what is the answer??????​

Answer:

CaCl2

CaO

Al2O3

AlCl3

You can do this by swapping the values of charges on the ions as below

eg- Ca2+ - 2

Cl- - 1

Now combine 1 Ca atom and 2 Cl atoms and it will give you CaCl2

Al 3+ - 3

O2- - 2

Combine 2 atoms of Al and 3 atoms of O

Al2O3

In the case of CaO, we have to make sure that the atoms are in their simplest ratio. So although we get Ca2O2, CaO is the correct form

The pH of the resulting solution after adding NaOH is approximately 3.92.

To answer your question, let's first recall the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

In this case, formic acid (HCOOH) is the weak acid (HA) and sodium formate (HCOONa) is its conjugate base (A-).

The pKa for formic acid is 3.74.

Before adding NaOH, the initial concentrations of formic acid and sodium formate are both 1.00 M.

When you add 400 mL of 0.500 M NaOH, the moles of NaOH added are:

0.400 L × 0.500 mol/L = 0.200 mol NaOH reacts with formic acid, producing sodium formate:

HCOOH + NaOH → HCOONa + H2O

This reaction consumes 0.200 mol of formic acid and produces 0.200 mol of sodium formate.

The new concentrations can be calculated as follows:

[HCOOH] = (1.00 mol - 0.200 mol) / 1.00 L = 0.800 M

[HCOONa] = (1.00 mol + 0.200 mol) / 1.00 L = 1.20 M

Now, we can plug these values into the Henderson-Hasselbalch equation:

pH = 3.74 + log(1.20/0.800) = 3.74 + 0.1761 = 3.92

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Answer:

B.) Na₂SO₄

Explanation:

Covalent bonds are bonds which involve electrons being shared between two atoms. These bonds only occur between nonmetals.

Ionic bonds are bonds which involve electrons being given or taken by two atoms. These bonds form between a metal and a nonmetal.

NH₃ contains nitrogen and hydrogen, two nonmetals. Therefore, this molecule only contains covalent bonds.

Na₂SO₄ has both covalent and ionic bonds. The covalent bonds are found within the polyatomic ion, SO₄²⁻. Sulfur and oxygen are both nonmetals, thus covalently combining. The ionic bond forms between the  Na⁺ and SO₄²⁻. Sodium (Na) is a metal which wishes to give up electrons to have a complete octet. SO₄²⁻ serves as the anion which it bonds with.

CH₃COOH contains hydrogen, carbon, and oxygen. All of these are nonmetals, thus the only bonds formed are covalent.

NaCl contains sodium, a metal, and chlorine, a nonmetal. Therefore, this compound is held together by an ionic bond.

Answer:

C) creating a relationship between mass and volume

g/L

g or gram is mass

L or Litres is for volume

Answer:

5.1 moles of P2O3 are required

Explanation:

Based on the reaction, 1 mole of diphosphorus trioxide, P2O3, reacts with 3 moles of water to produce 2 moles of phosphorous acid, H3PO3. That means the conversion factor is: 1mol P2O3 = 2mol H3PO3.

The moles of P2O3 required to produce 10.2mol H3PO3 are:

10.2mol H3PO3 * (1mol P2O3 / 2mol H3PO3) =

When the base sodium hydroxide dissolves in water, sodium and chloride ions are formed.

Any material that is slippery to the touch, tastes unpleasant, alters the colour of indicators (such as turning red litmus paper blue), combines with acids to generate salts, and facilitates certain chemical processes is considered a base in chemistry (base catalysis).

When the base sodium hydroxide dissolves in water, sodium and chloride ions are formed.

Since sodium hydroxide is an inert solid in itself, its ions do not impart any colour to the solution when it dissolves in water. So, a colourless solution is produced when sodium hydroxide is added to water.

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The molecule ammonium lauryl sulfate is a detergent because of its hydrophobic tail and a hydrophilic head which are responsible for its cleansing action.

