Please help anyone !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer:

Explanation:

It seems like you’re trying to count the number of atoms in some chemical formulas. Here’s the list of the number of each atom in the formulas you provided:

Formula 1: Н - 1 Formula 2: H - 1, O - 1 Formula 3: Н - 14 Formula 4: Н - 2, C - 3 Formula 5: I - 1 Formula 6: Н - 3 Formula 7: H - 2 Formula 8: Н - 2, C - 3, O - 1 Formula 9: Li - 1 Formula 10: Н - 2 Formula 11: Н - 2 Formula 12: H - 1 Formula 13: Н - 2, C - 2, O - 1 Formula 14: II

The volume of the balloon once they have cooled to the outside temperature of -34.0 c is 1.53 L.

Charles' law predicts the relationship between the volumes and the temperatures of a sample of an ideal gas at different conditions. For this equation to hold true, the number of molecules and the pressure must remain constant despite changes in the environment.

Determine the volume of the balloon outside, V2. We do this by applying Charles' law, such that we relate the volume, V, and the temperature, T, of a sample of gas as

V₁ /T₁ = V₂/ T₂

at two conditions. We are given the following values for the variables:

• V₁ = 1.90 L

T₁ = 22.0+ 273.15 = 295.15 K

T₂= 34.0+273.15= 239.15 K

We proceed with the solution.

V₁/T₁ = V₂/T₂

V₁ /T₁ × T₂ = V₂

1.90 L/295.15 K x 239.15 K = V₂

1.53 L =V₂

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The complete question is

Balloons for a New Year's Eve party in Fargo, ND, are filled to a volume of 1.90 L at a temperature of 22.0 degrees Celsius and then hung outside where the temperature is -34.0 degrees Celsius. What is the volume of the balloons after they have cooled to the outside temperature? Assume that atmospheric pressure inside and outside the house is the same.

The equilibrium constant is written as;

Keq = [tex][D]^2 [C]^5/[A] [B]^3[/tex]

The equilibrium constant's value is influenced by the reaction's chemical make-up and temperature.

The product of the product concentrations, each raised to the power of their stoichiometric coefficient, divided by the product of the reactant concentrations, each raised to the power of their stoichiometric coefficient, is known as the equilibrium constant.

The equilibrium constant is Keq = [tex][D]^2 [C]^5/[A] [B]^3.[/tex]

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A crucial trace mineral is chromium. Trivalent chromium, which is safe for people, and hexavalent chromium, which is toxic, are the two types.

Thus, Foods and dietary supplements both contain trivalent chromium. It might assist maintain normal blood sugar levels by enhancing the body's utilization of mineral.

Chromium is used by people to treat deficiencies. Additionally, it is used to treat bipolar disorder, diabetes, high cholesterol, and a variety of other conditions, but the majority of these uses are not well-supported by science.

Chromium by mouth doesn't help control blood sugar levels in people with prediabetes. Schizophrenia. Taking chromium by mouth doesn't affect weight or mental health in people with schizophrenia.

Thus, A crucial trace mineral is chromium. Trivalent chromium, which is safe for people, and hexavalent chromium, which is toxic, are the two types.

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(NH4)3PO4 is insoluble in water. The correct option is D

According to their chemical formula and ionic charges, ionic compounds generally follow a set of solubility laws that define their solubility patterns in water. These guidelines aid in determining whether an ionic compound will dissolve in water or not as well as if it will precipitate when combined with other ionic compounds.

Therefore, (NH4)3PO4 is the compound that is expected to be insoluble in water based on the solubility rules.

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After considering the given data and performing the evaluation regarding the convertion of energy units the answer derived is 6.61 x 10⁶ J = 1577.16 kcal.

