All substances are composed of which type of matter?
a. ions
b. mixtures
c. elements
d. molecules

19.) Single-displacement is the correct answer.

20.) I am not sure, but Single-displacement is incorrect. At least you can cancel that one out.

Boron is an element as it can not be broken down further by simple chemical processes.

It is defined as a substance which cannot be broken down further into any other substance. Each element is made up of its own type of atom. Due to this reason all elements are different from one another.

Elements can be classified as metals and non-metals. Metals are shiny and conduct electricity and are all solids at room temperature except mercury. Non-metals do not conduct electricity and are mostly gases at room temperature except carbon and sulfur.

The number of protons in the nucleus is the defining property of an element and is related to the atomic number.All atoms with same atomic number are atoms of same element.Elements are majorly classified according to their chemical properties.

Learn more about elements,here:

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Explanation:

A balanced chemical reaction is a reaction in which there is equal number of atoms on both reactant and product side. Also, the mass of atoms or compounds present on reactant side equals the mass of atoms or compounds present on product side.

For example, [tex]CaSO_{4} + 2HCl \rightarrow CaCl_{2} + H_{2}SO_{4}[/tex]

Molar mass of [tex]CaSO_{4}[/tex] = 136.14 g/mol

Molar mass of 2[tex]HCl[/tex] = [tex]2 \times 36.46[/tex] g/mol = 72.92 g/mol

Therefore, sum of molar mass of reactants = (136.14 + 72.92) g/mol

                                                                       = 209.05 g/mol

Molar mass of [tex]CaCl_{2}[/tex] = 110.98 g/mol

Molar mass of [tex]H_{2}SO_{4}[/tex] = 98.07 g/mol

Therefore, sum of molar mass of products = (110.98 + 98.07) g/mol

                                                                       = 209.05

Therefore, we can see that it is true that when you balance a chemical reaction, you are making sure that the law of conservation of matter is obeyed.

An increase in dissolved CO₂ in ocean water results in an increase in acidity (decrease in pH) of the water. This is due to the formation of carbonic acid that dissociates into hydrogen ions. This process has adverse effects on marine life, particularly animals such as corals and shellfish that rely on calcification to build their shells and skeletons.

When carbon dioxide (CO₂) dissolves in the ocean water, it reacts with water to form a weak acid called carbonic acid. This acid then dissociates to form hydrogen ions (H+) and bicarbonate ions (HCO3-). Increased concentration of these hydrogen ions decreases the pH of the water, thus increasing its acidity.

Ocean acidification is detrimental to many marine organisms, especially corals and shellfish. This is because more acidic water interferes with the process of calcification, which they use to build their calcium carbonate shells and skeletons.

The increasing levels of atmospheric CO₂ and rising ocean temperatures are contributing to the acidity of ocean waters, causing harm to marine animals and ecosystems in the process.

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Final answer:

Option d is the correct answer. If the volume of air in a pump doubles with temperature remaining constant, according to Boyle's Law, the pressure is halved which means the number of collisions per unit area is reduced by half.

Explanation:

When you draw back on the piston of a pump, causing the volume of air to double, and temperature remains constant, Boyle's Law applies here. Boyle's Law states that the pressure of a gas is inversely proportional to its volume, when temperature and the number of molecules are constant. Hence, when the volume doubles, the pressure is halved. This means that the number of collisions per unit area of the gas molecules against the walls of the piston is reduced by half because there is now twice the space for the gas molecules to move around in.

Considering the given options, the correct answer to what happens when the volume of air in the pump doubles is d. The number of collisions per unit area is reduced by one-half., given that temperature and the number of gas molecules remain constant.

Radioactive hydrogen will be found in glucose, and radioactive oxygen will be found in the oxygen gas produced during photosynthesis.

In photosynthesis, water (H₂O) is split into hydrogen (H) and oxygen (O₂) by the energy from sunlight. When you expose a photosynthesizing plant to water that contains both radioactive hydrogen (³H) and radioactive oxygen (¹⁸O), these elements will be incorporated into the products of photosynthesis in specific ways. The radioactive hydrogen (³H) will be found in glucose (C₆H₁₂O₆) as hydrogen atoms are part of the glucose molecule. The radioactive oxygen (¹⁸O) will be found in the molecular oxygen (O₂) released as a byproduct.