Though not shown, I presume that the property labelled as C must have to do with the cleasing action of the molecule ammonium lauryl sulfate.

The substance has a hydrophobic tail and a hydrophilic head which enables it to form miscles and these are the responsible for the cleansing action of the ammonium lauryl sulfate molecule.

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Answer:

Rate cup 1: 36.00 tsp/h.

Rate cup 2: 59.50 tsp/h.

Rate cup 3: 70.13 tsp/h.

Explanation:

Based on the given example, the units must be in tsp/hour. To do this we just have to divide tsp by seconds and then convert it to hours remembering that 1 hour equals 3600 seconds.

Let's calculate each case:

- Cup 1: 1 tsp, 1 minute 40 seconds. Remember that 1 minute equals 60 seconds, so 1 minute and 40 seconds equals 100 seconds. The calculation of the rate will look like this:

The rate of cup 1 is 36.00 tsp/h.

- Cup 2: 2 tsp, 2 minutes 1 second. As we saw before 1 minute equals 60 seconds, so 2 minutes equals 120 seconds. 2 minutes and 1 second equals 121 seconds. The rate of cup 2 will be calculated as:

The rate of cup 2 is 59.50 tsp/h.

- Cup 3: 3 tsp, 2 minutes 34 seconds. 2 minutes are the same that 120 seconds, so 2 minutes 34 seconds equals 154 seconds. The rate of cup 3 is calculated as follows:

The rate of cup 3 is 70.13 tsp/h.

The appropriate symbol for an isotope of potassium-39 using isotope notation would be written as "39K".

The isotope notation, also known as nuclear notation, is used to represent different isotopes of an element. It consists of the element's atomic number (Z), the element's symbol (A), and the mass number (X). The atomic number represents the number of protons in the nucleus of an atom and is used to determine the element to which an atom belongs. The mass number represents the total number of protons and neutrons in the nucleus of an atom. The symbol represents the element to which the isotope belongs.

In this case, 39 is the mass number (X) and K is the symbol of the element Potassium (A).

So, in isotope notation for Potassium-39 it would be represented as 39K.

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Answer:

did you have options, cause if you did chose something alond the lines of

Explanation:

A real gas is a gas that does not behave as an ideal gas due to interactions between gas molecules. A real gas is also known as a nonideal gas because the behavior of a real gas in only approximated by the ideal gas law.

Answer:

both iodine

Explanation:

The difference in values between microstates and the structure of liquid water is due to the fact that microstates refer to the different arrangements of water molecules at a molecular level, while the structure of liquid water refers to the overall arrangement of water molecules in a bulk phase.

The structure of liquid water is determined by the intermolecular forces between water molecules, which results in a unique arrangement of molecules that allows for the liquid state. Microstates, on the other hand, describe the various possible arrangements of individual water molecules within this overall structure. The number of possible microstates increases with the number of molecules in the system, while the overall structure of liquid water remains constant. Thus, while the structure of liquid water determines its physical properties, the microstates describe the statistical distribution of molecules within this structure.

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Answer:

268ml

Explanation:

What would the volume of a gas be at -150C, if it has volume of 693ml at 45.0C?

V and T are related as V =kT

first, change all C into K since these are gas laws!

45C =45+273 =318K

-150C =-150+273 =123K

now 693ml=kX318

so k=693/123 =2.18

so new V=2.18 X 123=268ml

Titration curves representing a strong acid titrated with a strong base typically exhibit certain characteristics.

Here's a general classification of the titration curve for this scenario:

In summary, the titration curve for a strong acid titrated with a strong base will exhibit an initial low pH, a steep slope near the equivalence point, a pH of approximately 7 at the equivalence point, a buffer region after the equivalence point, and a final pH determined by the excess strong base added.

Incomplete question:

Classify each titration curve as representing a strong acid titrated with a strong base, a strong base titrated with a strong acid, a weak acid titrated with a strong base, a weak base titrated with a strong acid, or a polyprotic acid titrated with a strong base. A. Strong acid/ strong base.B. Strong base/ strong acid.C. Weak acid/ strong base.D. Weak base/ strong acid.E. Polyprotic acid/ strong base.

