The density of this material is approximately

Answer:

The resulting solution is basic.

Explanation:

The reaction that takes place is:

First we calculate the added moles of HNO₃ and KOH:

As there are more KOH moles than HNO₃, the resulting solution is basic.

The resulting solution is basic.

• It is known that KOH is a base and HNO3 is an acid, so when they mix they undergo a neutralization reaction.  

• The reaction between there will be,  

HNO3 + KOH ⇔ KNO3 + H2O

Based on the given information,

• The volume of HNO3 is 12.5 ml and the molarity is 0.280 M, and the volume of KOH is 5 ml and the molarity is 0.920 M.

Now 1 mole of HNO3 completely reacts with 1 mole of KOH,

The millimoles of HNO3 is,

\(= Molarity * Volume (in ml)\\= 0.280 * 12.5\\= 3.5 mmol\)

The millimoles of KOH is,

\(= Molarity * Volume (in ml)\\= 0.920 * 5.0\\= 4.6 mmol\)

Now it can be seen that 3.5 millimoles of HNO3 completely reacts with 3.5 millimoles of KOH. Now we are left with 4.6-3.5 = 1.1 mmol of KOH.

Thus, KOH is in excess amount present in the solution, and as it is basic in nature, therefore, the resultant solution would be basic in nature.

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Answer:

A) water

Explanation:

When adding sugar to water, for example, the water (solvent) molecules are attracted to the sugar (solute) molecules. Once the attraction becomes large enough the water is able to pull individual sugar molecules from the bulk sugar crystals into the solution.

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Approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M.

To determine the mass of glucose (C6H12O6) required to prepare a 0.215 M solution in 5000 mL, we need to use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

First, let's convert the volume of the solution from milliliters (mL) to liters (L):

5000 mL = 5000/1000 = 5 L

Now, we can rearrange the formula to solve for moles of solute:

moles of solute = Molarity (M) x volume of solution (L)

moles of solute = 0.215 M x 5 Lmoles of solute = 1.075 mol

Since glucose (C6H12O6) has a molar mass of approximately 180.16 g/mol, we can calculate the mass of glucose using the equation:

mass of solute = moles of solute x molar mass of solute

mass of glucose = 1.075 mol x 180.16 g/mol

mass of glucose = 194.0 g (rounded to three significant figures)

Therefore, approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M. It's important to note that the molar mass of glucose used in this calculation may vary slightly depending on the level of precision required.

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Answer:

Examples of decomposition reactions include the breakdown of hydrogen peroxide to water and oxygen, and the breakdown of water to hydrogen and oxygen.

3 states

Matter can exist in three forms namely;

• Solid state

• Liquid state

• Gaseous state

From the given scenario, water is known to exist as ice cubes (which is the solid state). Also since the water Aon her skin evaporated the water changed to steam (gaseous state) in this case.

Also note that the ice cubes melted in the glass. This melted ice changes to liquid ,at this point. Therefore, we can conclude that the water in the scenario exists in three forms (Liquid, Solid and Gas).

The balanced equation for the reaction is as follows:

3 AI2O3 (s) + 3C (s) + 2CI2 (g) → 3AICI3 (s) + 3CO (g)

Combination reactions happen when two or more reactants come together to create a single product. In this instance, the reactants are chlorine gas, carbon, and aluminium oxide (AI2O3) (CI2). Aluminum chloride (AICI3) and carbon monoxide make up the final product (CO). Exothermic means that energy is released when the reactants come together in this reaction. Both heat and light are manifestations of this energy. The bonds between the atoms in the reactants are broken and new bonds between the atoms in the product are formed when the reactants combine. The reaction is exothermic because of the energy released during this process.

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Answer:

787.401575 inches

The sample contains 0.986 g of Carbon, 0.107 g of Hydrogen, 0.579 g of Nitrogen, and 0.328 g of Oxygen. This elemental composition matches that of MDMA and not caffeine. There are 0.0099 moles of MDMA present in the sample.

Nitrogen is a colorless, odorless, tasteless, and mostly inert diatomic gas at standard conditions, constituting 78% of Earth's atmosphere. It is the most abundant element in the atmosphere, and is chemically unreactive under normal conditions. It is an essential element in all forms of life, and is a major component in amino acids and proteins. Nitrogen is also found in all plant and animal tissues, and is a necessary nutrient for their growth.

