someone pls help rn!!

The experiment looks into how temperature impacts how quickly magnesium metal reacts with hydrochloric acid. The creation of hydrogen gas over time is used to determine the dependent variable, which is the reaction rate.

The experiment can be run at various temperatures to change the experiment's independent variable, which is the reaction mixture's temperature. Higher temperatures cause molecules to move more quickly, increasing the likelihood of reactant collisions and kinetic energy. By enabling more successful collisions between magnesium atoms and hydrochloric acid, can quicken the reaction and increase the amount of hydrogen gas produced. In contrast, it is anticipated that the reaction rate will be slower at lower temperatures since there will be less molecular mobility and fewer successful collisions.

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--The complete Question is, What is the effect of temperature on the rate of reaction between hydrochloric acid and magnesium metal?

The dependent variable in this experiment is the rate of reaction, which can be measured by monitoring the production of hydrogen gas over time. The independent variable is the temperature of the reaction mixture, which can be controlled and varied by conducting the experiment at different temperatures.--

Answer:

paper being shredded

Explanation:

Answer:

its A

Explanation:

Answer:

methane

Explanation: methane is obtained from the decaying of flora and fauna mostlyunder damp

Incomplete question. Answered from a general science perspective.

Answer:

a physical change

Explanation:

It is important that to remember that when the ice cream melts as a result of the heat from the sun, the chemical elements that were mixed together to form the tasteful ice do not change, rather only the physical state of the ice cream changes.

For example, rather than being in mostly solid form, because of the hot temperature of the day, it becomes a liquid (which is rightly a physical change).

Pond water is poured through a paper cone Filtration, Water containing fine sand is allowed to stand for an hour Sedimentation,Hot tea turns cold due to this method Cooling.

Filtration is a process used to separate particles and substances from a mixture or solution. It is commonly used in water treatment, pharmaceutical production, and many other industrial processes. Filtration works by passing the mixture or solution through a porous material or filter medium, which traps the particles and substances while allowing the fluid or air to pass through. Depending on the type of filter used, different sized particles can be filtered out. Some of the most common filtration methods are sedimentation, screening, and centrifugal filtration. Sedimentation involves the settling of particles in a water or other liquid, usually caused by gravity. Screening utilizes a mesh or sieve to separate particles of different sizes.

Pond water is poured through a paper cone Filtration, Water containing fine sand is allowed to stand for an hour Sedimentation,Hot tea turns cold due to this method Cooling,Water used in injections is prepared by this method Distillation,Method used at home for killing germs in drinking water is Boiling.

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Answer:

Condensation

Explanation:

Zoe is quite keen to have noticed what we call condensation. Air contains many components, one of those being water vapor. Like how sugar is soluble in water, water can be said to be "soluble" in air. Water will evaporate into the air to a certain extent. The higher the temperature of the air, the more water the air can hold. If the air has more water that it can hold (potentially because of a temperature decrease), the extra water will come out of the air. Zoe's water bottle was cold, and because the air around Zoe's bottle had cooled down, the air can not hold as much water as it could when it was warm, so the air deposited the extra water in the form of liquid water onto the bottle, giving the illusion that her bottle was sweating.

The amount, in mL, of the suspension that should be given to an infant weighing 17 lb will be 1.16 mL

0.8 mL of the liquid contains 80.0 mg of the drug.

The recommended dose is 15 mg per kg of body weight

The infant to be given the drug weighs 17 lb.

First, let's convert the weight of the infant to kg.

1 lb = 453.59 g

17 lb =       453.59 x 17/1

                   = 7711.03 g

1000 g = 1 kg

7711.03 g = 7711.03 x 1/1000

               = 7.711 kg

So, the baby's weight is 7.711 kg.

The drug dose for the baby can thus be calculated as:

15 mg x 7.711 = 115.67 mg

But 0.8 mL of the drug contains only 80.0 mg. How many mL will contain 115.67 mg?

                   0.8 x 115.67/ 80.0 = 1.16 mL

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Answer:

0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl

Explanation:

Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.

Then, the more concentrated acid or base will be more acidic or basic.

CH3COOH. Weak acid

NaOH. Strong base

H2SO4. Strong acid

NH3. Weak base.

