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Answer:

116 g

Explanation:

From the question given above, the following data were obtained:

Number of mole of calcium = 2.9 moles

Mass of calcium =.?

The mole and mass of a substance are related according to the following formula:

Mole = mass / molar mass

With the above formula, we can obtain the mass of calcium. This can be obtained as follow:

Number of mole of calcium = 2.9 moles

Molar mass of calcium = 40 g/mol

Mass of calcium =.?

Mole = mass / molar mass

2.9 = mass of calcium / 40

Cross multiply

Mass of calcium = 2.9 × 40

Mass of calcium = 116 g

Therefore, the mass of 2.9 moles of calcium is 116 g.

Answer:

The problem provides you with the solubility of potassium chloride,

KCl

, in water at

20

∘

C

, which is said to be equal to

34 g / 100 g H

2

O

.

This means that at

20

∘

C

, a saturated solution of potassium chloride will contain

34 g

of dissolved salt for every

100 g

of water.

As you know, a saturated solution is a solution that holds the maximum amount of dissolved salt. Adding more solid to a saturated solution will cause the solid to remain undissolved.

In your case, you can create a saturated solution of potassium chloride by dissolving

34 g

of salt in

100 g

of water at

20

∘

C

.

Now, your goal here is to figure out how much potassium chloride can be dissolved in

300 g

of water at this temperature. To do that, use the given solubility as a conversion factor to take you from grams of salt to grams of water

Answer:

720 mL

Explanation:

To answer this problem we'll use the definition of molarity:

With that in mind we can calculate the volume in liters:

Finally we convert liters to mL:

Thus 720 mL of 0.25 M NaOH would have 0.18 moles of NaOH.

Approximately 4.37 x 10^23 molecules of HCl would be required to form 34.52 g of MgCl₂.

To determine the number of molecules of HCl required to form 34.52 g of MgCl₂, we need to use the molar mass and stoichiometry of the balanced equation:

Mg(OH)₂(s) + 2 HCl(aq) → 2 H₂O(l) + MgCl₂(aq)

The molar mass of MgCl₂ is 95.21 g/mol.

First, we need to calculate the number of moles of MgCl₂ formed:

Moles of MgCl₂ = mass of MgCl₂ / molar mass of MgCl₂

Moles of MgCl₂ = 34.52 g / 95.21 g/mol

Moles of MgCl₂ = 0.363 mol

According to the balanced equation, the stoichiometric ratio between HCl and MgCl₂ is 2:1. Therefore, the moles of HCl required can be calculated as follows:

Moles of HCl = 2 * Moles of MgCl₂

Moles of HCl = 2 * 0.363 mol

Moles of HCl = 0.726 mol

To calculate the number of molecules, we need to use Avogadro's number, which is approximately 6.022 x 10^23 molecules/mol.

Number of molecules of HCl = Moles of HCl * Avogadro's number

Number of molecules of HCl = 0.726 mol * 6.022 x 10^23 molecules/mol

Number of molecules of HCl = 4.37 x 10^23 molecules

Therefore, approximately 4.37 x 10^23 molecules of HCl would be required to form 34.52 g of MgCl₂.

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The chemical formula of water is H2O, and the chemical formula of carbon dioxide is CO2. This means that water has 2 atoms of hydrogen and 1 atom of oxygen, which carbon dioxide has 1 atom of carbon and 2 atoms of oxygen. The only similar atom that the two molecules have in common is oxygen. CO2 has one more atom of oxygen than H2O. Hope this helps!

H2O is water. CO2 is CD.

Answer:

=303.2g

Explanation:

molar mass of Pb = 207.2g

atomic mass of S = 32g

atomic mass of O = 16g

Solution

            PbSO4

            = 207.2 + 32 + 4 (16)

              = 239.2 + 64

               = 303.2 g

The molar mass of  compound lead sulfate is  207.2 g.

Molar mass of a compound or a molecule is defined as the mass of the elements which are present in it.The molar mass is considered to be a bulk quantity not a molecular quantity. It is often an average of the of the masses at many instances.

Molar masses of an element are given as relative atomic masses while that of compounds is the sum of relative atomic masses which are present in the compound.

atomic mass of S = 32g

atomic mass of O = 16g

PbSO₄= 207.2 + 32 + 4 (16)= 239.2 + 64  = 303.2 g.

