Wind eroding rocks. Is it a chemical or physical change ?

Final answer:

The IUPAC name for CH3–CH2–C ≡ C–CH3 is 3-hexyne.

Explanation:

The IUPAC name for CH3–CH2–C ≡ C–CH3 is 3-hexyne.

In this compound, the longest carbon chain contains six carbons, so it is called a hexyne. The triple bond is between the third and fourth carbon, so the name includes the prefix 3-

Answer : There is large empty space between the molecule in a gaseous state as compared to the solid and liquid state.

Explanation :

Solid state : In this state, the molecules are arranged in regular and repeating pattern. The molecules are closely packed that means they are fixed and vibrate in place but they can not move from one place to another.  In the solid, there is no empty space present between the molecules.

Liquid state : In this state, the molecules are present in random and irregular pattern. The molecules are closely packed but they can move from one place to another.  In the liquid, there is a some empty space present between the molecules.

Gaseous state : In this state, the molecules are present in irregular pattern. The molecules are not closely packed and they can move freely from one place to another and spread out. In the gaseous, there is large empty space present between the molecules.

Final answer:

The solution with the greatest osmotic pressure depends on the total concentration of solute particles. While a 60% sucrose solution has a high concentration, the dissociation of magnesium sulfate in a 30% solution increases its total solute particle concentration, potentially giving it a higher osmotic pressure when considering its van 't Hoff factor.

Explanation:

The question asks which of the solutions has the greatest osmotic pressure: 30% sucrose, 60% sucrose, or 30% magnesium sulfate. Osmotic pressure is directly proportional to the concentration of solute particles in a solution. While a higher percentage indicates a more concentrated solution, the key factor in determining osmotic pressure is the number of solute particles in solution, not just the concentration of the solution itself.

Sucrose is a non-electrolyte and does not dissociate in water, meaning a 30% or 60% sucrose solution will provide a straightforward concentration of molecules. Magnesium sulfate, on the other hand, is an electrolyte and will dissociate into magnesium and sulfate ions, effectively increasing the number of solute particles in the solution.

Therefore, even though the sucrose solution at 60% is more concentrated than the 30% magnesium sulfate solution, the dissociation of magnesium sulfate into ions means that the 30% magnesium sulfate solution may actually have a greater total concentration of solute particles, leading to a higher osmotic pressure. However, to accurately determine which solution has the highest osmotic pressure, one must consider the molar concentration of particles (ions for magnesium sulfate and molecules for sucrose) and the van 't Hoff factor (i), which accounts for the dissociation of ions in solution.

Answer : Inorganic Chemist.

Explanation : When Joseph Priestly discovered and described the chemical properties of oxygen, then he would be considered as inorganic chemist.

As the physical and chemical properties of elements are studied under the branch of inorganic chemistry, therefore he would be considered as an inorganic chemist.

Final answer:

The probability of four 13C isotopes being adjacent to each other in a benzene molecule, given their natural molar abundance of roughly 1%, is extremely low at 1x10⁻⁸, due to the need for this specific arrangement to occur consecutively.

Explanation:

The question asks about the probability of four 13C isotopes being adjacent to each other in a benzene molecule. Given that the natural molar abundance of 13C is roughly 1%, the probability of any given carbon atom in benzene being 13C is 0.01. Since benzene has a ring structure with 6 carbon atoms, the probability of any specific sequence of four carbon atoms being all 13C can be calculated using the multiplication rule of probability.

To find the probability of four 13C isotopes in a row, we raise the single-event probability (0.01) to the fourth power because we need this event to happen four times consecutively for four specific carbon atoms:

Probability = 0.01⁴ = 0.00000001 or 1x10⁻⁸

This calculation shows that the probability is extremely low, reflecting the rare occurrence of such an event due to the low natural abundance of 13C.

In the table, the highlighted reactants NH3 and Ag2CO3 are Brønsted-Lowry bases, whereas the reactant in the electrolysis of a solution of strontium chloride reaction cannot be determined without more information.

In the table, we have the following reactions:

a) NH3 + HC104
b) Ag2CO3 + HNO3
c) electrolysis of a solution of strontium chloride

From the information given, we can determine the type of reactant for each reaction.

a) NH3 is a weak base and HC104 is a strong acid, so NH3 is a Brønsted-Lowry base.
b) Ag2CO3 is a weak base and HNO3 is a strong acid, so Ag2CO3 is a Brønsted-Lowry base.
c) In this reaction, the type of reactant is not specified, so we cannot determine if it is a Brønsted-Lowry acid or base.

