Ionic bonds, such as in sodium chloride, __________.

The element that would be a positive ion in a compound between Sr and Te would be Sr.

Elements form positive or negative ions based on the number of valence electrons their atoms contain and the tendency to either donate these electrons to form bonds or accept more electrons to form bonds.

The donation of acceptance of electrons will enable atoms to form an octet structure.

The number of valence electrons in Sr atoms is 2. In order to form an octet when bonding, these 2 electrons are readily donated rather than accepting more electrons. Thus, Sr ions are usually [tex]Sr^{2+[/tex].

The number of valence electrons in Te atoms is 6. Hence, instead of donating these electrons to form an octet, it will rather accept 2 more electrons. Thus, Te ions are usually [tex]Te^2^-[/tex].

In other words, Sr would be the positive ion in a compound.

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Final answer:

Strontium (Sr) would form a positive ion in a compound since it is an alkaline earth metal that readily loses electrons to become a cation, whereas tellurium (Te) tends to gain electrons to form an anion.

Explanation:

Between the elements strontium (Sr) and tellurium (Te), Sr would typically form a positive ion in a compound. This is because Sr is an alkaline earth metal and is found in Group 2 of the periodic table, where elements readily lose electrons to form cations with a +2 charge. On the other hand, Te is a metalloid in Group 16, known for gaining electrons to form anions with a -2 charge. Compounds are neutral, which means that the total number of positive charges must balance the total number of negative charges.

Additionally, Sr is lower down in its group than Ca, making it easier to remove an electron from Sr than from Ca. Hence, Ca has the higher ionization energy (IE) compared to Sr. When considering a potassium ion (K+), it would also be harder to remove an electron because it already has a positive charge. Therefore, the IE of K+ is significantly higher, especially since the next electron to be removed would come from a different electron shell.

Answer: D

Explanation:

Option D: 5

IMA is Ideal Mechanical advantage, for pulley system it is equal to the number of ropes present in the system. At ideal conditions, it is equal to the number of times, input force is getting multiplied. For example, If a pulley uses two ropes, its required mechanical force to displace the load will be divided into two or the distance that should be pulled will be multiplied by two (keeping total energy constant).

In the pulley system in diagram, there are total of 5 ropes in the system that is total strand that lead directly to the load thus, IMA for this pulley system will be 5.

Final answer:

To make 150.0 mL of a 1.70 M KBr solution, you need 30.33 grams of KBr. This is calculated by multiplying the molarity (1.70 M), the volume in liters (0.150 L), and the molar mass of KBr (119.00 g/mol).

Explanation:

To calculate the mass of KBr needed to make 150.0 mL of a 1.70 M solution, we need to use the formula:

Mass (g) = Molarity (M) × Volume (L) × Molar Mass (g/mol)

First, convert the volume from milliliters to liters by dividing by 1000:

150.0 mL = 0.150 L

Next, find the molar mass of KBr, which is the sum of the atomic masses of potassium (K) and bromine (Br). Potassium has an atomic mass of 39.10 g/mol, and bromine has an atomic mass of 79.90 g/mol:

Molar Mass of KBr = 39.10 g/mol + 79.90 g/mol = 119.00 g/mol

Now, we can calculate the mass of KBr:

Mass of KBr = 1.70 M × 0.150 L × 119.00 g/mol = 30.33 grams

Therefore, you should use 30.33 grams of KBr to make the solution.

Phospholipids form a bilayer when mixed with water, since the hydrophilic head is attracted to water, while the hydrophobic tails reject water and tend to aggregate with other hydrophobic molecules, this conflict is resolved with the formation of a lipid bilayer , which satisfies both parties.

They are those that have phosphoric acid in their composition, they are usually amphipathic: their molecules have a part that is soluble in water (that is, hydrophilic) and another that is not (hydrophobic).

Therefore, we can conclude that phospholipids are found in the active membranes of cells whose distinctive feature is they are usually amphipathic.

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Final answer:

Phospholipids form a bilayer in water due to their amphipathic properties, with hydrophilic heads facing the water and hydrophobic tails facing inward, creating a structure critical to cellular membranes.

Explanation:

Phospholipids spontaneously form a bilayer when mixed with water due to their amphipathic nature. Each phospholipid molecule has a hydrophilic (water-attracting) head and two hydrophobic (water-avoiding) tails.

In an aqueous solution, phospholipid molecules arrange themselves so that their hydrophilic heads are exposed to water, while their hydrophobic tails face each other inside the bilayer, away from the water.

