HURRY!!!! TIMED!!! WILL GIVE BRAINLIST!!!

A 10.0 g sample containing calcium carbonate (CaCO3) and an unreactive material was placed in excess hydrochloric acid, HCl. A reaction occurred, producing calcium chloride (CaCl2), water, and carbon dioxide. (a) write a balanced equation for the reaction (b) when the reaction was complete, 800 mL of carbon dioxide gas was collected. How many moles of calcium carbonate were used in the creation? (c) How many grams of CaCO3 were used? (d) If there was another contaminant in the sample that was not unreactive, would this have caused the percent yield of carbon dioxide to be higher, lower, or the same, explain your answer.

Answer:

The four substances recycled during photosynthesis and respiration are carbon dioxide, water, oxygen, and glucose.

So because you don't have all these choices, just pick, carbon, hydrogen, and oxygen, which is the last answer.

Thanks!

Mark me brainliest!

~\(MiniMeteorologist\\\)~

Answer:

carbon hydrogen and oxygen

Explanation:

Answer:
The answers are:

1. Gas particles act like tiny, solid spheres.
2. Gas particles are in constant, random motion.
4. Collisions are elastic, there is no energy lost as the particle hits the sides of the container.

Hope this helps!

According to the Kinetic Molecular Theory, the statements that are true of the gas particles are:

The Kinetic Molecular Theory actually states that the particles of gas are in motion constantly and their collisions are perfectly elastic.

This theory is actually used to give explanations for Boyle's and Charles' laws.

According to the Kinetic Molecular Theory, the collisions of the gas are elastic and are in constant motion.

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Answer:

The dirty dozen.

Explanation:

This is mainly seen in the control of hazardous waste and its regulations through treaty in the control of persistent organic pollutant(POPs). This regulation is seen to occur in 12 widely used persistent organic pollutants; these are seen to have the ability to absorb and store tissues that make them fatty especially in humans and also animals found in the higher trophic levels or in the food web. That being said, it can be noted that the dirty dozen can move or even attain hundreds, or thousands of levels towards the environ and also in other cases, transporting them through air or water as the case may be.

An explosión happens when asteroids collide.

When asteroids collide, little bits are expelled at orbital speeds that are slower than the original asteroids'.

When asteroids collide, the ejected pieces have speeds that are low compared to the original asteroid's orbital speed. As a result, the pieces' orbits are near to one another, creating a "family" of smaller asteroids.

A massive impact between two asteroids may have caused the Earth to enter an ice age some 466 million years ago. The asteroid that was destroyed by the cosmic collision between Mars and Jupiter, which was around 93 miles (150 km) broad, was covered in a dense cloud of dust that travelled across the inner solar system.

The asteroids that might destroy the world if they struck the Earth are exceedingly uncommon. They would most likely need to be around a kilometre apart.

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Option C, a more saturated fat is the product(s) for the hydrogenation of corn oil.

Organic molecules are frequently reduced or saturated by the process of hydrogenation, which is a chemical reaction between hydrogen (H2) and another substance or element. The process of incorporating hydrogen molecules into the various carbon-carbon bonds is known as hydrogenation. Saturated fats emerge from this process. Double bonds change into single bonds throughout this Saturated fats, and the melting point steadily increases.

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Answer: I would love to help but i cant see the table

Explanation:

Jana can modify her device by increasing the amount of zinc and decreasing the amount of copper sulfate to generate more heat, which will increase the water temperature within the desired range of 70°C to 100°C.

How could Jana modify her design to meet the temperature constraints?

By making these modifications, Jana can increase the water temperature to the desired range and ensure that her device is portable, safe, and effective.

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The reaction equations for acetic acid-acetate buffer reacting with an acid and a base are as follows:
Reaction with Acid:
\(CH_{3} COOH\) + HX → \(CH_{3} COOH\)  + \(X^{-}\)  + \(H^{+}\)
Reaction with Base:
\(CH_{3} COOH\)  + YOH → \(CH_{3}COO ^{-}\) + \(Y^{+}\) + \(H_{2}O\)

To explain how an acetic acid-acetate buffer reacts with an acid and a base, we can write the following reaction equations:

1. Reaction of the buffer with an acid:
Buffer: \(CH_{3} COOH\) (acetic acid) /  \(CH_{3}COO ^{-}\)  (acetate ion)
Acid: HX (general acid)

Reaction equation:
\(CH_{3}COO ^{-}\)  + HX → \(CH_{3} COOH\) +  \(X^{-}\)

In this reaction, the acetate ion (CH3COO-) reacts with the added acid (HX) to form acetic acid (CH3COOH) and the corresponding anion of the added acid ( \(X^{-}\)). This helps neutralize the added acid.