Thus, if you supply a plant with radioactive water, the radioactive H will show up in glucose, while the radioactive O will show up in the oxygen gas produced during photosynthesis.

Answer:

The average kinetic energy of the particles would reduce, subsequently leading to reduced average motion of the particles. If the decrease in temperature persists, the particles eventually change phase and move to a less mobile state of matter than the phase in which they currently are in.

Explanation:

The image for the question is missing and we couldn't find thay online.

But it isn't hard to imagine what the image would be.

It is said to be a picture depicting particles that are in two different phases (liquid and gas).

What would likely happen if the temperature of the particles were decreased?

All particles of matter are said to always be in a constant state of motion (random motion); with the motion very evident in particles in gaseous phase than particles in liquid phase, then particles in the solid phase.

A decrease in temperature for any type of particles in whichever state will result in a reduced kinetic energy of such particles. As the kinetic energy with which the particles moving with constant, random motion is directly proportional to the absolute temperature of the system in which such particles are contained in.

So, a decrease in temperature for both gaseous and liquid particles will result in reduced and reduced kinetic energy and subsequently a reduction in the average motion of the particles.

For the gaseous particles, if the decrease in temperature continues to a point, the gaseous particles lose enough kinetic energy to change phase from gaseous form to a phase (liquid phase) where the motion is more restricted, average motion reduces and kinetic energy of the particles drop too. This is condensation; changing from gaseous phase to liquid phase.

The liquid particles follow a similar course too, only that a continuous decrease in temperature will lead to reduced motion and kinetic energy until the liquid particles 'freeze' by changing phase into the solid phase where the average motion is much more reduced and limited to vibrational motion about a particular fixed point.

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Final answer:

Chromium oxide enhances stainless steel by forming a protective passivating layer that inhibits corrosion and increases the alloy's durability. Chromium contributes its corrosion resistance to stainless steel, ensuring long-lasting quality and protection against environmental elements.

Explanation:

Chromium oxide plays a crucial role in improving the properties of stainless steel. By alloying the iron with chromium, stainless steel gains significant resistance to corrosion. This is largely due to the chromium tending to migrate towards the surface, where it reacts with oxygen to form a passivating chromium oxide layer.

This oxide layer protects the iron from further corrosion, enhancing the durability and longevity of the stainless steel.

While pure chromium does not readily corrode, it bestows its resistive properties upon steel when used as an alloy. The created oxide layer is thin but very protective, acting as a barrier that blocks moisture, air, and other corrosive agents from reaching the iron.

In different environmental conditions, chromium can exist in several oxidation states, such as Cr(III) and Cr(VI). Cr(III) is relatively insoluble in water and low in toxicity, while Cr(VI) is considerably more toxic and soluble in water.

However, in stainless steel, chromium's positive contributions in the form of its corrosion resistance make it a valuable element for maintaining the material's quality and safety.

The chemical equation representing the reaction of silver nitrate with barium chloride:

[tex] 2AgNO_{3}(aq) + BaCl_{2}(aq)--> 2AgCl (s) + Ba(NO_{3})_{2}(aq) [/tex]

Given mass of barium chloride = 4.62 g

Moles of [tex] BaCl_{2} = 4.62 g BaCl_{2}*\frac{1 mol BaCl_{2}}{208.23 g BaCl_{2}} = 0.0222 mol BaCl_{2} [/tex]

Moles of AgCl = [tex] 0.0222 mol BaCl_{2} * \frac{2 mol AgCl}{1 mol BaCl_{2}} [/tex] = 0.0444 mol AgCl

Mass of AgCl = [tex] 0.0444 mol AgCl * \frac{143.32 g AgCl}{1 mol AgCl} = 6.36 g AgCl [/tex]

When you squeeze an air-filled balloon, there are fewer collisions of air molecules against the wall of the balloon.

When you squeeze an air-filled balloon, there are fewer collisions of air molecules against the wall of the balloon.

When pressure is applied to the balloon by squeezing it, the volume of the balloon decreases. As a result, the air molecules inside become more crowded, reducing the number of collisions with the balloon wall.

So, the correct answer is B. There are fewer collisions of air molecules against the wall of the balloon.