This results in the formation of a lipid bilayer, a structure that is fundamental to the plasma membrane of cells. In this bilayer, the hydrophobic tails interact only with each other, forming a barrier that separates the water and other materials on one side from those on the other side.

Furthermore, when phospholipids are heated in aqueous solutions, they tend to form small spherical structures called liposomes or micelles, with the hydrophilic heads outside and the hydrophobic tails inside, underscoring their self-assembling properties.

The gas that is having lowest density will diffuse faster than other gases. As lesser denser will diffuse more quickly than the one which is more denser. Density is directly proportional to mass of the substance that is more mass of the substance is , more denser it will be. So here the Gas A is having lesser molecular weight than the other gases so it will diffuse more quickly than the other gases.

Gas A; MW = 14.28 g/mol

Answer:

people walking on the sidewalk crush the calcium chloride pellets.

Explanation:

When calcium chloride pellets are crushed with feet it increases the surface area of the calcium chloride. As the surface area increase the number of collision between the reactant molecules increases. Here the reactant molecules are calcium chloride ions and water molecules. In the process of dissolution the calcium chloride ions get hydrated by the dipoles of the water molecules. As the frequency of collision increases it lowers the activation energy or threshold barriers to start a reaction.  

The half-life of an element that decays by 3.415% each day is approximately 20.27 days.

The half-life of an element can be calculated using the formula:

half-life = (ln(2)) / (decay constant)

where the decay constant is the rate at which the element decays per time period. In this case, the element decays by 3.415% each day. To find the decay constant, we convert the percentage to a decimal by dividing it by 100:

decay constant = 0.03415

Now we can calculate the half-life:

half-life = (ln(2)) / (0.03415)

A calculator can be used to find the value of ln(2), which is approximately 0.693:

half-life = 0.693 / 0.03415 = 20.27 days

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The equilibrium associated with [tex]\text{K}_{a3}[/tex] for [tex]\text{H}_3\text{PO}_4[/tex] is as follows:

 [tex]\boxed{\text{HPO}^{2-}_{4}(aq)+\text{H}_{2}\text{O}(l)_{\rightleftharpoons}^{{K}_{a3}}\text{PO}^{3-}_{4}(aq)+\text{H}_{3}\text{O}^{+}(aq)}[/tex]

Further Explanation:

Acids are substances that can donate hydrogen ion [tex]\text{H}^{+}[/tex] or proton in their aqueous solutions.

Classification of acids:

1. Monoprotic acids

Acids that are capable to donate single proton in solutions are monoprotic acids. For example, HCl is a monoprotic acid due to its tendency to donate only one proton in solutions.

2. Polyprotic acids

Acids that are capable to donate more than one proton in aqueous solutions are called polyprotic acids. For example, [tex]\text{H}_2\text{SO}_4[/tex] and [tex]\text{H}_3\text{PO}_4[/tex] are all polyprotic acids.

Acid strength is measured with the help of dissociation constant. Itis denoted by [tex]\text{K}_a[/tex]. Since [tex]\text{H}_3\text{PO}_4[/tex] contains three hydrogens in it, it can donate three hydrogen ions in solutions so it is triprotic acid.

Dissociation of [tex]\text{H}_3\text{PO}_4[/tex] occurs as follows:

[tex]\begin{aligned}\text{H}_3\text{PO}_4(aq)+\text{H}_2\text{O}(l)_{\rightleftharpoons}^{K_{a1}}\text{H}_2\text{PO}^{-}_{4}(aq)+\text{H}_3\text{O}^{+}(aq)\\\text{H}_2\text{PO}^{-}_{4}(aq)\text{H}_2\text{O}(l)_{\rightleftharpoons}^{K_{a3}}\text{HPO}^{2-}_{4}(aq)+\text{H}_{3}\text{O}^{+}(aq)\\\text{HPO}^{2-}_{4}(aq)+\text{H}_{2}\text{O}(l)_{\rightleftharpoons}^{K_{a3}}\text{PO}^{3-}_{4}(aq)+\text{H}_{3}\text{O}^{+}(aq)\end{aligned}[/tex]  

Where,

[tex]{{\text{k}}_{{\text{a1}}}}[/tex] is the first dissociation constant of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex].

[tex]{{\text{k}}_{{\text{a2}}}}[/tex] is the second dissociation constant of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex].

[tex]{{\text{k}}_{{\text{a3}}}}[/tex] is the third dissociation constant of [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex].