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The alkaline earth metals are beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). The elements have very similar properties because belong at same group

The answer is calcium and beryllium

The equation C2H4(g) + H2(g) + C2H6(g) → Caco3(s) + Cao(s) + CO2(g) shows an increase in entropy due to the formation of a gas as a product. Option A

In this equation, the reactants on the left-hand side consist of gases (C2H4 and H2), while the products on the right-hand side include a solid (Caco3) and a gas (CO2).

When a reaction involves a change from gaseous to solid or liquid states, there is typically a decrease in entropy because the particles become more ordered and constrained in the solid or liquid phase.

Conversely, when a reaction involves the formation of gases, there is generally an increase in entropy because gases have higher degrees of molecular motion and greater freedom of movement compared to solids or liquids.

In the given equation, the reactants include three gaseous compounds (C2H4, H2, and C2H6), and one of the products is a gas (CO2). Therefore, the overall entropy of the system increases during this reaction.

The equation Fe(s) + S(s) → FeS(s) does not show an increase in entropy. Both the reactants (Fe and S) and the product (FeS) are solids. Since solids have lower entropy compared to gases or liquids, the entropy of the system does not increase in this reaction. Option A

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Answer:

From the given equation, we can see that for every 2 moles of Al, we get 3 moles of H2

So, we can say the the number of moles of H2 is 3/2 times the number of moles of Al

We are given the number of moles of Al and we have to find the number of moles of H2

We have deduced the relationship:

Moles of Al * 3 / 2 = Moles of H2

Replacing the variables with given values

3.5 * 3 / 2 = Moles of H2

Moles of H2 = 5.25 moles

Answer:

T = 308.57 K

Explanation:

Given data:

Number of moles of K = 0.335 mol

Volume of vessel = 2.00 L

Pressure in vessel = 4.32 bar

R = 0.08314 L.bar /mol.K

Temperature of K = ?

Solution:

Formula:

PV = nRT

T = PV/nR

By putting values,

T = 4.32 bar ×2.00 L /0.335 mol× 0.08314 L.bar /mol.K

T = 8.64 bar.L / 0.028 L.bar /K

T = 308.57 K

Answer: chemistry question??

Explanation:

Answer:

Astronomers use radio waves to study a variety of celestial objects and phenomena, including stars, galaxies, quasars, pulsars, black holes, and the cosmic microwave background radiation. Radio telescopes can detect and analyze radio waves emitted by these objects, allowing astronomers to study their physical properties, such as their temperature, composition, and motion. Radio astronomy also plays an important role in the search for extraterrestrial life, as scientists use radio telescopes to search for potential signals from other civilizations in the universe.

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.

Answer:

127529 KJ

Explanation:

Since 1gallon = 3.78 L= 3780 ml

The density of C8H18= 0.703 g/mL

Density = mass/volume

Mass= Density × volume

Mass= 0.703 g/mL × 3780 ml

Mass= 2657.34 g

Molar mass of C8H18= 114 g/mol

Number of moles= mass/molar mass

Number of moles= 2657.34/114

Number of moles= 23.31 moles

Since 1 mole evolved=5471 kJ

23.31 moles = 5471 kJ × 23.31 = 127529 KJ

The reaction order and specific reaction rate constant can be determined by performing the kinetics experiment on irreversible polymerization A. Kinetic experiments can be used to investigate the rate and mechanism of chemical reactions. Chemical kinetics is the study of chemical reactions' speed and pathway.

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At equilibrium, the number of moles of \(Cl_2\) (g) will be 0.2025 mol.

1: Write the balanced chemical equation:

\(C_O\)(g) + \(Cl_2\)(g) ⟶ \(C_OCl_2\)(g)

2: Set up an ICE table to track the changes in moles of the substances involved in the reaction.

