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KH2PO4 is a more suitable choice as a buffer because it has a greater buffering capacity due to the presence of the weak acid and its conjugate base.

KHPo4 is not considered an effective buffer compared to KH2PO4 due to its limited buffering capacity. The effectiveness of a buffer is determined by the concentration and dissociation properties of its conjugate acid-base pair.

KH2PO4 is a salt composed of the weak acid H2PO4- and its conjugate base HPO4^2-. In an aqueous solution, KH2PO4 can dissociate to release H+ ions from the H2PO4- component, which acts as a weak acid, and the HPO4^2- component can accept H+ ions, acting as a weak base. This allows KH2PO4 to effectively resist changes in pH when small amounts of acid or base are added to the solution.

On the other hand, KHPo4 consists of the strong acid H3PO4 and the weak base HPO4^2-. H3PO4 fully dissociates in water, providing a large concentration of H+ ions, making it difficult for the HPO4^2- to effectively act as a base and maintain pH stability.

Therefore, KH2PO4 is a more suitable choice as a buffer because it has a greater buffering capacity due to the presence of the weak acid and its conjugate base.

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Answer:Your circulatory system delivers oxygen-rich blood to your bones. Meanwhile, your bones are busy making new blood cells. Working together, these systems maintain internal stability and balance, otherwise known as homeostasis

Explanation:

The two functional groups that the aminoacid has according to the picture are amine and carboxyl.

In chemistry and related areas, a functional group can be defined as a group of atoms bonded in a specific molecule that can affect the was the molecule reacts or the specific behavior of it.

In the case of the molecule presented, which is an amino acid, two functional groups can be identified:

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Answer:

yes

Explanation:

if u watch or look closely u will observe it

The CH₃ has resonance structures as it has odd number of hydrogen atoms.

In chemistry, resonance, also called mesomerism, is a way of describing bonding in certain molecules or polyatomic ions by the combination of several contributing structures ,also variously known as resonance structures or canonical structures into a resonance hybrid (or hybrid structure) in valence bond theory. It has particular value for analyzing delocalized electrons where the bonding cannot be expressed by one single Lewis structure.

Thus,  CH₃ has resonance structures as it has odd number of hydrogen atoms.

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They appear as the trait over a recessive allele. Statement C) is true about the dominant alleles.

Dominant alleles are genetic variants that, when present in an individual's genotype, are expressed phenotypically, meaning they determine the visible or observable traits. Dominant alleles are represented by capital letters, while recessive alleles are represented by lowercase letters in genetics.

In terms of inheritance, if an individual has at least one copy of the dominant allele, it will be expressed in the phenotype, regardless of the presence of a recessive allele. This is because dominant alleles exert their influence over recessive alleles, thus "dominating" their expression.

To illustrate this, let's consider a specific example using a trait controlled by a single gene with two possible alleles: dominant (A) and recessive (a). If an individual is homozygous dominant (AA), meaning they possess two copies of the dominant allele, the dominant trait will be expressed.

However, if an individual is homozygous recessive (aa), with two copies of the recessive allele, the recessive trait will be expressed since there are no dominant alleles to override it.

Therefore, dominant alleles appear as the trait over recessive alleles, regardless of the presence or absence of a recessive allele. The presence of even a single copy of the dominant allele is sufficient for its expression in the phenotype. Option C

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The enthalpy change for the given reaction is 1344 kJ/mol.

First, we need to write the balanced chemical equation for the given reaction:

HCN(g) + 2 H2(g) → CH3NH2(g)

To calculate the enthalpy change (ΔH) for this reaction, we need to subtract the total energy of the reactants from the total energy of the products. We can do this by calculating the energy required to break the bonds in the reactants and the energy released when new bonds are formed in the products.

Reactants:

HCN(g): 1 C-N bond (305 kJ/mol) + 1 C-H bond (413 kJ/mol) + 1 N-H bond (391 kJ/mol) = 1109 kJ/mol

2 H2(g): 4 H-H bonds (4 x 432 kJ/mol) = 1728 kJ/mol

Total energy of reactants: 1109 kJ/mol + 1728 kJ/mol = 2837 kJ/mol

Products:

CH3NH2(g): 1 C-H bond (413 kJ/mol) + 3 C-N bonds (3 x 305 kJ/mol) + 7 H-H bonds (7 x 432 kJ/mol) = 4181 kJ/mol

Total energy of products: 4181 kJ/mol

ΔH = (total energy of products) - (total energy of reactants)

ΔH = 4181 kJ/mol - 2837 kJ/mol

ΔH = 1344 kJ/mol

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Copper oxide, CuO has a relative formula mass of 80.

What is the Relative Formula mass?

This refers to the total of the relative atomic masses of the atoms in the numbers seen in the formula.