In order to alter joules (J) to kilocalories (kcal), the below conversion can be applied.
1 kcal = 4.184 kJ.
We start by, converting J to kJ by dividing by 1000:
6.61 x 10⁶ J = 6.61 x 10³ kJ
Next step we convert kJ to kcal by dividing by 4.184:
= 6.61 x 10³ kJ ÷ 4.184
= 1577.16 kcal (rounded to five significant figures)

1 joule (J) is the amount of energy needed to apply a force of 1 newton (N) over a distance of 1 meter (m).
1 kilocalorie (kcal), on the other hand, is described as the amount of energy required to increase the temperature of 1 kilogram (kg) of water by 1 degree Celsius (°C), which is equal to 4184 joules (J).
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Answer:

Explanation:

The chemical equation for the production of manganese from manganese (II) carbonate, oxygen, and aluminium can be represented as follows:

3MnCO3(s) + 3O2(g) + 4Al(s) → 3Mn(s) + 3CO2(g) + 2Al2O3(s)

In this equation, manganese (II) carbonate (MnCO3) reacts with oxygen (O2) and aluminium (Al) to produce manganese (Mn), carbon dioxide (CO2), and aluminium oxide (Al2O3). The equation is balanced with three molecules of manganese carbonate, three molecules of oxygen, and four molecules of aluminium reacting to produce three molecules of manganese, three molecules of carbon dioxide, and two molecules of aluminium oxide.

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We can use the combined gas law to solve this problem:

(P1V1/T1) = (P2V2/T2)

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

We are given that the initial pressure is P1 = 37.8 mm Hg and the initial volume is V1 = 245 mL. The initial temperature is T1 = 23.5°C, which we need to convert to Kelvin by adding 273.15:

T1 = 23.5°C + 273.15 = 296.65 K

We are also given that the final volume is V2 = 54 mL, and the final temperature is the temperature of the ice water, which is 0°C or 273.15 K.

Now we can solve for the final pressure, P2:

(P1V1/T1) = (P2V2/T2)

P2 = (P1V1T2) / (V2T1)

P2 = (37.8 mm Hg * 245 mL * 273.15 K) / (54 mL * 296.65 K)

P2 = 24.4 mm Hg

Therefore, the pressure of the gas in the smaller flask at the new temperature is 24.4 mm Hg.

Answer:

The main difference between practical work inside and outside a laboratory is that the practical work inside the lab includes good equipment and chemicals which are very advanced and the practical outside a laboratory is more about the safety of life.

Explanation:

Practicals are set up at stations with lab equipment and chemicals, where students can learn, and researchers can experiment and find different new things.

Thus, the practical work inside the lab includes lab equipment and chemicals, and the practical outside a laboratory is more about conserving nature.  

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Sure, I can explain the four types of nuclear decay and provide a model for each using Si-32 as an example.

Si-32 is a radioactive isotope of Silicon with 14 protons and 18 neutrons.

1. Alpha Decay:

In alpha decay, an unstable nucleus emits an alpha particle, which consists of two protons and two neutrons, reducing the atomic number by two and the mass number by four. This makes the resulting nucleus a different element.

Model: Si-32 → alpha particle + Mg-28

Explanation: Si-32 decays into an alpha particle (two protons and two neutrons) and becomes Mg-28.

2. Beta Decay:

In beta decay, a neutron is converted into a proton and an electron. The proton stays in the nucleus, and the electron is emitted as a beta particle. This increases the atomic number by one while keeping the mass number the same.

Model: Si-32 → beta particle + P-32

Explanation: Si-32 decays into a beta particle (an electron) and becomes P-32.

3. Gamma Decay:

Gamma decay occurs when an unstable nucleus emits high-energy photons called gamma rays. Unlike alpha and beta decay, gamma decay does not change the atomic number or mass number of the nucleus.

Model: Si-32 → Si-32 + gamma ray

Explanation: Si-32 emits a gamma ray but remains Si-32.

4. Electron Capture:

In electron capture, an unstable nucleus absorbs an electron from an inner shell, converting a proton into a neutron. This reduces the atomic number by one while keeping the mass number the same.

Model: Si-32 + electron → Al-32

Explanation: Si-32 captures an electron and becomes Al-32.

