Use the following chemical equation to answer the following questions:

2SO3(g)+Energy><2SO2(g)+O2(g)

Answer:

Idon't know if this helps but I think it is a linear structure and if I am wrong I am so sorry

Answer:

1. Participating in calcium homeostatis storage of calcium.

2. High capacity calcium (Ca) regulation and protection against herbivory

\(\large \boxed{\sf 2 \: functions \: of \: crystals \: are :- } \)

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\( \sf \: \underline{ Calcium \: oxalate \: (CaOx) \: crystals} \: are \: distributed \: \\\sf among \: all \: taxonomic \: levels \\ \sf\: of \: photosynthetic \: organisms \: from \\ \sf \: small \: algae \: to \: angiosperms \: and \: giant \: gymnosperms .\)

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\(\sf Bone \: is \: mostly \: made \: of \: \underline{mineral \: crystals} \: \\ \sf and \: the \: protein \: collagen. \: The \: mineral  \: crystals \: bone  \\ \sf\: provide \: strength \: and \: rigidity \: for \: the \: matrix \: upon \: \\ \sf \: and \: within \: which \: they \: are \: deposited.\)

Answer:

A harmless nuclear fallout

Explanation:

just did this

Answer:

The answer is B

Explanation:

Neap tides are a type of to tie occuring after spring tides. It occurs when moon and sun are at the right angles to each other. It produce low tidal ranges. Hence, phrase B is correct.

Tides are a geographical phenomenon taking place due to the gravitational pull by sun and moon on the ocean of earth. They are long period waves causing the ocean back and forth to the cost.

During new moons larger tides appears in earth and they are called spring tides. Spring tides occurs twice in a lunar month. During spring tides, moon appears full and earth is between sun and moon.

After seven days of spring tide, lower tides are produced when the moo and sun comes in right angle to each other. They are called the neap tides producing lower range of tidal waves . During this time moon will be half full thus for the first and last quarter of the moon. Hence, the option B neap tide produce low tidal ranges is correct.

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Answer:

C. 1.07 M.

Explanation:

Hello,

In this case, we can define the molarity of the bleach as shown below:

\(M=\frac{moles_{NaClO}}{V_{solution}}\)

In such a way, given the mass of bleach in a 1-L solution, we can compute the density:

\(\rho = \frac{1100g}{1L}=1100g/L =1.1 kg/L=1.1g/mL\)

In such a way, we can use the previously computed density to compute the volume of the solution, assuming a 100-g solution given the by-mass percent:

\(V_{solution}=100g*\frac{1mL}{1.1g} *\frac{1L}{1000mL} =0.091L\)

Afterwards, using the by-mass percent of bleach we compute the mass:

\(m_{NaClO}=100g*0.0725=7.25g\)

And the moles:

\(n_{NaClO}=7.25g*\frac{1mol}{74.44g} =0.097mol\)

Therefore, the molarity turns out:

\(M=\frac{0.097mol}{0.091L}\\ \\M=1.07M\)

Thus, answer is C. 1.07 M.

Regards.

Answer:

0.214 J/g.°C

Explanation:

Step 1: Given data

Step 2: Calculate the specific heat of the substance (c)

The specific heat of the substance is an intrinsic property of a substance. We can calculate it using the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

c = 1287.424 J / 752 g × 8 °C = 0.214 J/g.°C

Ammonia, carbon (iv) oxide and ethylene undergo translational, rotational and vibrational motion differently.

Ammonia (NH3):

Translational motion: NH3 can undergo translational motion as a whole molecule, i.e., it can move from one place to another.

Rotational motion: NH3 can undergo rotational motion around its center of mass, since it has a non-zero moment of inertia. It can rotate about any of its three perpendicular axes, leading to three possible rotational modes. These are known as the A, B, and C rotational modes, where A is the rotation about the z-axis, B is the rotation about the x-axis, and C is the rotation about the y-axis.

Vibrational motion: NH3 can also undergo vibrational motion, involving the stretching and bending of its bonds. Specifically, it has three vibrational modes: a symmetric stretching mode, an asymmetric stretching mode, and a bending mode.

Carbon (IV) oxide (CO2):

Translational motion: CO2 can also undergo translational motion as a whole molecule.

Rotational motion: CO2 can undergo rotational motion around its center of mass, but since it is a linear molecule (with a carbon atom in the center and two oxygen atoms on either side), its moment of inertia is zero about its principal axis. Therefore, it can only rotate about an axis perpendicular to its principal axis, which is known as the A rotational mode.

Vibrational motion: CO2 can undergo vibrational motion, involving the stretching and bending of its bonds. Specifically, it has two stretching modes (symmetric and asymmetric) and one bending mode.

Ethylene (C2H4):

Translational motion: Ethylene can also undergo translational motion as a whole molecule.