Initial:

\(C_O\)(g) = 0.3500 mol

\(Cl_2\)(g) = 0.05500 mol

\(C_OCl_2\)(g) = 0 mol

Change:

\(C_O\)(g) = -x

\(Cl_2\)(g) = -x

\(C_OCl_2\)(g) = +x

Equilibrium:

\(C_O\)(g) = 0.3500 - x mol

\(Cl_2\)(g) = 0.05500 - x mol

\(C_OCl_2\)(g) = x mol

3: Write the expression for the equilibrium constant (Kc) using the concentrations of the species involved:

Kc = [\(C_OCl_2\)(g)] / [\(C_O\)(g)] * [\(Cl_2\)(g)]

4: Substitute the given equilibrium constant (Kc) value into the expression:

1.2 x \(10^3\) = x / (0.3500 - x) * (0.05500 - x)

5: Solve the equation for x. Rearrange the equation to obtain a quadratic equation:

1.2 x \(10^3\) * (0.3500 - x) * (0.05500 - x) = x

6: Simplify and solve the quadratic equation. This can be done by multiplying out the terms, rearranging the equation to standard quadratic form, and then using the quadratic formula.

7: After solving the quadratic equation, you will find two possible values for x. However, since the number of moles cannot be negative, we discard the negative solution.

8: The positive value of x represents the number of moles of \(Cl_2\)(g) at equilibrium. Substitute the value of x into the expression for \(Cl_2\)(g):

\(Cl_2\)(g) = 0.05500 - x

9: Calculate the value of \(Cl_2\)(g) at equilibrium:

\(Cl_2\)(g) = 0.05500 - x

\(Cl_2\)(g) = 0.05500 - (positive value of x)

10: Calculate the final value of \(Cl_2\) (g) at equilibrium to get the answer.

Therefore, at equilibrium, the number of moles of \(Cl_2\) (g) will be 0.2025 mol.

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Answer:

a.) (aq)

Explanation:

(aq) means aqueous

so its in a solvent

Answer:

A

Explanation:

Atomics number = number of protons

The word "symptom" refers to a specific manifestation or indication of a condition, disease, or disorder that is experienced or observed by an individual.

Symptoms are subjective or objective changes in the body's normal functioning that may be recognized as abnormal, uncomfortable, or problematic. Symptoms can manifest in various ways depending on the nature of the underlying condition. They can be physical, such as pain, rash, cough, fever, or fatigue, indicating an illness or injury affecting the body. Symptoms can also be psychological, such as anxiety, depression, or confusion, reflecting disturbances in mental health.

Symptoms serve as important clues for medical professionals to identify and diagnose diseases or disorders. They provide valuable information about the nature, severity, and progression of an illness, helping healthcare providers formulate appropriate treatment plans. Additionally, symptoms may also be important for individuals to self-assess their own health status and seek appropriate medical attention.

It is essential to note that symptoms alone may not provide a definitive diagnosis, as they can overlap across different conditions. Further evaluation, including medical tests and examinations, is often necessary to confirm a diagnosis and determine the appropriate course of action.

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Answer:

A. Chromosome

Explanation:

Answer:A  A Chromosome

Explanation:

The excess reagent in the given reaction is Cl₂.

An excess reactant is a substance that is not wholly consumed or entirely reacted in a chemical reaction. It is also known as an excess reagent.

Let us consider the following reaction of the formation of sodium chloride.

2Na(s)+Cl₂(g) → 2NaCl(s)

Sodium metal reacts with diatomic chlorine gas. The above equation indicates that 2 moles of sodium will react with 1 mole of chlorine. If we have equivalent moles of sodium and chlorine,

Then sodium will be a limiting reactant, while chlorine will be an excess reactant.

Given,

ZrSiO₄ + 2Cl₂ → ZrCl₄ + SiO₂ + O₂

Cl₂ is the excess reagent.

Therefore, The excess reagent in the given reaction is Cl₂.

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Answer:

Au^3+(aq) +3e ------> Au(s). 1.50 V

Explanation:

When we construct the galvanic cell, our intention is to produce energy by spontaneous electrochemical reactions. In order to have a spontaneous electrochemical reaction, E°cell must be positive. The more positive the value of E°cell, the more spontaneous the reaction is.

E°cell= E°cathode - E°anode

If E°cathode= 1.50 V

E°anode= -0.25 V

E°cell= 1.50 -(-0.25)

E°cell= 1.75 V

Hence the process; Au^3+(aq) +3e ------> Au(s) yields the highest standard cell potential

As combustion always result into the formation of water, the other type of reactions that may result into formation of water are Acid-Base Neutralization Reactions and Hydrogen and Oxygen Reaction.

