Tenemos 9 gramos de aluminio que reacciona totalmente con ácido sulfúrico y origina sulfato de aluminio e hidrógeno ( gas )
a) cuantos gramos de sulfato de aluminio se forman
b) cuántos litros de hidrógeno se obtienen en condiciones normales

The dot in the image represents the unshared/ lone pair of electrons. The correct option is D. Dots are written around the element in an equation.

Unshared or lone electrons are those electrons, which are lone and not shared with another atom when making a covalent bond. These electrons are represented by dots. The valence electron help in making the valence complete of other atoms when forming a bond.

The diagram is of water, oxygen has six valence electrons, of which two are shared by hydrogen and four are lone electrons. The lone pairs will be attached to other hydrogen molecules.

Thus, the correct option is D, Unshared/lone pairs of valence electrons regarding the representation of the dots in the image given.

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A substance's form or appearance can be changed physically without affecting its chemical makeup.

Cutting fruits and vegetables, melting chocolate, and boiling eggs are examples of foods that go through physical changes.

For instance, slicing fruits and vegetables alter their size and shape, but not their chemical makeup.

Similar to melting chocolate, which transforms it from a solid to a liquid yet keeps its chemical makeup the same.

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Mass of KNO₃ : = 40.643 g

Given

28.5 g of K₃PO₄

Required

Mass of KNO₃

Solution

Reaction(Balanced equation) :

2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃

mol K₃PO₄(MW=212,27 g/mol) :

= mass : MW

= 28.5 : 212,27 g/mol

= 0.134

Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :

= 6/2 x mol K₃PO₄

= 6/2 x 0.134

= 0.402

Mass of KNO₃ :

= mol x MW KNO₃

= 0.402 x 101,1032 g/mol

= 40.643 g

Answer:

In order to find the two statements, we must first define what the enthalpy of formation and the enthalpy of reaction mean.

Enthalpy of formation:

The change in enthalpy when one mole of substance is formed from its constituent elemetns at standard state.

Enthalpy of reaction:

The change in enthalpy when a reaction occurs and the reactants and products are in their standard states.

Now, we check the statements. The true ones are:

The Hrxn for C(s) + O₂(g) → CO₂(g) is the same as Hf for CO₂

This is true because the formation of carbon dioxide requires carbon and oxygen in their standard states.

The Hf for Br₂(l) is 0 kJ/mol by definition.

Because the bromine is present in its standard state, the enthalpy of formation is 0.

The Hrxn for the reaction 1.5H₂(g) + 0.5N₂(g) → NH₃(g) is the same as the Hf for NH₃(g)

The reactants and products are present in their standard state, and the reaction is the same as the one occurring during the formation of ammonia.

The small stretches on a chromosome  of a deoxyribonucleic acid (DNA) molecule on a particular site is known as gene.

Gene is a fundamental, structural and operational component of heredity in living beings. They serve as a blueprints for synthesis of proteins. Genes are responsible for transfer of traits from parents to the progeny.

The sequence of nucleotide on a DNA strand determines the coding for various proteins. From a few hundred DNA bases to more than 2 million bases, genes range in size. Every human contains two copies of same gene, one from each parent.

Protein-coding genes make up only around 1% of DNA; the remaining 99% is non-coding genes. There are no protein-making instructions in non-coding part of DNA.

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Metallic elements exist in a solid-state and they are opaque, have a shiny surface, good conductors of electricity and heat, malleable and ductile, and are dense. The structure of metals is formed by atoms that are held together by metallic bonds. These atoms have loosely bound valence electrons that can be shared between the neighboring atoms.

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The percentage of daily kilocalories of the person that come from fat is 20%.

The diet of a person is composed mainly of carbohydrates (C), proteins (P) and fat (F), that is, the sum of the 3 sources amount to 100% of the consumed kilocalories. Given that 50% of the daily kilocalories of this person come from carbohydrates and 30% come from protein, the percentage that comes from fats is:

\(C + P + F = 100\%\\F = 100\% - C - P\\F = 100\% - 50\% - 30\% = 20\%\)

The percentage of daily kilocalories of the person that come from fat is 20%.

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Answer:

The pH of a solution with a (H+] = 0.80 M is  \(9.6^{-2}\)

Explanation:

pH = -[log H+]

Substituting the value of H+ ion concentration in the above equation, we get -

pH = -log [0.80]

pH= \(- (-0.096) = 0.096=9.6 e^{-2}\)

Answer:

Volume of one mole of any gas at NTP is

(b) 22.4 litre

Answer:

22.4L

Explanation:

One mole of any gas at STP weighs 22.4L.

STP stands for Standard Temperature and pressure

Large particles usually get smaller as a result of surface reactions. Both heating and freezing effects of tiny particles on the climate are seen.