Rotational motion: Ethylene can undergo rotational motion around its center of mass, and since it is a planar molecule with a non-zero moment of inertia, it can rotate about two perpendicular axes, which are known as the A and B rotational modes.

Vibrational motion: Ethylene can undergo vibrational motion involving the stretching and bending of its bonds. Specifically, it has four vibrational modes: a symmetric stretching mode, an asymmetric stretching mode, and two bending modes (one in the plane of the molecule and one perpendicular to the plane).

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The volume of the water is  5.11 mL.

Density is a measure of mass per unit of volume of a substance. It is a physical property of matter and is expressed in units of grams per cubic centimeter (g/cm^3).

The density of a substance determines how much of it will occupy a given space, and is a crucial factor in many physical and chemical phenomena, such as buoyancy, phase changes, and thermal conductivity.

We have that;

Volume of the water = ?

Density of the water = 0.9960 g/mL

Mass of the water = 5.097 g

Volume of the water = Mass/Density

= 5.097 g / 0.9960 g/mL

= 5.11 mL

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Answer:

B people control how plants or animals reproduce to get desired traits

Explanation:

if you get it correct mark meh brainliest or however you say it so ppl know its correct

Answer:

Volume = 3.4 L

Explanation:

In order to calculate the volume of CO₂ produced when 15.2 g of CaCO₃ is heated, we need to first write out the balanced equation of the thermal decomposition of CaCO₃:

CaCO₃ (s) + [Heat] ⇒ CaO (s) + CO₂ (g)

Now, let's calculate the number of moles in 15.2 g CaCO₃:

mole no. = \(\mathrm{\frac{mass}{molar \ mass}}\)

                 = \(\frac{15.2}{40.1 + 12 + (16 \times 3)}\)

                 = 0.1518 moles

From the balanced equation above, we can see that the stoichiometric molar ratios of CaCO₃ and CO₂ are equal. Therefore, the number of moles of CO₂ produced is also 0.1518 moles.

Hence, from the formula for the number of moles of a gas, we can calculate the volume of  CO₂:

mole no. = \(\mathrm{\frac{Volume \ in \ L}{22.4}}\)

⇒ \(0.1518 = \mathrm{\frac{Volume}{22.4}}\)

⇒ Volume = 0.1518 × 22.4

                 = 3.4 L

Therefore, if 15.2 g of CaCO₃ is heated, 3.4 L of CO₂ is produced at STP.

When the equilibrium constant is higher than one, it indicates that the reaction prefers to produce products, whereas if the equilibrium constant is less than one, it indicates that the reaction prefers to produce reactants. If the equilibrium constant is equal to one, the reaction proceeds in both directions equally.

In a chemical reaction, exothermic reactions are defined as reactions that release heat into their environment. It implies that heat is given off when reactants are converted to products. At equilibrium, an exothermic reaction continues to be exothermic, meaning that heat is given off even after the reaction reaches a state of equilibrium.There are two types of reactions: exothermic and endothermic.

A reaction is classified as exothermic if it releases heat, and endothermic if it absorbs heat. The direction of the reaction is determined by whether it is exothermic or endothermic. At equilibrium, the reaction is no longer moving forwards or backwards. It's also worth noting that reactions can be exothermic in one direction and endothermic in the other.

The equilibrium constant (K) is defined as the ratio of the concentration of products to the concentration of reactants in the chemical reaction equation. It is used to express how much of the products is generated by the reaction in comparison to the reactants.  the equilibrium constant aids in the identification of the direction in which the reaction will proceed at equilibrium.

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Answer:

balanced............

Based on the formula to determine the freezing point depression of the solvent, the freezing point of the solution is 263.85 K.

The freezing point of a substance is the temperature at which the liquid changes to solid without any further decrease in temperature occurring during the process.

The addition of solute substances in liquids usually lowers the freezing point of the liquid solvent.

The formula to determine the freezing point depression of solvent is given below:

where'

The molality of the given solution = moles of solute/kg of solvent

moles of solute = 46/92

mass of solvent = 100 g or 0.1 kg

Molality of solution = (46/92) / 0.1

Molality of solution = 5

for glycerine, i = 1

ΔT = ΔT = 1 * 1.86 * 5

ΔT = 9.3

The freezing point of the solution = 273.15 - 9.3

The freezing point of the solution = 263.85 K

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Answer:

It can be A or B

Explanation:

Atoms of the same element have the same atomic number so the number of protons (and electrons) are same.

To calculate the molarity of a solution, we need to determine the number of moles of the solute (Bal2) and then divide it by the volume of the solution in liters.