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When you squeeze an air-filled balloon, you reduce the volume for the air molecules which results in more frequent collisions against the interior of the balloon.

Yes, your answer is correct. When you squeeze an air-filled balloon, you are effectively reducing the volume available for the air molecules inside it. As a result, there will be a noticeable increase in the number of collisions the air molecules have against the wall of the balloon (option A). This is concisely explained by the kinetic theory of gases, which states that gases consist of a large number of molecules that are in constant, random motion. So, under compression, the space for these random movements is limited, and therefore, they collide more frequently with the balloon's interior.

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A reservoir is a body of water. Towns and municipalities use it to capture water and it then seeps through the ground which cleans it and then it gets filtered into this processing thing and it then gets shipped out in pipes to peoples houses. A reservoir is a broad term though. You can have a reservoir of a variety of things such as a reservoir of oil kind of like a surplus where you can store it but in most cases, people talk about water.

In a chemical reaction, the limiting reagent is the chemical being used up while the excess reactant is the chemical left after the reaction process.

Before calculating the limiting and excess reactant, it is important to balance the equation first by stoichiometry.

C25N3H30Cl + NaOH = C25N3H30OH + NaCl

Since the reaction is already balanced, we can now identify which is the limiting and excess reagent.

First, we need to determine the number of moles of each chemical in the equation. This is crucial for determining the limiting and excess reagent.

Assuming that there is the same amount  of solution X for each reactant

1.0 M NaOH ( X ) = 1.0 moles NaOH

1.00 x 10-5 M C25N3H30Cl ( X ) = 1.00 x 10-5 moles C25N3H30Cl

The result showed that the crystal violet has lesser amount than NaOH. Thus, the limiting reactant in this chemical reaction is crystal violet and the excess reactant is NaOH.

To identify the limiting reagent between sodium hydroxide and crystal violet, normally one would need the reaction equation and stoichiometry. Assuming a hypothetical 1:1 reaction, crystal violet would be the limiting reagent due to its much lower concentration compared to NaOH.

The question asks to identify the limiting reagent in a reaction between sodium hydroxide and crystal violet and calculate the amount of excess reagent remaining after the reaction is complete. Since the concentration of sodium hydroxide (NaOH) is 1.0 M and the concentration of crystal violet is 1.00E-5 M, we do not have the actual reaction equation or the stoichiometry to determine the limiting reagent directly. Normally, the limiting reagent is the one that will be completely consumed first during the reaction, based on the stoichiometry of the reaction. Typically, the number of moles of each reagent would be considered; the one with the fewer moles, per the reaction stoichiometry, would be the limiting reagent.

As an example, if the reaction were 1:1, then clearly crystal violet would be the limiting reagent due to its much lower concentration. To find out how much NaOH remains, we would subtract the moles of NaOH that reacted with crystal violet from the initial moles of NaOH. However, in absence of the reaction equation and assuming a hypothetical 1:1 reaction, all of the crystal violet would react, leaving behind an excess of NaOH. If we had information on the volumes used, we could calculate the actual amount of NaOH remaining using dimensional analysis as demonstrated by various examples provided.

2AlPO4 ( aq) + 3MgCl2 (aq) -> Mg3(PO4)2 (s) + 2AlCl3 (aq) 

Right answer is D
 

Answer : Yes, the given reaction is the complete balanced equation.

Explanation :

Balanced equation : It is defined as the number of individual atoms of an element present on the reactant side always be equal to the number of individual atoms of an element present on product side. That means the complete balanced equation are those which follows the law of conservation of mass.

When aluminum phosphate reacts with magnesium chloride in an aqueous solution then it gives magnesium phosphate and aluminium chloride as a product.

The complete balanced equation will be,

[tex]2AlPO_4(aq)+3MgCl_2(aq)\rightarrow Mg_3(PO_4)_2(s)+2AlCl_3(aq)[/tex]

By the stoichiometry, 2 moles of aluminium phosphate react with 3 moles of magnesium chloride to give 1 mole of magnesium phosphate and 2 moles of aluminium chloride.

Changes in temperature and changes in pressure are two circumstances that can cause a system in equilibrium to change the concentrations of its reactants or products.