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The number of moles of gas produced from the reaction of 2.00g of potassium with an excess amount of water is 0.025 moles.

The reaction of potassium with an excess amount of water is:

2K + 2H\(_2\)O \(\rightarrow\) 2KOH + H\(_2\)

To calculate the moles of hydrogen gas first we need to calculate moles of potassium in 2.00g

No. of moles = (mass) / (molecular mass)

The mass given is 2.00 g and the Molecular mass is 39.09 units

∴ No. of moles = (2) / (39.09) = 0.05

From the above reaction, we get that 2 moles of potassium give 1 mole of hydrogen gas. Thus, 0.05 moles of potassium gives 0.025 moles of hydrogen gas.

Therefore, the no. of moles of hydrogen gas produced is 0.025 moles.

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The cesium ion is the darker sphere in the unit cellThe coordination number of the cesium ion is 8. Option D.

The structure of cesium chloride adopts a primitive cubic lattice based on two atoms, with both atoms eight-coordinated. Cesium chloride is a type of unit cell, often erroneously called body-centered cubic. Cs remains in the center and is surrounded by eight Cl atoms forming a body-centered cubic structure.

Due to the central atom in the unit cell, CsCl is actually a non-close-packed structural type, although this is likely to lead to this misunderstanding. Cesium chloride forms a body-centered cubic lattice. The cesium and chloride ions are in contact along the diagonal of the unit cell. The unit cell has an edge length of 412 pm and a Cl- ion radius of 181 pm.

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A public good cannot be provided to one person without making it available to others as well.

A non-exclusive, non-rivalrous good is referred to as a public good. It cannot be given to just one person without also being made available to others. If one person uses it, it doesn't prevent others from doing so as well. Roads are an illustration of a public good.

Club goods and private goods are in opposition to public goods. An example of a public good is a club good. It is non-rivalrous but excludable. One example of a club good is paid streaming services. The service is not available to those who do not subscribe. However, each subscriber is equally valued by the service.

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The incorrect name-formula combination is FeCIO₄ -  Iron(I) perchlorate; option B.

The IUPAC nomenclature of naming compounds refers to the convention of naming compounds using the oxidation states of the elements present in the compound.

The IUPAC nomenclature of the given compounds is given below as follows:

A. Co(CIO₂)₂ - Cobalt(II) chlorite; the oxidation state of cobalt is +2

B. FeCIO₄ -  Iron(I) perchlorate; the formula of the compound is wrong because iron does not exist in the +1 state

C)Mn(CIO4)₂ -  Manganese(II) perchlorate;  the oxidation state of manganes is +2

D) Cr(CIO)₃​ - Chromium(III) hypochlorite ;  the oxidation state of chromium is +3

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Answer:c

Explanation:c

A 25.0 mL solution of Sr(OH)₂ is neutralized with 31.6 mL of 0.150 M HBr, then the concentration of the original Sr(OH)₂ solution is 0.189M.

Concentration of the solution will be calculated by using the below chemical reaction as:

M₁V₁ = M₂V₂, where

M₁ & V₁ is the molarity and volume of HBr solution and M₂ & V₂ is the molarity and volume of original Sr(OH)₂ solution.

On putting values on above equation by taking from question, we get

M₂ = (0.15)(31.6) / (25) = 0.189 M

Hence required molarity of Sr(OH)₂ solution is 0.189M.

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The answer is in the photo.

The molecular formula of fructose is C₆H₁₂O₆.

The molecular formula is derived in the following steps:

Step 1. the molality of the solution is determined first

Molality = Change in boiling point/Kb

Molality = (78.59 - 78.35)/1.2 = 0.2 molal

Step 2. Moles of solute is determined.

Moles of solute = molality * mass of solvent (kg)

moles = 0.2 * 0.325 = 0.065

Step 3. Molar mass of fructose = mass/moles

Molar mass of fructose = 11.7/0.065 = 180 g/mol

Step 4: Mole ratio of constituent elements

Carbon = 40% * 180/12 = 6

Hydrogen = 6.7% * 180/1 = 12

Oxygen = 53.3% * 180/16 = 6

Therefore, the molecular formula of fructose is C₆H₁₂O₆.

In conclusion, the molecular formula of fructose is derived from the molality of the solution.