These four types of nuclear decay can occur in radioactive isotopes, and they result in a change in the atomic number and/or mass number of the nucleus.

Answer:

The enthalpy of combustion of ethanol is -1430 kJ/mol, which means that for every mole of ethanol that is burned, 1430 kJ of heat is released.

To determine the amount of heat given off from the combustion of 1 dm³ of ethanol, we need to first calculate the number of moles of ethanol in 1 dm³.

1 dm³ is equivalent to 1000 cm³. Since the density of ethanol is 0.79 g/cm³, the mass of 1 dm³ of ethanol can be calculated as:

mass = density x volume

mass = 0.79 g/cm³ x 1000 cm³

mass = 790 g

To convert this mass to moles, we need to divide by the molar mass of ethanol:

moles = mass / molar mass

moles = 790 g / 46 g/mol

moles = 17.17 mol

Therefore, 1 dm³ of ethanol contains 17.17 moles of ethanol.

To calculate the heat given off from the combustion of 1 dm³ of ethanol, we can use the following equation:

heat = enthalpy of combustion x moles of ethanol

heat = -1430 kJ/mol x 17.17 mol

heat = -24,551 kJ

Therefore, the heat given off from the combustion of 1 dm³ of ethanol is -24,551 kJ, or approximately 24,551 kJ of heat is released.

Answer:

D

Explanation:

The special property of water is that it is able to be cohesive and adhesive due to their hydrogen bonds

The molecular formula of the strychnine, given that it is composed of 75.45% carbon, 6.59% hydrogen, 8.38% nitrogen, and the balance oxygen is C₂₁H₂₂N₂O₂

First, we shall obtain the empirical formula of compound. Details below:

Divide by their molar mass

C = 75.45 / 12 = 6.2875

H = 6.59 / 1 = 6.59

N = 8.38 / 14 = 0.5986

O = 9.58 / 16 = 0.59875

Divide by the smallest

C = 6.2875 / 0.5986 = 10.5

H = 6.59 / 0.5986 = 11

N = 0.5986 / 0.5986 = 1

O = 0.59875 / 0.5986 = 1

Multiply through by 2 to express in whole number

C = 10.5 × 2 = 21

H = 11 × 2 = 22

N = 1 × 2 = 2

O = 1 × 2 = 2

Thus, we can conclude that the empirical formula is C₂₁H₂₂N₂O₂

Now, we shall determine the molecular formula of strychnine. Details below

Molecular formula = empirical × n = mass number

[C₂₁H₂₂N₂O₂]n = 140.22

[(12×21) + (1×22) + (14×2) + (16×2)]n = 334

334n = 334

Divide both sides by 334

n = 334 / 334

n = 1

Molecular formula = [C₂₁H₂₂N₂O₂]n

Molecular formula = [C₂₁H₂₂N₂O₂]1

Molecular formula = C₂₁H₂₂N₂O₂

Thus, we can conclude that the molecular formula of strychnine is C₂₁H₂₂N₂O₂

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According to the ideal gas law, a gas's pressure is inversely related to its volume and directly proportionate to its temperature. So, if a gas sample's volume is reduced, the gas sample's pressure must also increase.

As a result, in order to determine the pressure of the gas sample under the specified circumstances, we must first determine the ratio of the two volumes before multiplying the starting pressure of the sample by that ratio.

We may get the ratio of the two volumes using the ideal gas law as follows: V2/V1 = (100.0 mL/0.75 L) x (273 K/25oC) = 8.02 As a result, the gas sample's pressure at 25 oC with a volume of 100.0 mL is 8.02 times higher than the sample's original pressure.

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Answer: Moss

Explanation: It's a flowerless plant

In comparison to towns located inland, cities close to water features like lakes or oceans typically experience cooler summer temperatures.

Since water has a higher specific heat capacity than land, this is the case. The quantity of energy needed to raise a substance's temperature by a specific amount is known as its specific heat capacity. Compared to land, raising the temperature of water requires more energy because water has a higher specific heat capacity.