2. Reaction of the buffer with a base:
Buffer: \(CH_{3} COOH\) (acetic acid) /  \(CH_{3}COO ^{-}\)  (acetate ion)
Base: YOH (general base)

Reaction equation:
\(CH_{3} COOH\)+ YOH →  \(CH_{3}COO ^{-}\)  + \(H_{2}O\) + \(Y^{+}\)

In this reaction, acetic acid (\(CH_{3} COOH\)) reacts with the added base (YOH) to form acetate ion ( \(CH_{3}COO ^{-}\) ), water (\(H_{2}O\)), and the corresponding cation of the added base (\(Y^{+}\) ). This helps neutralize the added base.

In both cases, the acetic acid-acetate buffer is able to maintain the pH of the solution by reacting with added acids or bases.

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The specific heat capacity of the unknown metal is 344 J/kg⁰C. And the type of metal that is the unknown sample is silver.

To answer this question, use the following formula:

                myCy ΔT= mₓ. Cₓ.ΔT

Information:

mₓ = Unknown mass of metal

ΔTₓ = temperature of unknown metal

my = Mass of water

Cy = Density of water

ΔTy = temperature of water

Given the values ​​for this question.

mₓ = 0.75 kg

ΔTₓ = (89.4⁰C – 0⁰C) = 89.4⁰C

Tₓ = 89.4⁰C

me = 1.25 kg

Cy = 4200J/kg⁰C

ΔTy = (24.4°C - 20.0°C) = 4.4°C

Then the specific heat capacity of the unknown metal is:

Cy ΔTy = Mₓ. Cₓ.ΔTx

Cₓ = Cy ΔTy/ Mₓ.ΔTx

Cₓ =1.25 x 4200 x 4.4/0.75 x 89.4

Cₓ = 23100/67.05

Cₓ = 344 J/kg⁰C

The specific heat of an unknown metal is silver.

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The KSP is approximately 1.30x10⁻¹⁴ if the pH of a saturated solution of M(OH)₂ is 10.648.

Hence, the correct option is a.

To calculate the KSP (solubility product constant) of M(OH)₂ using the given pH of the saturated solution, we need to convert the pH value to the hydroxide ion concentration ([OH⁻]). The hydroxide ion concentration can then be used to determine the concentration of the metal cation (M₂⁺) in the saturated solution.

The pH is defined as the negative logarithm of the hydrogen ion concentration ([H⁺]). Since M(OH)₂ is a strong base, it will dissociate completely in water, producing two hydroxide ions for every one M₂⁺ ion. Thus, the concentration of hydroxide ions will be twice the concentration of M₂⁺ ions.

Using the given pH, we can calculate the hydroxide ion concentration as follows

pOH = 14 - pH

pOH = 14 - 10.648

pOH ≈ 3.352

[OH⁻] = \(10^{(-pOH)}\)

[OH⁻] = \(10^{(-3.352)}\)

Now, since the concentration of hydroxide ions ([OH⁻]) is twice the concentration of M2⁺ ions in the saturated solution, we can write

[OH⁻] = 2[M₂⁺]

Now, let's substitute the value of [OH⁻] into the equation

\(10^{(-3.352)}\) = 2[M₂⁺]

Solving for [M₂⁺], we get

[M₂⁺] = \(10^{(-3.352)}\) / 2

Finally, we can calculate the KSP

KSP = [M₂⁺][OH⁻]²

KSP = ([M₂⁺])(2[M₂⁺])²

KSP = (\(10^{(-3.352)}\)/ 2)(2 × \(10^{(-3.352)}\))²

Evaluating the expression, the KSP is approximately 1.30x10⁻¹⁴

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To ensure safety while carrying out her experiment, Maria should follow proper laboratory protocols and take necessary precautions like handle equipment and glassware carefully,wear appropriate personal protective equipment (PPE),read safety data sheets (SDS) to understand the proper handling,never work alone in the laboratory,etc.

Here are some recommendations:

1.Wear appropriate personal protective equipment (PPE), including safety goggles, lab coat, and gloves, to protect her eyes, body, and hands from potential hazards.