The expression of [tex]{{\text{k}}_{{\text{a3}}}}[/tex] for [tex]{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}[/tex] is as follows:

[tex]{{\text{k}}_{{\text{a3}}}} = \dfrac{{\left[ {{\text{PO}}_4^{3 - }} \right]\left[ {{{\text{H}}_{\text{3}}}{{\text{O}}^ + }} \right]}}{{\left[ {{\text{HPO}}_4^{2 - }} \right]}}[/tex]  

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: acids, monoprotic acid, polyprotic acid, hydrogen ions, ka3, H3PO4, acid strength, dissociation constant, first dissociation constant, second dissociation constant, third dissociation constant.

Answer:

true

Explanation:

Answer:

C. IT FLOWS THROUGH VENEZUELA

Explanation:

correct asnwer

 the answer is combustion

The given equation represents a combustion reaction. The organic gas combines with atmospheric oxygen to form carbon dioxide and water .

Combustion is a type of reaction, in  which gaseous reactants combines with atmospheric oxygen to form carbon dioxide and water. In combustion reaction heat energy is released by the reaction system.

The reaction from which a heat energy is released is called an exothermic reaction. Thus, combustion is an exothermic reaction. The reaction in which heat is absorbed the system is called endothermic reaction.

In the given reaction, 2 moles of propanol reacts with 9 moles of oxygen forming 6 moles of carbon dioxide and 8 moles of water. This reaction is an example of combustion.

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Explanation:

Molar heat combustion:

Molar heat of combustion is defined as heat released when 1 mole of substance is burnt under the action of oxygen gas to give water and carbon-dioxide.It is measured in kiloJoules (kJ) per mole.This heat of energy can be positive or negative.

The molar heat of combustion is illustrated as the heat discharged when one mole of the component is burnt by reacting with oxygen gas to provide carbon dioxide and water. It is determined in kJ per mole. This heat of energy can be negative or positive. In case if the enthalpy is negative than the reaction is considered as an exothermic reaction, and if the enthalpy is positive than the reaction is considered as an endothermic reaction.

Bohr's model of the atom places a limitation on electrons by assuming they move in discrete orbits, but this model is not accurate and does not account for electron-electron interactions.

Bohr's model of the atom places a limitation on electrons in that it assumes they move in discrete orbits around the nucleus, with each orbit having a certain energy level. However, this model does not account for electron-electron interactions in atoms with more than one electron and does not accurately represent the true behavior of electrons in atoms. As quantum mechanics was developed, it revealed that electrons do not have well-defined orbits but instead exist as clouds of probability around the nucleus. This limitation of Bohr's model was an important step toward developing a more accurate understanding of atomic physics.

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The answer is plastic...

Hope this helps.

Final answer:

The potential first-order rate law is rate = k[A] or rate = k[B], and the potential second-order rate law is rate = k[A][B].

Explanation:

The potential first-order rate law for the given reaction aA + bB → cC + dD is rate = k[A] or rate = k[B]. This means that the rate of the reaction is directly proportional to the concentration of either reactant A or reactant B.

The potential second-order rate law for the given reaction is rate = k[A][B]. This indicates that the rate of the reaction depends on the product of the concentrations of both reactant A and reactant B.

Answer: Mass of the substance is 1.792 kg

Explanation: Density of a substance is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{volume}[/tex]

Given:

Volume = 560 ml

Density = 3.2 g/ml

[tex]Mass=Density\times volume[/tex]

[tex]Mass=3.2g/ml\times 560ml=1792g[/tex]

Mass of the substance is 1792 g or 1.792 kg.

Answer:   Determining the concentration of an unknown base

Explanation:   While performing a titration,below mentioned steps are followed in the written order -

a)  Finding out which pH indicator works - this is the important and the initial step

b)  After that , one will determine the molecular masses of the products formed in the reaction.

c)  After  determining the molecular mass, number of moles of the product are calculated in order to reach to the conclusion.

d) Thus the final step would be the determiation of the concentration of the unknown base.

Final answer:

After two half-lives of 5,000 years each, only 25% of the original radioactive isotope would remain.

Explanation:

If the half-life of a radioactive isotope is 5,000 years, after 10,000 years, two half-lives will have passed. According to exponential decay, after the first half-life, 50% of the original isotope remains. After the second half-life, another half of this remaining amount will have decayed, leaving only 25% of the original isotope. We can use the expression (1/2)n, where 'n' is the number of half-lives, to calculate the remaining amount of the isotope. In this case, it would be (1/2)2 = 1/4 or 25%.