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The balanced molecular equation for the neutralization reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH) is H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l). This showcases a typical neutralization reaction, where an acid reacts with a base to produce salt and water.

The neutralization reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH) results in the formation of water (H2O) and potassium sulfate (K2SO4). The reaction's balanced molecular equation is:

H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)

This equation is a clear representation of a neutralization reaction where an acid reacts with a base to produce salt and water. In this specific instance, the acid is H2SO4, the base is KOH, the salt is K2SO4, and, of course, the water is H2O.

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The balanced molecular equation for the neutralization reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH) is H2SO4 (aq) + 2KOH(aq) → K2SO4 (aq) + 2H2O(l). This reaction is a classic example of a neutralization reaction, where an acid reacts with a base to form salt and water.

The neutralization reaction between H2SO4 (sulfuric acid) and KOH (potassium hydroxide) in aqueous solution can be represented as follows:

H2SO4 (aq) + 2KOH(aq) → K2SO4 (aq) + 2H2O(l)

This resulting equation is balanced, meaning the number of atoms of each element is the same on both sides of the equation. In this case, a base (KOH) and an acid (H2SO4) react to form a salt (K2SO4) and water, which is characteristic of a neutralization reaction.

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boiling point is the temp at which a liquid's vapor pressure equals the external pressure. At higher altitudes the external pressure is lower so vapor pressure becomes equal to external pressure at a lower temperature which is why  boiling point decreases with increasing altitude.

 I hope this helps you

The process of Na→Na++ represents a sodium atom (Na) losing an electron to become a positively charged sodium cation (Na+). This can be a part of a reaction such as 2NaCl(aq) + 2H2O(l) -> 2NaOH(aq) + H2(g) + Cl2(g). The net ionic equation for this reaction is 2H2O(l) -> H2(g) + Cl2(g).

The chemical formula Na→Na++ indicates a sodium atom losing an electron to become a sodium cation with a positive charge. This is a part of a redox reaction that might occur in a chemical formula such as when sodium chloride (NaCl) reacts with water (H2O) to produce sodium hydroxide (NaOH), hydrogen gas (H2), and chlorine gas (Cl2).

The balanced molecular equation would be: 2NaCl(aq) + 2H2O(l) -> 2NaOH(aq) + H2(g) + Cl2(g)

The complete ionic equation would be: 2Na+(aq) + 2Cl−(aq) + 2H2O(l) -> 2Na+(aq) + 2OH−(aq) + H2(g) + Cl2(g)

The net ionic equation, which only focuses on the species that actually change and react, would thus be: 2H2O(l) -> H2(g) + Cl2(g)

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An 18 karat gold bracelet weighing 0.333 ounces contains approximately 2.16 × 10^22 atoms of gold. This calculation is based on converting the weight to grams, accounting for the 75% gold content of the bracelet, and using Avogadro's number.

To calculate how many gold atoms are in an 0.333 ounce, 18 k (karat) gold bracelet, we must first convert the mass of the bracelet into grams since the mass of gold is typically measured in grams. We know that 18 karat gold is comprised of 75% gold by mass. Once the mass of pure gold is identified, we can use Avogadro's number to determine the amount of atoms in that mass.

First, 0.333 ounces is approximately 9.43 grams (1 ounce = 28.3495 grams). Since the bracelet is 18k gold or 75% gold, the mass of pure gold would be 75% of 9.43 grams, which equals approximately 7.0725 grams of gold.

We know from the periodic table that the atomic mass of gold (Au) is approximately 197 g/mol, which means that 1 mole of gold has a mass of about 197 grams. Using this, we can determine the number of moles of gold in our bracelet:

7.0725 grams Au × (1 mole Au / 197 grams Au) = 0.0359 moles Au

Avogadro's number tells us that 1 mole of any substance contains approximately 6.022 × 1023 atoms. So:

0.0359 moles Au × (6.022 × 1023 atoms/mole) = approximately 2.16 × 1022 atoms of gold.

Therefore, an 18 karat gold bracelet weighing 0.333 ounces, which corresponds to 75% gold by mass, contains approximately 2.16 × 1022 atoms of gold.