2.Familiarize herself with the properties and potential risks associated with the chemicals she will be working with. Read safety data sheets (SDS) to understand the proper handling, storage, and disposal procedures.

3.Work in a well-ventilated area or use a fume hood to minimize exposure to any fumes or gases that may be generated during the experiment.

4.Handle equipment and glassware carefully to prevent accidents or breakages. Ensure that all equipment is in good condition and appropriate for the experiment.

5.Follow proper techniques for measuring, mixing, and transferring chemicals to prevent spills or splashes. Use appropriate containers and labels to clearly identify and store chemicals.

6.Be aware of emergency procedures, including the location of safety equipment such as fire extinguishers, eyewash stations, and emergency showers.

7.Never work alone in the laboratory. Inform a supervisor or lab mate about the experiment and maintain communication throughout the process.

By following these safety measures, Maria can minimize risks and ensure a safe working environment during her experiment.

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when ionic salts are dissolved in water, then the temperature of the resulting solution increases.

Amount of the temperature increases depending on the amount of salt dissolved in the water.

Ionic salts and different ionic compounds like will dissolve quicker the hotter the water it's far dissolved in. this is due to the fact hot temperatures make atoms pass quicker and the quicker they circulate, the easier they come apart.

The procedure of dissolving is endothermic whilst much less strength is launched when water molecules “bond” to the solute than is used to tug the solute apart. because less power is released than is used, the molecules of the answer flow more slowly, making the temperature lower.

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The value of electric flux is 2.45N.m²/C.

A flat sheet of paper of area 0.130 m² is oriented so that the normal to the sheet is at an angle of 56 ∘ to a uniform electric field of magnitude 18 N/C.

The electric flux through the paper is given by;

ϕ=E.A.cosθϕ

= Electric fluxE

= Electric field intensityA

= Area of the surfaceθ

= Angle between electric field intensity and normal to the surface

Given,E = 18 N/C A = 0.130 m² θ = 56°

The electric flux through the paper is :

ϕ=E.A.cosθϕ

= (18 N/C)(0.130 m²)cos56°ϕ

= 2.45 N.m²/C

The electric flux through any closed surface is proportional to the charge enclosed by the surface. The electric flux through an open surface can be calculated by multiplying the electric field intensity with the area of the surface and the cosine of the angle between the electric field intensity and the normal to the surface.

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Complete question:

A flat sheet of paper of area 0.130 m2 is oriented so that the normal to the sheet is at an angle of 56 ∘ to a uniform electric field of magnitude 18 N/C .Find the amount of elctric flux passing through the sheet?

The most likely result of millions of metal atoms crowding together so that molecular orbitals become combined is the  formation of bands in a crystal and is therefore denoted as option B.

This is referred to as a three dimensional description of the most likely location of an electron around an atom.

In a scenario where millions of metal atoms crowding together so that molecular orbitals become combined the energy levels of their valence electrons are modified and a collection of large number of closely spaced energy levels are formed thereby resulting in the  formation of bands in a crystal.

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Answer:

B: the formation of bands in a crystaL

Explanation:

Prediction of the exact energy difference between a secondary and tertiary carbocation is impossible.

It is not possible to predict the exact energy difference between a secondary and tertiary carbocation with certainty, as it can vary based on several factors such as the structure and stability of the carbocations, solvent effects, and other environmental factors.

However, it is generally accepted that tertiary carbocations are more stable than secondary carbocations, which in turn are more stable than primary carbocations.

This stability is due to hyperconjugation and other stabilizationizing mechanisms, which contribute to a decrease in the energy of the system.

The exact energy difference between secondary and tertiary carbocations can therefore be less than 27 kcal/mol, but the exact value would depend on the specific circumstances.

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2 Cr2O3 (s) = 4 Cr(s) + 3 O2(g)

The minimum temperature above which the given reaction will be product-favored (spontaneous) is 298 K.

For any chemical reaction to occur, the change in Gibbs free energy (∆G) is calculated. When the value of ∆G is negative, the reaction is spontaneous in nature, and when it is positive, the reaction is non-spontaneous. If the value of ∆G is zero, then the reaction is in equilibrium, and there is no spontaneous change.

The formula to calculate Gibbs free energy is:

∆G = ∆H - T∆S

Where ∆H is the enthalpy change of the reaction, T is the temperature, and ∆S is the entropy change of the reaction. If the temperature is increased, then the value of T∆S will also increase, and this will result in the decrease of ∆G. This can lead to the spontaneous nature of the reaction.