Two circumstances that can cause a system in equilibrium to change the concentrations of its reactants or products are changes in temperature and changes in pressure. Let's take temperature as an example. If the temperature of a system at equilibrium is increased, the reaction will shift in the direction that consumes heat, causing the concentrations of reactants and products to change. Similarly, if the temperature is decreased, the reaction will shift in the direction that releases heat, resulting in changes in concentrations as well.

The correct option is B: 14/35 × 100.

To determine the percent composition by mass of nitrogen in NH₄OH, we need to use the molar mass of each constituent element and the total formula mass:

(Mass of Nitrogen / Molar Mass of NH₄OH) × 100

Using the given values:

(14 g/mol / 35 g/mol) × 100 = 40%

Thus, the correct option is B: 14/35 × 100.

The  atomic  number of the atom represent by the model is

8 (answer B)

 Explanation.

 The atomic number  of atom is 8 because the   model below has  a total  number of 8 electrons.

 that  is;  2 electrons in the  inner energy level  and 6 electrons in the outermost energy level which make  a total of 2+6= 8 electrons

Answer: The the atomic number of this atom is 8.

Explanation: The particles with zero charge represents neutrons, The particles with negative charge represents electrons and the particles with positive charge represents protons.

The number of particles with negative charge are eight in number, thus there are eight electrons. The number of particles with positive charge are eight in number, thus there are eight protons. Thus it is a neutral atom.

The number of electrons represent the atomic number for a neutral atom. Thus the element has atomic number of 8.

From the equation kb = 1.8 x 10^-5

Therefore;

pKb = - log 1.8 x 10^-5 = 4.7

Moles NH3 in 75 ml = (75/1000) L x 0.200 M=0.0150

Moles HNO3 in 19 ml = (19/1000) L x 0.500 M= 0.0095

The net reaction is

NH3 + H+ = NH4+

Moles NH3 in excess = 0.0150 - 0.0095 =0.0055

Moles NH4+ formed = 0.0095

Total volume = 75.0 + 19.0 = 94.0 mL = 0.094 L

[NH3]= 0.0055/ 0.094 L=0.0585 M

[NH4+] = 0.0095/ 0.094 L = 0.1011M

pOH = pKb + log [NH4+]/ [NH3]

= 4.7 + log 0.1011/ 0.0585

= 4.938

pH = 14 - pOH

= 14 – 4.94

 =9.06

pH after the addition of 19.0 ml of HNO₃ : 9.018

The pH value of a reaction between strong acid HNO₃ and weak base NH₃ can be estimated from the rest of the reaction product

1. If the remainder of the reaction results obtained the remaining strong acid HNO₃, the pH is sought from the concentration of [H⁺] using the formula

[H⁺] = a. M

a = valence of acid / amount of H⁺ released

M = acid concentration

2. When strong acids and weak bases react, they form salts that are acidic, calculating the pH using the pH hydrolysis formula

[tex]\displaystyle [H +]=\sqrt{\frac{Kw}{Kb}.M }[/tex]

where

M = concentration of salt anion

3. If the remainder of the reaction results are a weak base remaining and the salt, the solution will form a base buffer solution and search for pH using the formula base buffer pH

[tex]\displaystyle [OH-]=Kb\times\frac{weak\:base\:mole}{salt\:mole\times valence}[/tex]

We count the moles of each reactant:

NH₃ mole = 75 ml x 0.2 M = 15 mlmol

mole HNO₃ = 19 ml x 0.5 M = 9.5 mlmol

NH₃ + HNO₃ ---> NH₄NO₃

15        9.5    

9.5      9.5            9.5

5.5       0              9.5

so there are remaining weak bases NH₃ = 15 - 9.5 = 5.5 mmol

Then a buffer solution is formed

[OH⁻] = Kb x [weak base mole] / [salt mole x valence]

[tex]\displaystyle [OH-]=1.8.10^{-5}\times\frac{5.5}{9.5\times 1}\\\\pOH=-log\:1.042.10^{-5}\\\\pOH=4.982\\\\pH=14-pOH\\\\pH=9.018[/tex]

the pH of a solution

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the pH of a 2.0 M solution of HClO4

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Keywords : pH, acid, base, HNO₃,NH₃

3.65

Given:

Question:

What would be the final ph if 0.0100 moles of solid NaOH were added to this buffer?

The Process:

Step-1

Let us prepare all the moles of substances.