Given:

Mass of Bal2 = 413 g

Volume of solution = 750 ml = 0.75 L

1. Calculate the number of moles of Bal2:

First, we need to convert the mass of Bal2 to moles using its molar mass. The molar mass of Bal2 can be calculated by summing the atomic masses of boron (B) and iodine (I):

Molar mass of Bal2 = (atomic mass of B × 1) + (atomic mass of I × 2)

Molar mass of Bal2 = (10.81 g/mol × 1) + (126.90 g/mol × 2)

Molar mass of Bal2 = 10.81 g/mol + 253.80 g/mol

Molar mass of Bal2 = 264.61 g/mol

Now we can calculate the number of moles of Bal2:

Moles of Bal2 = Mass of Bal2 / Molar mass of Bal2

Moles of Bal2 = 413 g / 264.61 g/mol

Moles of Bal2 ≈ 1.561 mol

2. Calculate the molarity of the solution:

Molarity (M) = Moles of solute / Volume of solution (in liters)

Molarity (M) = 1.561 mol / 0.75 L

Molarity (M) ≈ 2.081 M

Therefore, the molarity of the solution is approximately 2.081 M.

The molarity of the solution is approximately 1.408 M as to calculate the molarity of a solution, one must need to know the number of moles of the solute and the volume of the solution in liters.

The molar mass of BaI₂ is:

Ba (barium) atomic mass = 137.33 g/mol

I (iodine) atomic mass = 126.90 g/mol

Molar mass of  BaI₂ = (Ba atomic mass) + 2 × (I atomic mass)

= 137.33 + 2 × 126.90

= 137.33 + 253.80

= 391.13 g/mol

Given that the mass of BaI₂ is 413 g,

Number of moles = Mass / Molar mass

= 413 g / 391.13 g/mol

= 1.056 moles

Volume of solution = 750 ml = 750/1000 = 0.75 L

Finally, one can calculate the molarity of the solution using the formula:

Molarity = Number of moles / Volume of solution

= 1.056 moles / 0.75 L

= 1.408 M

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Answer: Industrial processes including mining and manufacturing historically have been leading causes of soil pollution. Industrial areas typically have much higher levels of trace elements and organic contaminants. This is due to intentional and unintentional releases from industrial processes directly into the environment, including to the soil, adjacent water bodies, and the atmosphere.

Explanation:

Answer:

Explanation:

Molal freezing point depression constant of butanol Kf = 8.37⁰C /m

ΔTf = Kf x m , m is no of moles of solute per kg of solvent .

mol weight of butanol = 70 g

235.1 g of butanol = 235.1 / 70 = 3.3585 moles

3.3585 moles of butanol dissolved in 4.14 kg of water .

ΔTf = 8.37 x 3.3585 / 4.14

= 6.79⁰C

Depression in freezing point = 6.79

freezing point of solution = - 6.79⁰C .

Answer:

The correct answer is 1332 KJ.

Explanation:

Based on the given information,  

ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.  

Now the balanced equation is:  

C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)

ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2

ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)

ΔH°rxn = -1411.9 KJ

Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.  

Now based on the balanced chemical reaction,  

ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2

ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)

ΔS°rxn = -267.68 J/K or -0.26768 KJ/K

T = 25 °C or 298 K

Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get

ΔG = -1411.9 - 298.0 × (-0.2677)

ΔG = -1332 KJ.  

Thus, the maximum work, which can obtained is 1332 kJ.  

For this section:

(d) To determine which acid would be completely dissociated when dissolved in water, consider the strength of the acid and its ability to ionize completely. Strong acids are those that ionize completely in water, while weak acids only partially dissociate.

Among the given options:

(H O)₂SO is sulfuric acid (H₂SO₄), which is a strong acid and completely dissociates in water.

(H O)IO₃ is iodic acid (HIO₃), which is also a strong acid and completely dissociates in water.

(H O)₂SO₂ is not a known acid and cannot be evaluated for dissociation.

HCO₂H is formic acid (HCOOH), which is a weak acid and only partially dissociates in water.

Therefore, the acids that would be completely dissociated when dissolved in water are (H O)₂SO and (H O)IO₃.

(e) To estimate the pH of the given solutions formed by titration, compare the moles of the acid and base used in the reaction.

(i) For the titration of 25.0 mL of 0.10 M aqueous NaOH with 25.0 mL of 0.10 M aqueous HCl, the reaction between a strong acid (HCl) and a strong base (NaOH) forms a neutral salt (NaCl) and water. The resulting solution would have a pH of 7, indicating neutrality.

(ii) For the titration of 25.0 mL of 0.10 M aqueous NaOH with 25.0 mL of 0.10 M aqueous acetic acid, consider the ionization of acetic acid and the formation of its conjugate base. The pKa of acetic acid is given as 4.76.

Since the volumes and concentrations of the acid and base are equal, we have a 1:1 mole ratio between them. This means that half of the acetic acid will be neutralized, and the remaining half will be in the form of the conjugate base (acetate ion, CH₃COO-). The resulting solution will be a buffer solution with a pH close to the pKa of acetic acid, which is 4.76.