The summer sun warms both land and water, but due to land's lower specific heat capacity, land warms up more quickly than water. As a result, communities farther from water bodies tend to be hotter than cities closer to water.

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The net ionic equation of the reaction is as follows:

4 Al3+(aq) + 12 NO3-(aq) + 3 Ag(s) = 4 Al(s) + 12 Ag+(aq) + 12 NO3-(aq)

Ions which remain in their ground state and do not take part in the reaction are called spectator ions. The net ionic equation cancels out these ions, which are present on both the reactant and product sides of the equation.

Spectator ions, which can be found on both the reactant and product sides, but are not included in the finished reaction from the net ionic equation. The [tex]NO^3^-[/tex] ions are spectator ions in this example, thus taking them out of the equation. The net ionic equation makes up the rest of the species.

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If we label the compounds ABCD from left to right;

A - 12.10

B - 15.90

C - 12.66

D - 12.35

A molecule or compound's acidity is quantified by the pKa, which is the negative logarithm (base 10) of the acid dissociation constant (Ka). The lower the pKa value, the stronger the acid; it reflects a compound's propensity to give a proton (H+) in a solution.

The compound that has the highest number of attachment of the most electronegative elements would have the greatest pKa.

The justification of the answer above is that, seeing that the compound labelled B has three highly electronegative atoms hence it would have the most or the highest pKa of about  15.90 among the other compounds. The other compounds A, C and D have fewer electronegative atoms attached and thus a lower pKa as shown

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The sample of the unknown metal has the mass of the 135 grams. The sample cools from the 100.5 °C to the 35.5 °C, and it releases the 7500 joules of the energy. The specific heat of the sample is 0.854 J/g °C.

Th mass of the metal = 135 g

The initial temperature = 100.5 °C

The final temperature = 35.5 °C

The heat energy releases = - 7500 J

The heat energy is expressed as :

Q = mc ΔT

Where,

The m is mass of the metal = 135 g

The c is the specific heat capacity = ?

The Q is heat energy releases = - 7500 J

The ΔT is the change in the temperature = final temperature - initial temperature.

The ΔT is the change in the temperature = 35.5 - 100.5

The ΔT is the change in the temperature =  - 65 °C

The specific heat capacity, c = Q / m ΔT

The specific heat capacity, c = - 7500 / 135 × - 65

The specific heat capacity, c = 0.854 J/g °C

The specific heat capacity of metal is 0.854 J/g °C.

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The density of the gas is  0.0340 g/L, moles of gas in the bulb is 0.00124 mol and apparent molar mass is 111.3 g/mol.

To determine the density of the gas, use the ideal gas law:

PV = nRT

where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature.

Since the volume and temperature are constant:

(P/n) = constant

Therefore, the density (ρ) of the gas is given by:

ρ = (m-m₀)/V = (Δm)/V

where m = mass of the bulb filled with the gas, m₀ = mass of the evacuated bulb, and Δm = m - m₀ is the mass of the gas.

Substituting the given values:

Δm = 22.651 g - 22.513 g = 0.138 g

V = 4.050 L

ρ = 0.138 g / 4.050 L = 0.0340 g/L

To find the number of moles of gas in the bulb, use the equation:

n = PV/RT

Substituting the given values:

n = (0.0250 atm)(4.050 L) / (0.0821 L·atm/mol·K)(289.2 K) = 0.00124 mol

Finally, to find the apparent molar mass of the gas, use the equation:

M = m/n

where M = molar mass of the gas and m = mass of the gas.

Substituting the given values:

M = 0.138 g / 0.00124 mol = 111.3 g/mol

Therefore, the density of the gas is 0.0340 g/L, there are 0.00124 mol of gas in the bulb, and the apparent molar mass of the gas is 111.3 g/mol.

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the answer is 2.5 according to me

The chemical molecule CrO2 is also known as chromium(IV) oxide or chromic acid. It has the molecular formula CrO2 and is an inorganic substance.