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Answer:

(B). it's metallic bonding

Answer: The wavelength of the x-ray wave is \(10^{-10}m\)

Explanation:

To calculate the wavelength of light, we use the equation:

\(\lambda=\frac{c}{\nu}\)

where,

\(\lambda\) = wavelength of the light  = ?

c = speed of x-ray= \(3.0\times 10^8m/s\)

\(\nu\) = frequency of x-ray = \(3.0\times 10^{12}MHz=3.0\times 10^{18}Hz= 3.0\times 10^{18}s^{-1}\)     \((1Hz=1s^{-1})\)

Putting in the values:

\(\lambda=\frac{3.0\times 10^8m/s}{3.0\times 10^{18}s^{-1}}=10^{-10}m\)

Thus the wavelength of the x-ray wave is \(10^{-10}m\)

The wavelength of the x-ray should be considered as the 10^-10m.

Here the final equation should be considered

So,

wavelength = c / v

Here

The frequency of an x-ray wave is 3.0 x 1012MHz.

Its wave speed is 3.0x 108m/s

So,

= 3.0x 108m/s/ 3.0*10^18

= 10^-10m.

hence, The wavelength of the x-ray should be considered as the 10^-10m.

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Answer:

it's A. The water in the pond reflects light from the Sun, making a mirror image

Explanation:

i just took the test

Answer:

Hydrogen H 3.330% Carbon C 26.450% Oxygen O 35.234%

Explanation:

Answer:

NH4OH

Explanation:

NH4OH is a weak base because it partially dissociates into NH + and OH− ions in aqueous solutions and the amount of OH− ions produced is low.

Answer:

Evaporation is the process by which water changes from a liquid to a gas or vapor. Evaporation is the primary pathway that water moves from the liquid state back into the water cycle as atmospheric water vapor.

Answer:

The process of turning from liquid into vapour is evaporation.

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Based on the mass of 1 mole of nitrogen atoms, 7.55 x 10²³ atoms of nitrogen have a mass of 17.56 grams.

The number of atoms in 1 mole of nitrogen = 6.02 * 10²³ atoms

The mass of 1 mole of atoms of nitrogen = 14 g

Hence, 6.02 * 10²³ atoms has a mass of 14.01 g

Mass of 7.55 x 10²³ atoms = 14.01 g/mol * 7.55 x 10²³ / 6.02 * 10²³

The mass of 7.55 x 10²³ = 17.56 g

So,7.55 x 10²³ atoms of nitrogen have a mass of 17.56 grams.

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Answer:

X = ¹₀n

Explanation:

The mass number is the total amount of neutrons and protons in the element. The atomic number is the total protons in the element. The mass number is the superscript and the atomic number is the subscript. The mass and atomic numbers must be balanced on both sides of the equation. The element can be identified by the atomic number.

In the event that we have an atomic number of zero, there is no element this represents, so the letter "n" should be used.

²³₁₁Na + ¹₁H  →  ²³₁₂Mg + X

Mass Number:

Reactants = Products

23 + 1 = 23 + ?

24 = 23 + ?

1 = ?

Atomic Number:

Reactants = Products

11 + 1 = 12 + ?

12 = 12 + ?

0 = ?

The complete equation is:

²³₁₁Na + ¹₁H  →  ²³₁₂Mg + ¹₀n

Answer:

Components present of sodium bicarbonate

Sodium bicarbonate is formed by mixing carbon, sodium, hydrogen and oxygen molecules. This mixture, also known as baking soda, is actually a type of salt.

Explanation:

Answer:

will be 9.7 Liters

Explanation:

Answer:

\({ \rm{C _{7} H _{8} + 9O _{2} → 7CO _{2} + 4H _{2}O : { \tt{combustion}}}} \\ \\ {\rm{2NaHCO _{3} → Na_{2}CO _{3} + H_{2}O + CO_{2} : { \tt{decomposition}}}} \\ \\ { \rm{CH _{4} → C + 2H _{2} O :{ \tt{decomposition}} }} \\ \\ { \rm{P _{4}O _{10} + 6H _{2}O → 4H _{3} PO _{10} :{ \tt{combination}} }} \\ \\ { \rm{FeSO _{4} → FeO + SO _{3} : { \tt{decomposition}} }} \\ \\ { \rm{CaCO _{3(s)} → CO _{2(g)} + CaO _{(s)} : { \tt{decomposition}}}}\)

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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