molar mass

n = 0.160/142 g/mole

n = 0.0011 mole of P2O5

One mole of P2O5 has five mole of oxygen.

Using conversion factor

= 0.0011 mole of P2O5 x [ 5 mole of oxygen/ 1 mole of P2O5]

= 0.0055 mole of oxygen

Total number of atoms = 0.0055 x 6.02 x 10^23 atom/mole

                                      = 3 x 10^21 atoms

Thus, the total number of atoms in 0.160g of P2O5 is 3 x 10^21 atoms.

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The mass of carbon dioxide is obtained as 236 g of carbon dioxide.

We know that the first step when we are trying to solve any type of question that involves stoichiometry is that we have to obtain the balanced reaction equation. This would show us the kind of interaction that is able to take place in order to obtain the results that we seek.

The reaction equation is given as;

C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O

If 1 mole of propane occupies 22.4 L

x mole of propane occupies 40.2 L

x = 1 mole  * 40.2 L /22.4 L

x = 1.79 moles

Now;

1 mole of propane produced 3 moles of carbon dioxide

1.79 moles of propane produced 1.79 moles * 3 moles/1 mole

= 5.37 moles of carbon dioxide

Mass of carbon dioxide =5.37 moles * 44 g/mol = 236 g of carbon dioxide.

The mass of carbon dioxide is obtained as 236 g of carbon dioxide.

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Answer:

250 cm ^3

Explanation:

Answer:

the answer is A. Chlorine combines with sodium to form table salt

The vapour pressure of methanol at 57.2°C is 66.3 kpa.

The enthalpy of vaporization for methanol,

              =35.2 kj/mol = 35.2×1000 j/mol                  

at temperature T2=  64.7°C = 64.7 + 273 = 337.4 K

temperature T1 = 57.2 °C = 57.2 + 273 = 330.2 K

vapour pressure P2 = 101.3 kpa

vapor pressure P1 = ?

Clausius - Clapeyron equation is geven as :

ln ( P1 / P2 ) =  ΔH / R ( 1 / T2 - 1 / T1 )

ln ( P1 / 101.3 kpa ) = 35200 j/mol / 8.314 j/Kmol × ( 1/337.4 K - 1/330.2 K)

ln ( P1 / 101.3 kpa)  =  [tex]e^{-0.423}[/tex]

P1 / 101.3 kpa = 0.655

P1 = 0.655 × 101.3 kpa

P1 = 66.3 kpa

Hence, The vapor pressure at 57.2°C is 66.3 kpa.

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The atoms of covalent compounds are connected by covalent bonds. Sharing one or more pairs of valence electrons results in the formation of a covalent bond.

How to name covalent compound-

1. Give the elemental name of the non-metal that is on the periodic table that is farthest to the left.

2. Give the other non-metal's name and a suffix ending, such as -ide.

3. To denote the quantity of an element in a molecule, use the prefixes mono-, di-, tri-, etc.

4. Mono is understood and not written if it is the initial prefix.

Names of covalent compounds according to the question-

1. CO₂ - Carbon Dioxide

2. CO - Carbon Monoxide

3. N₂O₄ - Dinitrogen tetraoxide

4. PCl₃ - Phosphorus trichloride

5. CBr₄ - Tetra bromomethane

6. P₂O₅ - Phosphorus pentoxide

7. PCl₅ - Phosphorus pentachloride

8. NF₃ - Nitrogen Trifluoride

9. SO₂ - Sulfur dioxide

10. SO₃ - Sulfur trioxide

11. SCl₆ - sulfur hexachloride

12. SiCl₄ - Silicon tetrachloride

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The electron is in the 5s orbital.

The term orbital is the region in space where there is a high probability of finding an electron. There are four quantum numbers that define an orbital and these are;

Principal quantum number

Orbital quantum number

Spin quantum number

Magnetic quantum number

Let us now look at the sets of quantum numbers as given, a look at the principal and the orbital quantum numbers (m and l) tell us that the electron is in the 5s orbital.

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What is meant that mass cannot be created nor destroyed in the law of conservation of mass is that, matter can take part in a chemical reaction, but the initial mass of the matter (reactants) will be equal to the final mass of the matter (products formed).

The law of conservation of mass states that the mass can neither be created nor destroyed but it can be transformed from one form to another.

Examples of law of conservation of mass;

So, in the law of conservation of mass, the sum of the masses of the reactants in a chemical reaction and products formed in the process remains unchanged.