In the solid state, CrO2 exists as a solid with a layered structure, and it is considered a cationic compound. The cation present in CrO2 is chromium(IV) ion, denoted as Cr4+.

On the other hand, the anion present in CrO2 is oxide ion, denoted as O2-. The oxidation state of oxygen in this compound is -2.

So, the cations present in CrO2 are Cr 4+ ions, and the anions present are O2 -2 ions.

In CrO2, the cation present is Chromium (Cr) with a charge of +4, and the anion present is Oxygen (O) with a charge of -2.

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All of the equation-related claims are not entirely true. The appropriate chemical formula should be:

Fe(OH)3 + 3NH4Cl = FeCl3 + 3NH4OH

Because the total mass of the reactants and products are equal, as well as the number of each type of atom in each of the reactants and products, mass is conserved in this balanced equation. Depending on the stoichiometric coefficients in the balanced equation, there may or may not be an equal amount of molecules in the reactants and products.

Iron(III) hydroxide (Fe(OH)3) and ammonium chloride (NH4Cl) are the products of the chemical reaction between iron(III) chloride (FeCl3) and ammonium hydroxide (NH4OH).

The coefficients (the numbers in front of the chemical formulae) must be changed to make sure that the number of each type of atom is the same on both sides of the equation in order to ensure that the equation is balanced. The coefficients in this instance are:

Fe(OH)3 + 3NH4Cl = FeCl3 + 3NH4OH

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The complete table for the phase changes would be as follows:

Phase changes occur when a substance changes from one phase to another. When a significant amount of energy is gained or lost, this process takes place.

Phase change also depends on elements like pressure and temperature.

There are six ways a substance can change between these three phases; melting, freezing, evaporating, condensing, sublimation, and deposition.

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We need 21 grams of potassium chloride to make a 50 mL saturated solution at 40 degrees Celsius. It's important to note that if the temperature or volume of the solution were to change, the amount of solute needed to make a saturated solution would also change, as solubility is dependent on both temperature and volume.

According to the solubility table, the solubility of potassium chloride at 40 degrees Celsius is 42 grams per 100 mL of water. This means that we can dissolve 42 grams of potassium chloride in 100 mL of water at 40 degrees Celsius to make a saturated solution.

To make a 50 mL saturated solution, we can use the following formula:
mass of solute = (volume of solution x solubility)/100
mass of solute = (50 x 42)/100
mass of solute = 21 grams
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The solubility of silver chloride (AgCl) can be represented by the following equilibrium equation:

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

In a saturated solution of AgCl, the concentration of Ag+ ions is equal to the solubility product constant (Ksp) for AgCl at that temperature. Since the concentration of Ag+ ions in the solution is given as 5.36 x 10^-6 M, we can write:

[Ag+] = 5.36 x 10^-6 M

According to the stoichiometry of the equilibrium equation, the concentration of chloride ions ([Cl-]) is also equal to the concentration of Ag+ ions, as one mole of AgCl dissociates to yield one mole of Ag+ ions and one mole of Cl- ions. Therefore:

[Cl-] = 5.36 x 10^-6 M

So, the concentration of chloride ions in the resulting solution is 5.36 x 10^-6 M.

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Answer:

When the gas is compressed, its molecules come closer and internal energy of gas is increased and the number of collisions will also increase. As the gas is compressed, the work done on it shows up as increased internal energy, which must be transferred to the surroundings to keep the temperature constant.

Statement that shows that the total energy of a system remains constant and is conserved would confirm the law of conservation of energy.

What is law of conservation?

The law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. Therefore, any statement that shows that the total energy of a system remains constant and is conserved would confirm the law of conservation of energy.

Here are some examples of statements that confirm the law of conservation of energy:

All of these statements confirm the law of conservation of energy by showing that the total energy of a system is conserved over time.

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Complete question is: "The total energy of a system remains constant and is conserved" statement would confirm the law of conservation of energy.