In the given reaction:

2 Cr2O3 (s) = 4 Cr(s) + 3 O2(g)

The enthalpy change of the reaction is:

∆H = [4 × enthalpy of Cr(s)] + [3 × enthalpy of O2(g)] - [2 × enthalpy of Cr2O3 (s)]

The value of enthalpy of Cr(s), O2(g), and Cr2O3 (s) can be taken from the standard enthalpy of formation table.

The entropy change of the reaction can be calculated as:

∆S = [4 × entropy of Cr(s)] + [3 × entropy of O2(g)] - [2 × entropy of Cr2O3 (s)]

The value of entropy of Cr(s), O2(g), and Cr2O3 (s) can be taken from the standard entropy table.

When the temperature is increased, the value of ∆S increases, and hence the value of T∆S will also increase. This will result in the decrease of ∆G, and the reaction will become spontaneous in nature.

So, the minimum temperature is 298 K.

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Answer:

84%

Explanation:

Fossil Fuels Supply 84 Percent Of Worlds Electricity

The balanced chemical equation is given as:

Cu(NO₃)₂ → CuO + 2 NO₂ + O₂

To balance the equation, we'll start by counting the atoms on each side:

Cu(NO₃)₂ → CuO + NO₂ + O₂

On the left-hand side (LHS):

Cu: 1

N: 2

O: 6

On the right-hand side (RHS):

Cu: 1

N: 1

O: 3

To balance the nitrogen (N) atoms, we need 2 NO₂ molecules on the right-hand side. This means the coefficient in front of NO₂ should be 2:

Cu(NO₃)₂ → CuO + 2 NO₂ + O₂

Now let's check the oxygen (O) atoms:

On the LHS: 6 O atoms from Cu(NO₃)₂

On the RHS: 2 O atoms from CuO + 2 O atoms from NO₂ + 2 O atoms from O₂

So we have a total of 6 O atoms on the LHS and 6 O atoms on the RHS. Therefore, the equation is already balanced.

The balanced chemical equation is:

Cu(NO₃)₂ → CuO + 2 NO₂ + O₂

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Answer:

Homogeneous

Explanation:

Answer:

i think people who take vitamins

Answer:

its d;grocery stores sell legumes

Explanation:

I just took the test

The concentration of free Ba2+ in the solution is 4.51 x\(10^{-8}\) M at pH 9.00.

Ba2+ + EDTA4- ⇌ [Ba(EDTA)]2-

The stability constant for this complex can be found in a table of stability constants, and for Ba(EDTA)2-, it is typically given as log Kf = 8.90.

At pH 9.00, the concentration of H+ ions in the solution is 10^-5 M, which we can use to calculate the concentration of OH- ions using the Kw expression:

Kw = [H+][OH-] = 1.0 x \(10^{-14}\)

[OH-] = Kw/[H+] = 1.0 x \(10^{-9}\) M

[Ba(EDTA)]2- ⇌ Ba2+ + EDTA4-

The equilibrium constant for this reaction can be expressed in terms of the stability constant as follows:

Kd = 1/Kf = 10^(-8.90)

At equilibrium, we can define the concentration of free Ba2+ as [Ba2+] and the concentration of the [Ba(EDTA)]2- complex as [Ba(EDTA)].

Then, we can write the mass balance equation as:

[Ba(EDTA)] + [Ba2+] = 0.060 M

[Ba2+] = [Ba(EDTA)] * Kd

[Ba2+] = (0.060 M - [Ba2+]) * Kd

Solving for [Ba2+], we get:

[Ba2+] = 4.51 x \(10^{-8}\)M

Concentration refers to the ability to focus one's attention and mental effort on a particular task or activity. It involves directing one's cognitive resources toward a specific goal or objective while ignoring distractions or irrelevant information. The level of concentration can vary depending on the nature of the task and the individual's cognitive abilities.

For example, tasks that require sustained attention, such as studying or reading, may require a higher level of concentration than tasks that are more automatic, such as walking or eating. Concentration is essential for effective learning, problem-solving, and decision-making. It enables individuals to process information more efficiently, retain it for longer periods, and make connections between different pieces of information. It also helps to reduce errors and improve performance.