Weathering can be classified as either physical or chemical. There is a wide variety in soil particle size due to variations in weathering processes.The overall surface area increases when the size is reduced more. This is significant since most chemical changes start at an object's surface.

The Earth's materials are the four major elements that make up the earth's crust. The earth is made up of minerals, rocks, soil, and water. Earth materials include the necessary building components for life, agriculture, and industry.

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The antiseptic may affect the results of tests for bilirubin uric acid is Povidone-iodine.

Polyvinylpyrrolidone (povidone, PVP) and elemental iodine combine chemically to form the stable compound known as povidone-iodine. On a dry basis, it has a 9.0% to 12.0% available iodine content. H. A. Shelanski and M. V. Shelanski found this particular complex in 1955 at the Industrial Toxicology Laboratories in Philadelphia.

The compound was discovered to be less harmful to mice than tincture of iodine during in vitro tests to exhibit anti-bacterial efficacy. Clinical tests on humans revealed that the medication outperformed competing iodine formulations. Povidone-iodine was marketed right away and has since taken over as the most widely used iodine antiseptic.

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Povidone-iodine is an antiseptic that may affect the results of tests for bilirubin and uric acid. Povidone-iodine is a commonly used antiseptic that can be applied topically to the skin to reduce the risk of infection during medical procedures. However, it can also interfere with certain laboratory tests, including tests for bilirubin and uric acid.

Bilirubin is a yellow pigment that is produced when red blood cells break down, and it is excreted by the liver. Uric acid is a waste product that is produced when the body breaks down purines. Both bilirubin and uric acid can be measured through laboratory tests to help diagnose certain medical conditions.

However, povidone-iodine can interfere with these tests by reacting with the substances being tested, leading to inaccurate results. It is important for healthcare professionals to be aware of the potential for povidone-iodine to affect laboratory test results and take appropriate measures to minimize this risk, such as avoiding its use immediately prior to testing.

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Explanation:

Density = mass / volume

ρ = 640 g / (4 cm)³

ρ = 10 g/cm³

Answer:

transfer energy to the surroundings

Explanation:

Exo means exit

The standard entropy of vaporization of ethanol at its boiling point 352K is 115.056 J/K-mol.

Given,

Enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 × 10³ J/mol

Entropy of vaporization of ethanol boiling point = 352 K

Determine the standard entropy of vaporization of ethanol,

We get here standard entropy of vaporization of ethanol that is expessed as,

Standard entropy of vaporization of ethanol ΔS = ΔH/T  .............1

Here ΔH is enthalpy of vaporization of ethanol and  T is the temperature.

Substitute the values in equation 1

Standard entropy of vaporization of ethanol ΔS = ΔH/T

                                                                                = [40.5 × 10³] / 352

                                                                                = 40500/352

                                                                                = 115.056 J/K-mol.

Hence, the standard entropy of vaporization of ethanol at its boiling point 352K is 115.056 J/K-mol.

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Answer:

1 quarter=2.3

X quarter=384403

384403/2.3=167,131.7391304348

THE ANSWER IS 167,131.7391304348

Answer:

sentence 2, sentence 2, sentence 1

Explanation:

it was right for me on edge 2020

Answer:

1. Fred visited Nick after his graduation.

2. After Nick's graduation, Fred visited him.

1. Marie told Sarah that she makes great spaghetti.

2. Marie told Sarah, "You make great spaghetti."

1. John broke the fence with his bike.

2. John's bike hit a fence and it broke.

Answer:

The answer is

Explanation:

To convert the volume from cm³ to mL we use the conversion

1 cm³ = 1 mL

So from the question

if 1 cm³ = 1 mL

Then 150 cm³ = 150 mL

Hope this helps you

The complete string representing the order of orbital filling without blank spaces is: \(1s^22s^22p^63s^23p^64s^23d^1{0}4p^65s^24d^105p^66s^24f^145d^{10}6p^67s^25f^146d^{10}7p^6\)

To express the order of orbital filling as a string without blank spaces, we follow the Aufbau principle, which states that electrons fill orbitals starting from the lowest energy level and move upward. The order of orbital filling is as follows:\(1s^22s^22p^63s^23p^64s^23d^1{0}4p^65s^24d^105p^66s^24f^145d^{10}6p^67s^25f^146d^{10}7p^6\)

Let's break it down step by step:

1s^2: The 1s orbital can hold up to 2 electrons.

2s^2: The 2s orbital can also hold up to 2 electrons.

2p^6: The 2p orbitals can hold up to 6 electrons.

3s^2: The 3s orbital can hold up to 2 electrons.

3p^6: The 3p orbitals can hold up to 6 electrons.

4s^2: The 4s orbital can hold up to 2 electrons.