Final answer:

The volume of liquid in the graduated cylinder is estimated by reading where the bottom of the meniscus falls on the scale, with typical measurements to the nearest 0.1 mL. In the example provided, estimating the volume as 21.6 mL is reasonable based on the description of the meniscus position.

Explanation:

To measure the volume of a liquid in a graduated cylinder, one must carefully read the measurement at the bottom of the meniscus, which is the curve formed by the liquid surface. For accurate measurement, the cylinder should be at eye level, and the scale should be mentally divided based on the smallest marked division, which typically allows considering smaller subdivisions, usually to one-tenth of a milliliter (0.1 mL).

In this scenario, if the meniscus lies between the 21 and 22 mL marks, closer to the 22-mL mark, an estimate of 21.6 mL would be reasonable. The digits 21 are certain, being the last clear markings, and the '.6' is an estimate that is influenced by how the meniscus appears relative to the 21 and 22 mL markings.

Measuring in Graduated Cylinders

It is essential to understand that different people might provide a slightly different estimation, as some might perceive the meniscus to be equally distant between two markings (estimating the tenth place as 5), while others might see it closer to the following mark (estimating as 7). Since the tenth-place digit is already an estimate, going any finer such as hundredths place would be pointless.

Answer:

-6.5

Explanation:

The mass of [tex]\( 2.1 \times 10^{24} \)[/tex] atoms of tungsten (W) is [tex]641.03 \text{ g}} \)[/tex].

To calculate the mass of [tex]\( 2.1 \times 10^{24} \)[/tex] atoms of tungsten (W), we'll follow these steps:

1. Find the molar mass of tungsten (W):

The molar mass of tungsten (W) is approximately [tex]\( 183.84 \)[/tex] g/mol.

2. Convert atoms to moles:

Use Avogadro's number to convert the number of atoms to moles:

[tex]\[ \text{Number of moles} = \frac{\text{Number of atoms}}{\text{Avogadro's number}} \][/tex]

Avogadro's number is [tex]\( 6.022 \times 10^{23} \)[/tex] atoms/mol.

[tex]\[ \text{Number of moles} = \frac{2.1 \times 10^{24}}{6.022 \times 10^{23}} \]\[ \text{Number of moles} = 3.487 \text{ moles} \][/tex]

3. Calculate the mass:

Now, multiply the number of moles by the molar mass to find the mass in grams:

[tex]\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]\[ \text{Mass} = 3.487 \text{ moles} \times 183.84 \text{ g/mol} \]\[ \text{Mass} = 641.03 \text{ g} \][/tex]

Therefore, the mass of [tex]\( 2.1 \times 10^{24} \)[/tex] atoms of tungsten (W) is [tex]641.03 \text{ g}} \)[/tex].

A hot saturated solution, when filtered using a Büchner funnel, may result in the precipitation of the solute as the solution cools down. This is due to the decreased solubility at cooler temperatures.

If a hot saturated solution was filtered using vacuum filtration with a Büchner funnel, some of the solutes could precipitate out of the solution as it cools down. This is because the solubility of most solutes decreases as the temperature decreases. Thus, as the hot solution comes in contact with the cooler Büchner funnel and cools down, the solute's ability to remain dissolved lessens, leading to the precipitation of the solute. The precipitate could end up on the filter in the Büchner funnel, potentially clogging it and affecting the filtration process.

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Vacuum filtration with a Büchner funnel for a hot saturated solution will cause rapid cooling, leading to crystal formation on the filter paper or in the stem, thus clogging the setup.

A crystal formation occurs on the filter paper or in the stem and as a result the setup is clogged.

To avoid this, ensure the funnel is pre-warmed and maintain a hot temperature during the filtration process.

The maximum amount of O2 that can be obtained from 227 g of nitroglycerin is 907 grams. The percent yield of O2 in this reaction is 0.72%.

Nitroglycerin decomposes according to the following reaction:

2 C3H5N3O9 → 3 N2 + 5 H2O + 7 CO + 7 C

To determine the maximum amount of O2 that can be obtained from a given amount of nitroglycerin, we need to calculate the molar mass of nitroglycerin and find the stoichiometric ratio between O2 and nitroglycerin. The molar mass of nitroglycerin is approximately 227 g/mol. From the balanced equation, we can see that 2 moles of nitroglycerin produce 7 moles of O2. Therefore, if we have 227 grams of nitroglycerin, we can calculate the amount of O2 produced as follows:

(227 g nitroglycerin) * (7 moles O2 / 2 moles nitroglycerin) * (32 g O2 / 1 mole O2) = 907 g O2

So, the maximum amount of O2 that can be obtained from 227 g of nitroglycerin is 907 grams.