This refers to the relative mass of atoms when compared with the carbon-12 atoms Ar. Relative formula mass is a measure of relative mass, thus it has no unit.

O=16 and Cu=64

copper oxide, CuO=  (1 * 64) + (1 * 16)

                              = 64      +  16

                               =80

Where,

1 = The subscript value of the element present.

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The volume of a 0.781 M KOH solution required to make 3.73 L of a solution is 6.012 × 10-¹²L.

The volume of a solution can be calculated using the following formula:

CaVa = CbVb

Where;

According to this question, a 0.781 M KOH solution is required to make 3.73 L of a solution with pH of 11.90.

The concentration of a solution with a pH of 11.90 is 0.000000000001259 or 1.259 × 10-¹²M.

0.781 × Va = 1.259 × 10-¹² × 3.73

0.781Va = 4.696 × 10-¹²

Va = 6.012 × 10-¹²L

Therefore, 6.012 × 10-¹²L is the volume of the solution.

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Answer:

13.22 L

Explanation:

Given data:

Volume of ammonia = ?

Mass of ammonia = 10.0 g

Temperature = standard = 273.15 K

Pressure = standard = standard = 1 atm

Solution:

First of all we will calculate the number of moles of ammonia.

Number of moles = mass/molar mass

Number of moles = 10.0 g/ 17 g/mol

Number of moles = 0.59 mol

one mole of any substance at standard temperature and pressure occupy 22.41 L . Thus, 0.59 moles of ammonia will occupy,

0.59 mol × 22.4 L / 1 mol

13.22 L

Answer: 12.4L

Explanation:

mw NH3 18

1 mole = 22.4L

10g = 10/18 moles

Explanation:

If we compare its solubility products without any calculation then, Magnesium hydroxide is more soluble than compound A and C. Magnesium hydroxide is less soluble than compound D.

- The solubility product of magnesium hydroxide and zinc carbonate is same so it is not possible to determine whether it is more or less soluble than compound B

Based on the given data, the relationship between rate and concentration is exponential.

A proposed rate law for the reaction based on the given data is:

Collision theory suggests that the rate of a chemical reaction is proportional to the frequency and energy of collisions between the reactant molecules.

As the concentration of iodate ions decreases, the frequency of collisions between reactant molecules decreases, which leads to a decrease in the rate of the reaction.

At higher temperatures, the kinetic energy of the reactant molecules increases, which increases the frequency and energy of collisions between reactant molecules.

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Answer:

___________

Poem with Compound Words

------------------------------------------

whenever there's a full moon,

I cannot overlook,

some alterations in my ways,

and changes in my look.

my werewolf hair grows everywhere,

my werewolf teeth get long,

my eyesight gets much keener and

I'm muscular and strong.

I get to roam around outside,

the moonlight makes me howl,

these otherworldly sound effects

mean I am on the prawl.

I see the moon is round and full,

I'm moonstruck by the sight,

I've made some telltale changes-so, you'd best stay in tonight.

Explanation:

I HOPE IT HELPS YOUR ENGLISH SUBJECT ;) ★

When dissolved in water, most Group 1 metal salts can be described as strong electrolytes.

When Group 1 metal salts are dissolved in water, they can be described as strong electrolytes. This is because Group 1 metals, such as lithium (Li), sodium (Na), potassium (K), and so on, readily lose their outermost valence electron to form positive ions (cations). These cations then dissociate completely in water, separating from the anions to which they were originally bonded.

The dissociation of Group 1 metal salts in water results in the formation of positively charged metal ions and negatively charged non-metal ions (anions). These ions are free to move and conduct electric current, making the solution a good conductor of electricity. The complete dissociation of Group 1 metal salts in water and the presence of freely moving ions make them strong electrolytes.

Strong electrolytes are substances that ionize completely or almost completely in solution, producing a high concentration of ions. This is in contrast to weak electrolytes, which only partially ionize and produce a lower concentration of ions.

In summary, when Group 1 metal salts are dissolved in water, they form strong electrolytes due to their ability to dissociate completely into ions, leading to a high concentration of freely moving ions in the solution, thus enabling efficient electrical conductivity.

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Answer:

what

Explanation:

Answer:

Still 30°−60°−90°

Explanation:

Angles do not change if rotated.

Angles do not change they can only rotate in a circular motion if the angles in a 30°−60°−90° triangle after it is rotated clockwise at 45°.

The angles are the distance between two lines that are attached at one point and they can vary in shapes like triangle and square or circle and rectangle.

The square has the handle of equality for all the sides rectangle has two opposite side angles equal and the circle It has only one angle triangle consisting of 3 angles which are fixed and cannot be changed.

Therefore, if the angles in a 30°−60°−90° triangle after it is rotated clockwise at 45°Angles do not change they can only rotate in a circular motion.