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Explanation:

The given intermediate chemical reactions do not lead to a single overall chemical equation for smog. Instead, they represent a combination of various reactions that contribute to the formation of different components of smog, such as nitrogen oxides (NOx) and ozone (O3).

The first reaction given is:

N2 + 3O2 + 2O3 + 2O → 8NO + 4NO2 + 2O3

This reaction represents the formation of nitrogen oxides and ozone from the reaction of nitrogen and oxygen with ozone and oxygen radicals. Nitrogen oxides and ozone are major components of smog.

The second reaction given is:

N2 + 3O2 + O3 + O → 9NO + 3NO2 + 2O3

This reaction also represents the formation of nitrogen oxides and ozone, but with a different stoichiometry.

The third reaction given is:

N2 + 3O2 → 2NO + O2

This reaction represents the direct formation of nitrogen oxides from the reaction of nitrogen and oxygen.

The fourth reaction given is:

N2 + 2O2 → 2NO2

This reaction represents the direct formation of nitrogen dioxide from the reaction of nitrogen and oxygen.

Overall, the formation of smog is a complex process that involves the interaction of various chemical reactions and environmental factors. Therefore, there is no single overall chemical equation that describes the formation of smog.

3 components of an expressway are :

Acceleration Lane

Merging Area

Slow  Lane

Safety design features of an expressway:

Safety elements on the freeways include lane separators and shoulders. Broad, clearly indicated lanes: Highway lanes are broad and feature distinct markings. The inside edge of paved shoulders often has elevated or grooved patterns called "rumble strips" that are typically found alongside major highways.

What are expressways designed for?

Instead of the current practice of every business or establishment connecting the main roadway, access restricted roads have a limited number of designated entry and exit points. Coming onto the main highway is done via segregated service roads.

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Answer:

Sorry, I don't know.

Explanation:

Molar mass of Cu = 6.45x10^25 atoms.

The molar mass of Cu = 63.55g/mol.

Mass of Cu= ?

Thus, Mass of cu= Molar mass of Cu / The molar mass of Cu

                             = 6.45x10^25 atoms/ 63.55g/mol.

                              = 0.176 g

The chemical element copper has the atomic number 29 and the letter Cu, which comes from the Latin word cuprum.

It is an extremely high thermal and electrical conductivity metal that is soft, malleable, and ductile. Pure copper has a pinkish-orange tint when it is first exposed to the air.

Thus, Molar mass of Cu = 6.45x10^25 atoms.

The molar mass of Cu = 63.55g/mol.

Mass of Cu= ?

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Answer:

Explanation:

A. Tabular form of the formation of the compound formed between X and Y (carbon and oxygen):

| Element | Electronic Configuration |

|---------|-------------------------|

| X       | 1s² 2s² 2p⁶ 3s²        |

| Y       | 1s² 2s² 2p⁶ 3s² 3p⁵   |

B. Formation of the compound:

The compound formed between X and Y is carbon dioxide (CO2). Carbon (X) has an electronic configuration of 1s² 2s² 2p⁶ 3s², and oxygen (Y) has an electronic configuration of 1s² 2s² 2p⁶ 3s² 3p⁵.

Carbon has 4 valence electrons in its outermost energy level, while oxygen has 6 valence electrons in its outermost energy level. In order to achieve stability, carbon needs to gain 4 electrons, while oxygen needs to gain 2 electrons.

To form the compound CO2, carbon will share electrons with two oxygen atoms. Carbon will share 2 electrons with each oxygen atom, resulting in a double bond between carbon and each oxygen atom.

The formation of the compound can be represented as follows:

O = C = O

2. Drawing the formation of the compound:

In text format, the formation of the compound CO2 can be represented as:

     O

    //

   C

    \\

     O

Here, the central carbon atom (C) is bonded to two oxygen atoms (O) through double bonds. The structure of carbon dioxide is linear, with the carbon atom in the center and the oxygen atoms on either side.