(f) To predict the products formed from mixing the given reagents, we need to consider the reactions and the possible chemical reactions that occur.

(i) 3I₂ + PBr₃: This reaction involves the combination of iodine (I₂) with phosphorus tribromide (PBr₃). The balanced equation is:

3I₂ + 2PBr₃ → 6IBr + P₂

(ii) 4LiH + GaCl₃: This reaction involves the combination of lithium hydride (LiH) with gallium trichloride (GaCl₃). The balanced equation is:

4LiH + GaCl₃ → 4LiCl + GaH₃

(iii) NH₃ + BCl₃: This reaction involves the combination of ammonia (NH₃) with boron trichloride (BCl₃). The balanced equation is:

NH₃ + BCl₃ → NH₃BCl₃

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If we add one or two hydrogen ions to a polyatomic ion that has a 3-charge, as the phosphate ion (PO₄3-), it will still be a polyatomic ion. (Three H+ would entirely cancel out the 3-charge, turning it into a neutral molecule and removing it from the category of polyatomic ions.

As a result, the carbonate ion has 2 more electrons than protons due to its negative charge. The doubly bonded oxygen in the carbonate ion is neutral, whereas each single bonded oxygen has a negative charge. This is the cause of the total charge of "-2," then.

An essential component of the atmosphere of stars like the Sun is the hydrogen anion.

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Answer:

FALSE

Explanation:

false. kinematics equations are used for many purposes.it is used to drive equations and to find the motion of an object. but it is used more in physics rather than maths.

Answer:

s+11r-15

Explanation:

ok ther only one solution n2 is ÷ 2h×2 is ok but the problem is that the standard form of the square is not yet

Air escape from a tire when tire valve is opened because the pressure from the weight of the car is forcing the already pressurized air out .

When air is filled in tires, it get into more compact area than the outer atmosphere, therefore particles in tire are close to each other and exert pressure continuously on tire walls to get out of it. Thus, material of the tire need to be pressure resistant as much possible.

After opening the tire valve, air starts leaving with huge sound as strain outside the tire is weaker than that inside the tire. Molecules strike with the same force on larger area and pressure starts diminishing.

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Note: The question given on the portal is incomplete. Here is the complete question.

Question: Air leaves a tire when the tire valve is opened because

A. the pressure outside the tire is lower than the pressure inside the tire.

B. the pressure outside the tire is greater than the pressure inside the tire.

C. the temperature is higher outside the tire than inside the tire.

D. there are more gas particles outside the tire than inside the tire.

The reaction time decreases as the temperature increases of the reaction mixture increases.

A sample record of results is:

An increase in temperature leads to an increase in reaction rate or a decrease in reaction time.

The increase in temperature provides more thermal energy to the reactant molecules, which leads to an increase in the average kinetic energy of the molecules. As a result, more reactant molecules have sufficient energy to overcome the activation energy barrier and undergo successful collisions, leading to an increased reaction rate.

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You would need to take 5 mL of the 12 M acid and then make up to the 2000 mL mark.

The dilution formula is a mathematical expression used to calculate the new concentration of a solution . The dilution formula can be used to calculate the volume of a solution required in dilution. To use the formula, you must ensure that the units of concentration and volume are consistent

The final concentration of the solution is = Antilog (-1.5) = 0.03 M

Then;

12 * V = 2 * 0.03

Where V is the initial volume of the acid.

V = 0.005 L or 5 mL

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a. The Mass of water driven off = 0.15 g

b. Moles of anhydrate = 0.00257 moles

c. Moles of water driven off is 0.00833 moles

d. There are 3 moles of water within the crystalline structure of one molecule of the hydrated salt.

e. The molecular formula of the hydrated salt will be X.3H₂O

a. The mass of water driven off from the hydrated salt is:

Mass of water driven off = 0.5 g - 0.35 g

Mass of water driven off = 0.15 g

b. Molecular mass of salt = 136 g/mol

moles of anhydrate = 0.35/136

Moles of anhydrate = 0.00257 moles

c. Moles of water driven off = mass/molar mass

molar mass of water = 18 g/mol

Moles of water driven off = 0.15/18

Moles of water driven off = 0.00833 moles

d. Moles of water within the crystalline structure of one molecule of the hydrated salt is determined by converting to whole number mole ratio by dividing with the smallest ratio,

Salt to water ratio = 0.00257 /0.00257  :  0.00833/0.00257

Salt to water ratio = 1 : 3

Therefore, there are 3 moles of water within the crystalline structure of one molecule of the hydrated salt.

e. Assuming the anhydrous salt is X, the molecular formula of the hydrated salt will be X.3H₂O

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