Thus, what is meant that mass cannot be created nor destroyed in the law of conservation of mass is that, matter can take part in a chemical reaction, but the initial mass of the matter (reactants) will be equal to the final mass of the matter (products formed).

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Answer:

The law of conservation of mass states that the mass can neither be created nor destroyed but it can be transformed from one form to another.

Explanation:

Empirical formula is C4H7.

Empirical formula- A formula that lists the relative amounts of the constituent components in a compound without specifying the number or configuration of atoms.

When a hydrocarbon (CxHy) burns, CO2 and H2O are produced.

(Note: It mentions that a hydrocarbon is burned, which suggests that oxygen isn't present.)

moles of C in the compound: 44 g x 1 mol C/mole and 27.8 g CO2 x 1 mol CO2. = 0.632 moles of carbon

9.96 g H2O x 1 mol H2O/18 g x 2 mol H/mol H2O Equals 1.11 moles of H in the molecule.

We can calculate the mass and verify that it adds up to 8.70 g to rule out the presence of oxygen in the original molecule.

mass C=0.632 mol C times 12 g/mol equals 7.58 g C

H mass equals 1.11 mol H = 1.11 g H x 1 g/mol

Total mass Equals 8.69 g.

We may divide both by the lowest value (0.632), which will give us the lowest whole number of moles.

0.632/0.632 = 1.0 moles C

H = 1.11/0.632 = 1.75 moles.

Now we can multiply them by 4 to get H's whole number, which is-

C = 4 moles.

H = 7 moles

∴Empirical equation: C4H7

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The orbitals where you can find each of the bonds;

a - sp3

b - sp3

c - sp2

d- sp3

The term hybridization is the term that is used to describe the mixture of atomic orbitals which can be used to combine with other orbitals of relevant energy.

Let us recall that in some cases, the energy that is required in bonding could only be obtained by the hybridization of orbitals in which orbitals are mixed. The mixed orbitals are orbitals that are very close to each other in energy and help in the formation of bond.

Now let us try to write the orbitals where you can find each of the bonds;

a - sp3

b - sp3

c - sp2

d- sp3

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The spontaneous fission reaction is:  

₉₈ Cf ²⁵² → ₅₄ Xe ¹⁵⁰+ ₄₄ Ru ¹⁰⁸ +4 ₀n¹

Given,

Californium-252 undergoes spontaneous fission to produce xenon-140, ruthenium-108, and an unknown number of neutrons.

The Nuclear reaction is

₉₈ Cf ²⁵² → ₅₄ Xe ¹⁵⁰+ ₄₄ Ru ¹⁰⁸ +4 ₀n¹

Spontaneous fission reaction always happens in heavier elements, that is the element with atomic numbers more than 89 and mass numbers greater than 230.

The nuclei of these elements are unstable and decay by radioactive decay in different forms like beta decay, positron, etc. At some percentage of the time, they decay by spontaneous fission.

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Answer:

Methanol

Explanation:

Methanol is a compound and compounds are pure substances formed by the combination of elements; they can be decomposed by ordinary chemical means. The other answers are incorrect because they are elements and elements consists of only one kind of atom and they cannot be broken down into a simpler type of matter by either physical or chemical means.

Answer:

A sentence fragment is a sentence that is missing either its subject or its main verb.

Explanation:

Incorrect: Went to the store yesterday.

Incorrect: After the classes, the library. My life nowadays

To tolerate biting forces that average 28,000 psi, modern dental restorative materials have strengths in excess of dna psi.

Deoxyribonucleic acid is a polymer made of two polynucleotide chains that coil around one another to form a double helix and which contains the genetic material necessary for the growth, development, and reproduction of all known organisms as well as a large number of viruses. The two types of nucleic acids are DNA and RNA. One of the four major types of macromolecules that are necessary for all known forms of life is nucleic acids, along with proteins, lipids, and complex carbohydrates (polysaccharides).

Due to the fact that the two DNA strands are made up of more straightforward monomeric units known as nucleotides, they are referred to as polynucleotides. Each nucleotide consists of a phosphate group, a sugar called deoxyribose, and one of the four nucleobases that contain nitrogen: cytosine (C), guanine (G), adenine (A), or thymine (T).

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Explanation:

When magnesium reacts with oxygen, it produces light bright enough to blind you temporarily. Magnesium burns so bright because the reaction releases a lot of heat. As a result of this exothermic reaction, magnesium gives two electrons to oxygen, forming powdery magnesium oxide (MgO).