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Answer:

C15 H31 O4 S

Explanation:

molecular formula is also the same because the value of "n" is 1

The molality of the sugar solution is 0.034 mol/kg.

The molality and molarity of a chemical solution are equivalent for aqueous solutions of covalent molecules, such as sugar.

the empirical formula of the compound  that contains 1. 04 g k, 0. 70 g cr, and 0. 86 g o. is K0.5Cr1O2.

The empirical formula of a compound is the simplest whole-number ratio of atoms of each element in the compound. To find the empirical formula of a compound, you need to determine the number of moles of each element present in the sample.

First, you need to convert the masses of each element to moles:

For potassium (K), 1.04 g / 39.10 g/mol = 0.0264 moles

For chromium (Cr), 0.70 g / 51.99 g/mol = 0.0134 moles

For oxygen (O), 0.86 g / 16.00 g/mol = 0.0538 moles

Next, divide each number of moles by the smallest number of moles to get the simplest whole-number ratio:For potassium (K), 0.0264 moles / 0.0264 moles = 1

For chromium (Cr), 0.0134 moles / 0.0264 moles = 0.5

For oxygen (O), 0.0538 moles / 0.0264 moles = 2.0

Therefore, the empirical formula of the compound is K0.5Cr1O2. This represents the simplest whole-number ratio of atoms of each element in the compound.

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Answer:

0.491 mol NaCl

Explanation:

Convert from grams to moles using the molar mass of NaCl:

28.7 g NaCl x 1 mol NaCl/58.44 g NaCl = 0.491 mol NaCl

The molar mass (58.44 g) can be found from adding the molar masses of the individual elements (Na and Cl); the g NaCl will cancel out (dimensional analysis), leaving you with mol NaCl.

Answer:

hola como esats UwU

Explanation:

yo bien y tu? UwU

Answer:

B. the number of protons added to the number of neutrons in the nucleus.

Explanation:

Sana makatulong

Answer:

1. Some combination of ions form a solid precipitate because it is not favorable for the ions to become solvated (dissolved). Large and lowly charged ions tend to form precipitates, especially metals such as lead, barium, and silver.

2.
a. Pb(NO3)2 + 2KI -> 2KNO3 + PbI2

b. PbI2 is a precipitate because no other combinations of cations and anions will make an insoluble compound. KI, KNO3, and Pb(NO3)2 are all soluble.

c.

\(Pb^{2+}(aq) + 2NO_3^{-}(aq) + 2K^+(aq) + 2I^-{aq} = > PbI_2(s) + 2NO_3^{-}(aq) + 2K^+{(aq)}\\\\\)

is the ionic equation. Spectator ions are NO3- and K+

d.

\(Pb^{2+}(aq)+ 2I^-{aq} = > PbI_2(s) \\\)  is the net ionic equation

ask questions in comments if you have any

Answer:

the answer is A

I made a chart for AP chem if you want to refer to it.

The reaction that has a positive change in entropy

spontaneous will be " At all temperatures for exothermic reactions".

A spontaneous reaction is one where the production of products is favored under the conditions of the reaction.

A reaction wherein energy will be emitted in the term of light as well as heat has been known as that of an exothermic reaction.

The free energy change will always be negative and even the reaction has always been spontaneous if such reaction was exothermic (H equals negative) as well as the entropy S was positive.

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The ∆H of solution of NaOH is 46.8 kJ/mol.

First, we need to calculate the amount of heat absorbed by the solution:

q = m × c × ∆T

where q is the heat absorbed (in Joules), m is the mass of the solution (in grams), c is the specific heat capacity of the solution (in J/(g °C)), and ∆T is the change in temperature (in °C).

In this case, the mass of the solution is the sum of the mass of NaOH and the mass of water:

m = 7.59 g + 80.0 g = 87.59 g

The change in temperature is:

∆T = 48.0 °C - 25.0 °C = 23.0 °C

Substituting the values, we get:

q = 87.59 g × 4.184 J/(g °C) × 23.0 °C = 8,878 J

Next, we need to convert the heat absorbed into the enthalpy change of solution (∆H). The enthalpy change of solution is the heat absorbed per mole of solute. The number of moles of NaOH is:

n = m/M

where M is the molar mass of NaOH, which is 40.00 g/mol.

n = 7.59 g / 40.00 g/mol = 0.1898 mol

Therefore, the enthalpy change of solution is:

∆H = q/n = 8,878 J / 0.1898 mol = 46,780 J/mol = 46.78 kJ/mol

The H of a NaOH solution, rounded to three significant numbers, is 46.8 kJ/mol.

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According to the given statement one mole of potassium has a mass of 39.1 g.