Acid-Base Neutralization Reactions:

A neutralisation reaction is a chemical process in which an acid and a base combine to produce salt and water as the end products.

H⁺ ions and OH⁻ ions combine to generate water during a neutralisation reaction. Acid-base neutralisation is the most common type of neutralisation reaction.

Example: Formation of Sodium Chloride (Common Salt):

HCl + NaOH → NaCl + H₂O

Hydrogen and Oxygen Reaction:

Water vapour is created when hydrogen gas (H₂) and oxygen gas (O₂) are combined directly. This reaction produces a lot of heat and releases a lot of energy.

Example: 2 H₂ + O₂ → 2 H₂O

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Answer:

A multicellular organism develops from a single cell (the zygote) into a collection of many different cell types, organized into tissues and organs. Development involves cell division, body axis formation, tissue and organ development, and cell differentiation (gaining a final cell type identity).

Explanation:

Ideal gas law is valid only for ideal gas not for vanderwaal gas. To solve such, we need to know the relation between rate of effusion and molar mass of gases. Therefore, the new volume V₂ is 18L.

Ideal gas equation is the mathematical expression that relates pressure volume and temperature. an ideal gas on the walls of the container is inversely proportional to the volume occupied that gas.

Mathematically, Boyle's law can be written as

P₁V₁ = P₂V₂

P₁=initial pressure=342 torr

P₂=final pressure=1710 torr

V₁ = initial volume= 29.0 L

V₂=final pressure=?

Substituting all the given values in the above equation, we get

V₂ = P₁V₁/P₂

V₂ = (342 × 29.0)/1710

V₂ = 18L

Therefore, the new volume V₂ is  18L.

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The total supply is not known, hence, cannot be determined.

If each pound of the crystal contains 2 crystals, then 8 pounds of crystal will contain:

8 pounds x 2 crystals/pound = 16 crystals

So, Jessie and co need to make 16 crystals before Gustavo Fring comes.

However, without knowing the amount of their current supply or the time frame available before Gustavo Fring arrives, it is impossible to determine how much of the supply that can be used to make the 16 crystals.

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Answer:

............................................................................................................................

Explanation:

because ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

Answer: Cells from the smallest to the largest of mammals often seem to be “one size fits all.” Now a closer look reveals that whether a cell lives in an elephant, mouse or something in between can make a big difference in its life

Explanation:

Considering the Hess's Law, the enthalpy change for the reaction is 221.8 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: H₂ (g) + F₂ (g) → 2 HF (g)     ∆H° = -79.2 kJ/mol

Equation 2: C (s) + 2 F₂ (g) → CF₄ (g)     ∆H° = 141.3 kJ/mol

Equation 3: 2 C(s) + 2 H₂ (g) → C₂H₄ (g)     ∆H° = -97.6 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

First, to obtain the enthalpy of the desired chemical reaction you need one mole of C₂H₄ (g) on reactant side and it is present in first equation. Since this equation has one mole of C₂H₄ (g) on the product side, it is necessary to locate it on the reactant side (invert it).

When an equation is inverted, the sign of ΔH° also changes.

Now, you need 2 moles of CF₄ (g) on the product side. The second equation has 1 mole of CF₄ (g) on the product side, so it is necessary to multiply it by 2 to obtain 2 moles of CF₄ (g).

Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.

Finally, you need 4 moles of  HF (g) on the product side. The first equation has 2 moles of  HF (g) on the product side, so it is necessary to multiply it by 2 to obtain 4 moles of the compound.

Since the equation is multiply by 2, the variation of enthalpy also is multiplied by 2.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 H₂ (g) + 2 F₂ (g) → 4 HF (g)     ∆H° = -158.4 kJ/mol

Equation 2: 2 C (s) + 4 F₂ (g) → 2 CF₄ (g)     ∆H° = 282.6 kJ/mol

Equation 3: C₂H₄ (g) → 2 C(s) + 2 H₂ (g)     ∆H° = 97.6 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g)     ΔH°= 221.8 kJ/mol

Finally, the enthalpy change for the reaction is 221.8 kJ/mol.

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