3d^10: The 3d orbitals can hold up to 10 electrons.

4p^6: The 4p orbitals can hold up to 6 electrons.

5s^2: The 5s orbital can hold up to 2 electrons.

4d^10: The 4d orbitals can hold up to 10 electrons.

5p^6: The 5p orbitals can hold up to 6 electrons.

6s^2: The 6s orbital can hold up to 2 electrons.4f^14:

The 4f orbitals can hold up to 14 electrons.

5d^10: The 5d orbitals can hold up to 10 electrons.

6p^6: The 6p orbitals can hold up to 6 electrons.

7s^2: The 7s orbital can hold up to 2 electrons.

5f^14: The 5f orbitals can hold up to 14 electrons.

6d^10: The 6d orbitals can hold up to 10 electrons.

7p^6: The 7p orbitals can hold up to 6 electrons.

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A. The stoichiometry of the balanced equation should be used to calculate the amount of product NO. We can conclude from this equation that 1 mole of nitrogen reacts to produce 2 moles of NO. Therefore, if 10 liters of nitrogen is completely reacted, 20 liters of nitrogen gas will be evolved.

B. The mole ratio from the balanced equation can be used to determine how many liters of nitrogen can react with four moles of oxygen at STP (standard temperature and pressure).

According to the equation, 1 mole of nitrogen and 1 mole of oxygen combine to form 2 moles of NO. So four moles of oxygen would require four moles of nitrogen. 4 moles of oxygen would require 4 * 22.4 = 89.6 liters of nitrogen because 1 mole of any ideal gas takes up 22.4 liters at STP.

C. To calculate the amount of NO formed from 32 g of oxygen at STP, we must convert the mass of oxygen into moles. The molar mass of oxygen (O2) is about 32 g/mol. Consequently, 1 mole of oxygen is equal to 32 grams. 1 mole of oxygen reacts to produce 2 moles of NO in the balanced form of the equation. Therefore, 32 grams of oxygen will result in 2 moles of NO. Noting that 2 moles of NO will take up 2 * 22.4 = 44.8 liters at STP, where 1 mole of any ideal petrol takes up 22.4 liters.

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The molecule that has a polar covalent bond among the available possibilities is (c). H-F.

When there is an unequal distribution of electrons between two atoms in a covalent connection, a polar covalent bond develops, leaving one atom with a partial positive charge and the other with a partial negative charge. The reason for this unequal sharing is typically because the atoms involved have different electronegativities.

Fluorine (F) has an electronegativity value of 3.98 in the case of H-F, while hydrogen (H) has an electronegativity value of 2.2. Fluorine pulls the shared electrons in the covalent bond more strongly than hydrogen because fluorine is substantially more electronegative than hydrogen. As a result, fluorine has a partial negative charge (-) and hydrogen has a partial positive charge (+).

As a result, the hydrogen-fluorine bond has a polar structure, with the hydrogen atom having a partial positive charge and the fluorine atom having a partial negative charge. The molecule is also polar because of this polarity.

The other possibilities, H-H, F-F, and Cl-Cl, on the other hand, include bonds between atoms of the same element. Since there is no difference in electronegativity in these situations, the electrons are shared evenly and nonpolar covalent bonds are the outcome.

In conclusion, because hydrogen and fluorine have very different electronegativities, the H-F molecule has a polar covalent bond. The other choices use covalent nonpolar bonding.

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2 moles of H2 react with 1 mole of O2 gas, according to the balanced reaction. the stoichiometric ratio of the moles of H2 to the moles of O2 is 2:1 as a result.

When all of the reactants are consumed and none are left after the chemical reaction has finished, it is said to have occurred in a stoichiometric chemical reaction.

Stoichiometry can be summed up as the process of calculating the products and reactants of a chemical reaction. It primarily has to do with numbers. Stoichiometry is a fundamental concept in chemistry that makes it easier to use balanced chemical equations to ascertain the amounts of reactants and products.

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Under the condition of a muscle at rest, the Golgi tendon organ (GTO) would be very active, but the muscle spindle would not be very active.

(a) A muscle at rest: When a muscle is at rest, the Golgi tendon organ (GTO) would be highly active. The GTO is located at the junction between the muscle and tendon and is sensitive to changes in muscle tension. During rest, there is minimal tension in the muscle, and the GTO detects this low tension. In response, the GTO sends inhibitory signals to the muscle, preventing excessive contraction.

On the other hand, the muscle spindle is not very active when the muscle is at rest. The muscle spindle is responsible for detecting changes in muscle length. Since the muscle is not being stretched or experiencing significant movement at rest, the muscle spindle is not actively sending signals to the nervous system.

In summary, during muscle rest, the Golgi tendon organ is highly active due to low muscle tension, while the muscle spindle is not very active since there is no significant stretch or movement.