To calculate the percent yield, we compare the actual yield (6.55 g) to the theoretical yield (907 g) and calculate the percentage:

Percent Yield = (Actual Yield / Theoretical Yield) * 100%

Percent Yield = (6.55 g / 907 g) * 100% = 0.72%

The number of moles of calcium ions present in a 50.00 mL water sample with hardness of 75.0 ppmis [tex]\boxed{0.00003747{\text{ mol}}}[/tex].

Further Explanation:

The hardness of water is calculated with the use of following formula:

[tex]{\text{Hardness}} = \dfrac{{{\text{Mass}}}}{{{\text{Volume}}}}[/tex]                                                                      …… (1)

Since hardness in given problem occurs due to presence of [tex]{\text{CaC}}{{\text{O}}_{\text{3}}}[/tex], the above formula involves mass of [tex]{\text{CaC}}{{\text{O}}_{\text{3}}}[/tex] and volume of [tex]{\text{CaC}}{{\text{O}}_{\text{3}}}[/tex].

Rearrange equation (1) to calculate mass.

[tex]{\text{Mass}} = \left( {{\text{Hardness}}} \right)\left( {{\text{Volume}}} \right)[/tex]                                                       …… (2)

In case of water, 1 ppm is approximately equivalent to 1 mg/L. Therefore hardness becomes 75.0 mg/L.

Volume is given in mL. It is to be converted into L. The conversion factor for this is,

[tex]1{\text{ mL}} = {10^{ - 3}}{\text{ L}}[/tex]  

Therefore volume of water sample can be calculated as follows:

[tex]\begin{aligned}{\text{Volume of water sample}} &= \left( {50.00{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\&= 0.05{\text{ L}} \\\end{aligned}[/tex]  

Substitute 75.0 mg/L for hardness and 0.05 L for volume in equation (2) to calculate mass of [tex]{\text{CaC}}{{\text{O}}_3}[/tex].

[tex]\begin{aligned}{\text{Mass of CaC}}{{\text{O}}_{\text{3}}} &= \left( {75.0{\text{ mg/L}}} \right)\left( {0.05{\text{ L}}} \right) \\&= 3.75{\text{ mg}} \\\end{aligned}[/tex]  

Mass of [tex]{\text{CaC}}{{\text{O}}_3}[/tex] is to be converted from mg to g. The conversion factor for this is,

[tex]1{\text{ mg}} = {10^{ - 3}}{\text{ g}}[/tex]  

Therefore mass of [tex]{\text{CaC}}{{\text{O}}_3}[/tex] can be calculated as follows:

 [tex]\begin{aligned}{\text{Mass of CaC}}{{\text{O}}_{\text{3}}} &= \left( {3.75{\text{ mg}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ g}}}}{{1{\text{ mg}}}}} \right) \\&= 0.00375{\text{ g}} \\\end{aligned}[/tex]

The formula to calculate moles of [tex]{\text{CaC}}{{\text{O}}_3}[/tex] is as follows:

[tex]{\text{Moles of CaC}}{{\text{O}}_{\text{3}}} = \dfrac{{{\text{Mass of CaC}}{{\text{O}}_{\text{3}}}}}{{{\text{Molar mass of CaC}}{{\text{O}}_{\text{3}}}}}[/tex]                                                    …… (3)

Substitute 0.00375 g for mass of [tex]{\text{CaC}}{{\text{O}}_{\text{3}}}[/tex] and 100.08 g/mol for molar mass of    [tex]{\text{CaC}}{{\text{O}}_{\text{3}}}[/tex] in equation (3).

[tex]\begin{aligned}{\text{Moles of CaC}}{{\text{O}}_{\text{3}}} &= \frac{{{\text{0}}{\text{.00375 g}}}}{{{\text{100}}{\text{.08 g/mol}}}} \\&= 0.00003747{\text{ mol}} \\\end{aligned}[/tex]  

Therefore number of calcium ions present in the given water sample is 0.00003747 mol.

Learn more:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: CaCO3, mass, volume, hardness, molar mass, moles, mass of CaCO3, 3.75 mg, 0.00375 g, 75.0 ppm, 50.00 mL, 0.00003747 mol