The calculated answer is 1.99 x 10/21 x Au

Gold has a molar mass of 196.96657 g/m. The mass of 1 mole of gold atoms, which is commonly given in grams, is referred to as the molar mass of gold.

Divide the specified mass by the molar mass of gold, which is 196.966569 g/mol, to determine the moles.

The term "atomic weight" refers to an atom's overall weight. With the addition of a small amount from the electrons, it is roughly equal to the sum of the protons and neutrons. The number of neutrons an atom has has a significant impact on the stability of the nucleus and, consequently, the radioactivity of the atom. The average mass of an element's atoms expressed in atomic mass units is known as its atomic mass (amu, also known as daltons, D). Gold has a molar mass of 196.96657 g/m. The mass of 1 mole of gold atoms, which is commonly given in grams, is referred to as the molar mass of gold.

Divide the specified mass by the molar mass of gold, which is 196.966569 g/mol, to determine the moles.

(Atomic weight in g/mol on the periodic table).

0.00330 mol = 0.650 g Au = 196.966569 gmol Au Gold atoms 1 mole of atoms is equal to 6.02210/23 atoms.

double the estimated mol Au by several.

6.022×10/23 atoms1mol.

0.00330mol Au = 6.022 x 10/23 x 1 mol

= 1.99 x 10/21 x Au

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After 90 years amount of  Cs-137 isotopes has a half-life of 30 years. would remain is equal to = 0.125 grams.

An important and useful factor of radioactive decay is half of life, which is the amount of time it takes for one half of a radioactive isotope to decay. The half of-life of a selected radioactive isotope is constant; it is unaffected by way of situations and is independent of the initial quantity of that isotope.

The half of life of a chemical reaction can be defined because the time is taken for the awareness of a given reactant to attain 50% of its preliminary awareness. The half of-life of a response is the time required for the reactant's attention to decrease to at least one-half of its initial price. The half of-life of a first-order response is consistent this is associated with the charge consistent for the reaction. t1/2 = 0.693/k

Since all radioactive reaction follows first-order reaction, The rate of reaction does not depend on initial concentration.

calculation:-

1 gram Cs-137 at T = 0

after the first half,

⇒0.5 gram Cs-137 at T = 30

After the second half,

⇒0.25 gram Cs-137 at T = 30.

After the third half,

⇒0.125 gram Cs-137 at T = 90 years.

Hence, after 90 years amount of  Cs-137 would remain is equal to = 0.125 grams.

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Comparing consumer products on the basis of their components like use, safety, quality, and cost is necessary as it creates awareness about the product among consumers.

Consumer Products - It is a kind of product that is bought by individuals for personal use or consumption. These are the end products or final goods reached to the end-user i.e consumer by retailer via various distribution channels. Ex- grocery, clothing, etc.

Marketing strategies are built based on different market segmentation and types of consumer products available to the target market.

Types of Consumer Products

There are four types of consumer products, they are:-

Convenience Products

Shopping Products

Specialty Products

Unsought Products.

Hence, it is necessary to compare consumer products on basis of the components like use, safety, quality, and cost. It also increases the competition as well as creates awareness among consumers.

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Major periodic trends include atomic radius, ionization energy, electron affinity, electronegativity, valency and metallic character.

Explanation:

Cs(s)+HBr(aq)=H(g)+CsBr(aq)

sodium hydroxide (NAOH) a.k.a causic soda commonly used in labs​.

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Answer: Sodium hydroxide (NaOH), also known as caustic soda or lye, is a highly versatile substance used in a variety of manufacturing processes.

Explanation: For Example: Sodium hydroxide is also the most common base used in chemical laboratories, being able to test for quite a number of cations (this is called Qualitative Inorganic Analysis), as well as to provide alkaline mediums for some reactions that need it, such as the Biuret test.

Answer:

The structure has a six-carbon ring which is represented by a hexagon and it includes 3-double bonds.

We need to add 7.5 mL of distilled water to the 1% m/V HCI solution available.

The percent concentration is one of the common units of concentration that could be used to describe the amount of substance. Let us now try to obtain the volume needed.

Given that;

C1V1=C2V2

C1 = initial concentration

V1 = Volume of the stock solution

C2 = final concentration

V2 = volume of the new solution

Thus;

V1 = C2V2/C1

V1 = 0.25 * 10/1

= 2.5 mL

Hence, the volume of distilled water needed =  10mL - 2.5 mL

= 7.5 mL

Thus, we need to add 7.5 mL of distilled water to the 1% m/V HCI solution available.

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