Answer:

yes

Explanation:

The identity of element E in the nuclear reaction 238/92U → 234/90E + 4/2He is thorium (Th). The main ans for balancing nuclear reactions is to ensure that the total mass and atomic numbers are conserved on both sides of the reaction.

In this case, the total mass on the left side (238) is equal to the sum of the masses on the right side (234 + 4). Similarly, the total atomic number on the left side (92) is equal to the sum of the atomic numbers on the right side (90 + 2). Therefore, the only element that satisfies these conditions and has an atomic number of 90 is thorium (Th). This explanation confirms that the identity of element E in the nuclear reaction is thorium (Th).

In the nuclear reaction 238/92U → 234/90E + 4/2He, the identity of element E is Thorium (Th). To find the identity of element E, you need to look at the atomic number provided (90). Element 90 on the periodic table is Thorium (Th). In this reaction, Uranium (238/92U) decays into Thorium (234/90Th) and an alpha particle (4/2He).

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The compounds in their order of increasing positive character of carbon are: CH₃Li → CH₃F → CH₄ → CH₃NH₂ → CH₃OH.

The positive character of the carbon atom refers to its electropositive nature or its ability to attract electrons towards itself in a chemical bond. This property of the carbon atom can be influenced by the other atoms or groups bonded to it, as well as their electronegativity, size, and electronic configuration.

The positive character of the carbon atom can be determined by various methods, including chemical reactions, spectroscopic measurements, and theoretical calculations.

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Reaction equations contains symbols which show the physical state of the reactants and products.

A chemical equation is an expression that shows the interaction of reactants to yield products. Usually, symbols such as (s), (l), (g), and (aq) are used to show the state of the reactants and products.

The following are the respective meanings of these symbols;

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What is the similarity between chlorine and bromine?

Chlorine and bromine have very similar chemical properties and are best explained by the fact that they both belong to the category of halogens.

Similarities shared by halogens:

Chemical elements are represented as periods and groups on a periodic table. Chemical elements with comparable properties make up each category. Halogens are group 17 elements that are primarily non-metallic.

In the case of halogens, each element's atom possesses seven valence electrons, \(ns^{2} np^{5}\), in which n stands for the energy level. The electron configuration of chlorine is \([Ne]3s^{2} 3p^{5}\), and the electron configuration of bromine is \([Ar]3d^{10}4s^{2} 4p^{5}\).

Because of comparable electron configurations, elements of the same group have similar properties. The number of valence electrons, which are located in the outermost (highest energy) s and p orbitals, is particularly similar.

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Answer:

agua 52,g

Explanation:

por que tiene un baso

The mass of water produced is 792 grams by the combustion of 568 grams of decane.

Given:

Combustion of 568 grams of decane with 2979 grams of oxygen.

\(2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O\)

To find:

The mass of water produced by combustion of 568 grams of decane.

Solution:

Mass of decane = 568 g

Moles of decane :

= \(\frac{568 g}{142 g/mol}=4 mol\)

Mass of oxygen gas = 2976 g

Moles of oxygen gas:

= \(\frac{2976 g}{32 g/mol}=93 mol\)

\(2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O\)

According to reaction, 2 moles of decane reacts with 31 moles of oxygen, then 4 moles of decane will react with:

\(=\frac{31}{2}\times 4mol=62\text{ mol of}O_2\)

But according to the question, we have 93.0 moles of oxygen gas which is more than 62 moles of oxygen gas.

So, this means that oxygen gas is present in an excessive amount. Which simply means:

According to reaction, 22 moles of water is produced from 2 moles of decane, then 4 moles of decane will produce:

\(=\frac{22}{2}\times 4mol=44\text{mol of }H_2O\)

Mass of 44 moles of water ;

\(=44mol\times 18g/mol=792g\)

792 grams of water is produced by the combustion of 568 grams of decane.

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