Each element's atomic weight is listed in a periodic table along with how many grams (or an Avogadro number) are needed to make one mole of atoms. Since potassium has an atomic weight of 39.0983 g/mole, a mole of potassium weighs 39.1 g.

Avogadro's number, which is equal to 6.02214076 * 10²³, is the quantity of units in one mole of any material (defined as its molecular weight in grams). Depending on the substance and the nature of the reaction, the units could be electrons, atoms, ions, or molecules.

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The expected lattice energies, in increasing order, are MgO> MgBr2> KBr.  

The ranking is related that the lattice energy being directly proportional to the strength of the ionic bond. If the bond is strong, the lattice energy would be high.

If the intermolecular force is strong, then the energy requires to break the bond would be greater. Thus, the lattice energy would be high.

MgO has the lowest lattice energy because it is the smallest and has the weakest attractive force between the ions. MgBr2 has a slightly higher lattice energy because the bromine ions are larger and have a stronger attractive force than the oxygen ions. KBr has the highest lattice energy because the potassium ions are larger than the bromine ions and have a stronger attractive force.

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The enthalpy of combustion for 1kg of urea is -1223525.84 J/mol.

Urea is a compound that is used in fertilizers and in some plastics.The enthalpy of combustion for urea is the amount of energy that is released when urea is burned. In order to calculate the enthalpy of combustion for 1kg of urea, we need to use the information that is provided to us in the question. Let us start by writing down the balanced equation for the combustion of urea: CO(NH2)2 + 3/2 O2 → CO2 + 2H2O + N2
The balanced equation shows that 1 mole of urea reacts with 1.5 moles of oxygen gas to produce 1 mole of carbon dioxide, 2 moles of water, and 1 mole of nitrogen gas. The enthalpy change for this reaction is equal to the amount of energy that is released when 1 mole of urea is burned.
The heat of combustion (ΔHc) of urea is -632.6 kJ/mol. This means that 632.6 kJ of energy is released when 1 mole of urea is burned. We know that 12g of urea raised the temperature of water by 30°C. We can use this information to calculate the amount of energy that was released when 12g of urea was burned.
The specific heat capacity of water is 4.18 J/g°C. This means that it takes 4.18 J of energy to raise the temperature of 1 gram of water by 1°C. Therefore, it takes 4.18 x 1000 = 4180 J of energy to raise the temperature of 1 kg of water by 1°C.
We know that 12g of urea raised the temperature of water by 30°C. Therefore, the amount of energy that was released when 12g of urea was burned is:
Energy = mass x specific heat capacity x temperature change
Energy = 0.012 kg x 4180 J/kg°C x 30°C
Energy = 1497.6 J
We can now use this information to calculate the enthalpy of combustion for 1kg of urea:
Enthalpy of combustion = energy released / moles of urea burned
Enthalpy of combustion = 1497.6 J / (0.012 kg / 60.06 g/mol)
Enthalpy of combustion = - 1223525.84 J/mol
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The necessary volume of H₂ gas to obtain 5.0 g of propane (C₃H₈) is approximately 12.60 liters.

To determine the necessary volume of H₂ gas to obtain 5.0 g of propane (C₃H₈), we need to use the molar ratio between H₂ and C₃H₈, as well as the density of H₂ gas.

First, let's calculate the molar mass of propane (C₃H₈):

C: 12.01 g/mol

H: 1.01 g/mol

Molar mass of C₃H₈ = (3 * C) + (8 * H) = (3 * 12.01) + (8 * 1.01) = 44.11 g/mol

Next, we can determine the number of moles of propane (C₃H₈) using its mass:

Number of moles = Mass / Molar mass

Number of moles of C₃H₈ = 5.0 g / 44.11 g/mol ≈ 0.1134 mol

Now, we can establish the molar ratio between H₂ and C₃H₈ from the balanced chemical equation:

C₃H₈ + 5H₂ → 3CH₄

According to the balanced equation, 5 moles of H₂ are required to produce 1 mole of C₃H₈.

Therefore, the number of moles of H₂ required can be calculated as:

Number of moles of H₂ = 5 * Number of moles of C₃H₈ = 5 * 0.1134 mol = 0.567 mol

Finally, we can determine the necessary volume of H₂ gas using the ideal gas law equation:

Volume = (Number of moles * Gas constant * Temperature) / Pressure

Given that the density of H₂ is 0.09 g/L, we can convert it to moles per liter:

Density = Mass / Volume

0.09 g/L = 2 g/mol / Volume (since the molar mass of H₂ is 2 g/mol)

Solving for Volume:

Volume = 2 g/mol / 0.09 g/L ≈ 22.22 L/mol

Now, we can calculate the necessary volume of H₂ gas:

Volume of H₂ = Number of moles of H₂ * Volume per mole

Volume of H₂ = 0.567 mol * 22.22 L/mol ≈ 12.60 L

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