[tex]\boxed{ \ n = MV \ }[/tex]  

Moles of HCOOH =  

[tex]\boxed{ \ 0.600 \ \frac{mol}{L} \times 100 \ ml = 60 \ mmol \ }[/tex]

Moles of HCOONa =  

[tex]\boxed{ \ 0.300 \ \frac{mol}{L} \times 100 \ ml = 30 \ mmol \ }[/tex]

Moles of NaOH =

[tex]\boxed{ \ 0.0100 \ moles = 10 \ mmol \ }[/tex]

Step-2

Let use the ICE table (in mmol).

                  [tex]\boxed{ \ HCOOH_{(aq)} + NaOH_{(s)} \rightarrow HCOONa_{(aq)} + H_2O_{(l)} \ }[/tex]

Initial:                  60                  10                   30                   -

Change:             -10                 -10                 +10                +10

Equlibrium:       50                   -                    40                  10

NaOH as a strong base acts as a limiting reagent.

The remaining HCOOH as a weak acid and HCOONa salt forms an acidic buffer system.

The HCOONa salt has valence = 1 according to the number of HCOO⁻ ions as a weak part, i.e.,  [tex]\boxed{ \ HCOONa \rightleftharpoons HCOO^- + Na^+ \ }[/tex]

HCOOH and HCOO⁻ are conjugate acid-base pairs.

Step-3

To calculate the specific pH of a given buffer, we need using The Henderson-Hasselbalch equation for acidic buffers:

[tex]\boxed{ \ pH = pK_a + log\frac{[A^-]}{[HA]} \ }[/tex]

where,  

[tex]\boxed{ \ pH = pK_a + log\frac{[HCOO^-]}{[HCOOH]} \ }[/tex]

[tex]\boxed{ \ pH = -log(1.8 \times 10^{-4}) + log \Big(\frac{40}{50}\Big) \ }[/tex]

[tex]\boxed{ \ pH = 4 - log \ 1.8 - 0.0969 \ }[/tex]

[tex]\boxed{ \ pH = 4 - 0.2553 - 0.0969 \ }[/tex]

[tex]\boxed{ \ pH = 3.65 \ }[/tex]

Thus, the pH of this buffer equal to 3.65.

_ _ _ _ _ _ _ _ _ _

What if we calculate the buffer pH value before the addition of NaOH?

Moles of HCOOH =  60 mmol

Moles of HCOONa =  30 mmol

[tex]\boxed{ \ pH = pK_a + log\frac{[HCOO^-]}{[HCOOH]} \ }[/tex]

[tex]\boxed{ \ pH = -log(1.8 \times 10^{-4}) + log \Big(\frac{30}{60}\Big) \ }[/tex]

[tex]\boxed{ \ pH = 4 - log \ 1.8 - 0.301 \ }[/tex]

[tex]\boxed{ \ pH = 4 - 0.2553 - 0.301 \ }[/tex]

[tex]\boxed{ \ pH = 3.44 \ }[/tex]

Thus, the initial pH of this buffer is 3.44. This proves the nature of the buffer that keeps the pH value relatively unchanged with the addition of a strong electrolyte.

In a phase change diagram, the arrows representing an increase in kinetic energy of particles are those showing endothermic transitions: melting, vaporization, and sublimation, which are marked by purple arrows.

Phase changes involve a change in kinetic energy of particles. When particles transition from solid to liquid, liquid to gas, or solid to gas, their kinetic energy increases. These changes are represented by the purple arrows in the diagram.

In a phase change diagram, the arrows that represent changes where  the kinetic energy of the particles increases are those that indicate heating: from solid to liquid, liquid to gas, and solid to gas. These transitions are endothermic, meaning they require an input of energy from the surroundings. When a substance changes from a solid to a liquid (melting), from a liquid to a gas (vaporization), or directly from a solid to a gas (sublimation), the particles gain kinetic energy, allowing them to move more freely and overcome the intermolecular forces holding them together in a more ordered state.

On the phase diagram, these changes are denoted by the purple arrows. Conversely, transitions from a less ordered state to a more ordered state (freezing, condensation, and deposition), indicated by the green arrows, are exothermic. This means energy is released as particles lose kinetic energy and move into a more stable, ordered state.