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The second buret is a necessary component of this titration experiment as it allows you to accurately measure the amount of HCl needed to reach the endpoint. It is also necessary to accurately calculate the amount of NaOH delivered in the reaction.

The purpose of the second buret in this experiment is to measure the amount of hydrochloric acid (HCl) needed to reach the endpoint of the titration. This is necessary because the concentration of the hydrochloric acid is unknown. By using a second buret to measure the HCl, it allows you to accurately titrate the NaOH solution until the solution in the Erlenmeyer flask changes color, indicating the endpoint of the titration. This measurement also allows you to calculate the amount of NaOH delivered in the reaction. In order to use a second buret for the experiment, it should be filled with the HCl solution and placed above the Erlenmeyer flask. To start, you should open the valve at the top of the buret, allowing the HCl to begin to flow into the Erlenmeyer flask. Then, you should slowly add the HCl until the solution in the flask changes color, which indicates the endpoint of the titration. After that, you should record the FINAL HCl reading from the buret and calculate the HCl delivered (mL) by subtracting the initial reading from the final reading.

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Answer:

4:D 5:D

Explanation:

The mass (1.055×10⁻²² g) of a single copper atom is the same as  1.055×10⁻²⁵ Kg (3rd option)

To know which options are correct, we shall convert 1.055×10⁻²² g to the units given in the options. Details below:

1.055×10⁻²² g to mg

1 g = 1000 mg

Therefore

1.055×10⁻²² g = 1.055×10⁻²² × 1000

1.055×10⁻²² g = 1.055×10⁻¹⁹ mg

1.055×10⁻²² g to μg

1 g = 10⁶ μg

Therefore

1.055×10⁻²² g = 1.055×10⁻²² × 10⁶

1.055×10⁻²² g = 1.055×10⁻¹⁶ μg

1.055×10⁻²² g to Kg

1 g = 0.001 Kg

Therefore

1.055×10⁻²² g = 1.055×10⁻²² × 0.001

1.055×10⁻²² g = 1.055×10⁻²⁵ Kg

1.055×10⁻²² g to ng

1 g = 10⁹ ng

Therefore

1.055×10⁻²² g = 1.055×10⁻²² × 10⁹

1.055×10⁻²² g = 1.055×10⁻¹³ ng

Considering the options given, we can see that 1.055×10⁻²² g is the same as 1.055×10⁻²⁵ Kg (3rd option)

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It is important to mark the solvent level on the chromatography paper as soon as you remove it from the petri dish in a chromatography of food dyes lab because if the solvent level is not marked as soon as possible, the solvent front can evaporate causing the results to be inaccurate.

Chromatography is a laboratory technique for separating a mixture into its individual components. The mixture is dissolved in a solvent and then placed in contact with a stationary phase. The components of the mixture are then separated based on their individual interactions with the stationary phase and the solvent. Chromatography of food dyes is a lab that is used to separate different food dyes that are present in a sample.

The sample is placed on chromatography paper which is then placed in a petri dish containing a solvent. As the solvent moves up the chromatography paper, the different dyes in the sample are separated based on their individual interactions with the paper and the solvent.

In a chromatography of food dyes lab, it is important to mark the solvent level on the chromatography paper as soon as it is removed from the petri dish because the solvent front can evaporate causing the results to be inaccurate. If the solvent front evaporates, the distance traveled by the different dyes will be shorter, making it appear as though they are less separated than they actually are.

By marking the solvent level as soon as possible, the distance traveled by the different dyes can be accurately measured, and the results will be more accurate.

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The reason why it is important to mark the solvent level on the chromatography paper as soon as you remove it from the petri dish is that the solvent level must be measured to calculate the Rf value. The Rf value is a way to quantify how far a particular compound travels in chromatography.

It is calculated as the distance traveled by the compound divided by the distance traveled by the solvent.The chromatography of food dyes lab is a experiment that aims to identify the dyes used in food products by using paper chromatography. The procedure includes: Cut a strip of chromatography paper and mark the solvent level using a pencil as soon as you remove it from the petri dish; prepare the chromatography solvent by mixing rubbing alcohol with water; then, spot the dyes on the chromatography paper using toothpicks or capillary tubes.

Afterwards, place the paper in the petri dish containing the solvent, making sure that the dyes do not touch the solvent, and cover it. Allow the solvent to travel up the paper until it reaches the solvent level mark. Once the solvent level has reached the mark, remove the paper from the petri dish and allow it to dry before analyzing the results.

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