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1.63×10^-18J is the energy for the transition of an electron from the level n=1 to the level  n= 2 of a hydrogen atom.

According to the Bohr equation for hydrogen atom,

Energy, ∆E= -RH[(1/nf2 ) - (1/ni2)]

Where RH is the Rydberg constant and its value is 109737.3cm-1, 'nf' and 'ni' are integers that represent the energy levels.

1). ni= 1 and nf= 2, substituting these values in Bohr equation,

∆E = -109737.3cm-1[(1/22). -(1/12)

= -109737.3cm-1[(1/4) -1]

= -109737.3cm-1[0.25-1]

= -109737.3cm-1(-0.75)

= 82302.975cm-1

=82302.975×1.986×10-23J

=1.63×10-18J

(where 1 cm-1= 1.986×10-23J)

Here the electron jumps from n=1 to n=2, so it must absorb a photon of light. Therefore this is an absorption process.

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The E2 reaction of 3-bromo-2,3-dimethylpentane with methoxide ion can produce four different alkene products, and their reactivity order is 3 > 1 > 2 > 4, with 3-methyl-2-pentene being the major product formed.

The E2 (elimination bimolecular) reaction of 3-bromo-2,3-dimethylpentane with methoxide ion can produce four different alkene products. The reaction occurs when the base removes beta hydrogen and a leaving group, in this case, the bromine atom, from the carbon chain. The four possible alkene products formed are:

To rank the alkene products according to the amount that would be formed, we need to consider Zaitsev's rule, which states that the most substituted alkene is the major product. Therefore, the order of reactivity for the alkene products is as 3 > 1 > 2 > 4

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Rhenium (IV) ion is represented as Re4+.Iridium (IX) ion is not possible as the maximum possible element state for iridium is +VIII.Platinum (II) ion is represented as Pt2+. Technetium (VII) ion is represented as Tc7+.

The net charge is expressed in superscript directly just after chemical structure of the molecule/atom in the chemical equation for just an ion. The doubly charge cation is denoted as 2+ rather than +2 because the size of the net charge was written before the sign.

A metal ion occurs whenever a metal gives up the outermost electrons to obtain a full octet. Calcium (Ca) will become a cation when it loses two electrons from of the outermost energy shell. There will be a +2 charge on it because it lost 2 electrons. With more protons than electrons, it has a positive charge (+2).

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The reactants that will form products will be \(2CuI_2 + Br2\).

Elements at higher positions in the activity series will be able to displace those in lower positions in solutions.

Thus:

\(2CuI_2 + Br2 --- > 2CuBr +2 I_2\) because Br is higher than I in the activity series.

Cl cannot displace F in \(Cl2 + AlF_3\) because F is higher than Cl in the activity series.

Also, Cl is higher than Br in the activity series, thus, the reaction \(Br_2 + NaCl\) cannot form products.

I cannot displace F, thus, \(CuF_2 + I_2\) cannot form products.

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Percent yield is defined as the 100 times the ratio of actual yield and theoretical yield. It ca be calculated by finding theoretical yield of the reaction and come as 76.6%.

2Ca₃(PO₄)₂(s) + 6SiO₂(s) → 6CaSiO₃(l) + P₄O₁₀(g)

P₄O₁₀(g) + 10C(s) → P₄(g) + 10CO(g)

We need to find out the moles of Ca₃(PO₄)₂ and SiO₂ to find limiting reactant. With the help of limiting reactant we can find moles of P₄O₁₀ = Moles of P₄. We must convert  moles of P₄ to mass using Molar mass (P₄ = 123.895g/mol):

Moles Ca₃(PO₄)₂ -Molar mass: 310.1767g/mol-

293.5g (1mol / 310.1767g) = 0.9462moles

Moles SiO₂ -Molar mass: 60.08g/mol-:

378.5g  (1mol / 60.08g) = 6.30 moles

For the complete reaction of 6.30 moles of SiO₂ there are required:

6.30 moles SiO₂  (2 moles Ca₃(PO₄)₂ / 6 moles SiO₂) =

2.10 moles Ca₃(PO₄)₂. Now there are just 0.9462 moles, Ca₃(PO₄)₂ is limiting reactant

Moles of P₄O₁₀ = Moles P₄:

0.9462moles of  Ca₃(PO₄)₂ (1mol P₄O₁₀ / 2 mol Ca₃(PO₄)₂)

= 0.4731 moles of  P₄O₁₀ = Moles P₄.

The mass is calculated as:

0.4731 moles P₄  (123.895g / 1mol) = 58.6g = Theoretical yield.

Thus, Percent yield is calculated as:

44.9g / 58.6g ×100 = 76.6%

So, 76.6% is percent yield of P₄

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