HCl. Strong acid

The less acid (More basic):

Strong base, weak base, weak acid, diluted strong acid, undiluted strong acid

Answer:

please find the attached file.

Explanation:

Answer: Electrons are taken up by \(PbO_2\) and they are lost by \(HCl\)

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously. It is also called the reaction where the exchange of electrons takes place.

An oxidation reaction is defined as the reaction in which a chemical species loses electrons takes place. In this reaction, the oxidation state of a substance gets increased.

A reduction reaction is defined as the reaction in which a chemical species gains electrons takes place. In this reaction, the oxidation state of a substance gets reduced.

For the given chemical reaction:

\(PbO_2+4HCl\rightarrow PbCl_2+Cl_2+2H_2O\)

The half-reactions for this redox rection follows:

Oxidation half-reaction:  \(2HCl\rightarrow ClO_2 + 2e^-\)

Reduction half-reaction:  \(PbO_2+2e^-\rightarrow PbCl_2\)

Hence, electrons are taken up by \(PbO_2\) and they are lost by \(HCl\)

At pOH value  of 6.0 the pH value of the following solution is 8.0 and the concentration of [\(H^{+}\) ] ion is \(10^{-8}\)

In this question we will apply the formula

      pH +pOH = 14 . . . . . . . . . . . . .(1)

where pH = concentration of [\(H^{+}\) ] ion

pOH = concentration of [\(OH^{-}\) ] ion

As per the question

pOH =6.0

Putting the value of pOH in equation (1) we get the value of pH

              pH + 6.0 =14

             pH = 14 -6.0

             pH  = 8.0

 The value of pH if the pOH value is 6.0 is 8.0

To find the concentration of \(H^{+}\) ion we will use the following formula

This is calculated by the formula

                                 [\(H^{+}\)} = \(10^{-pH}\)

where we will write the values of pH

Hence the concentration of [\(H^{+}\)} ion is \(10^{-8}\)

Therefore at  pOH of 6.0 the pH value of the following solution is 8.0 and the concentration of [\(H^{+}\) ] ion is \(10^{-8}\)

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The complete question is -

What is the pH value and concentration of [\(H^{+}\) ] ion of the following if the pOH value of the solution is 6.0 ?

Intermediate is a  high energy substance that exists for a short time in The process of the reaction.

Transition state : The point where bonds break or form.

1. Intermediate : An intermediate is a molecule that is formed from two or more reactants and then reacts further to give products.

2. Transition state : A very short-lived configuration of atoms at a local energy maximum in a reaction-energy diagram.

3. Activation energy : The minimum energy required to cause a process (Such as a chemical reaction) to occur.

4. Rate determining step : The slowest step of a chemical reaction that Determines the speed (rate) at which the overall reaction proceeds.

5. Exothermic : A chemical reaction that involves the release of energy in The form of heat or light.

6. Catalyst : A substance that changes the mechanism, but is not itself Changed by the reaction.

7. Chemical Kinetics : Governs the rate of a reaction (speed of the Reaction).

8. Thermodynamic control : Governs the extent of a reaction (Spontaneous).

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The mass of NaCl in 44mL of a 1.6M solution is 4.1grams.

The mass of NaCl can be calculated by using the following formula;

Molarity = no. of moles ÷ volume

Molarity is the concentration of a substance in solution, expressed as the number moles of solute per litre of solution.

no of moles = 1.6M × 0.044L = 0.07moles

The mass of the sodium chloride solution can be calculated by multiplying the molar mass of the salt by the number of moles present in the solution.

The molar mass of NaCl solution can be calculated as follows;

M.M = 23g/mol + 35.5g/mol = 58.5g/mol

mass of NaCl = 58.5 × 0.07 = 4.1grams.

Therefore, 4.1grams is the mass of NaCl solution.

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The mass of water that can be heated from 15.0° C to 43.1° C by the addition of 1282J is 10.9grams.

The mass of water can be calculated using the following expression;

Q = mc∆T

Where;

According to this question, a sample of water can be heated from 15.0° C to 43.1° C by the addition of 1282J. The mass of the substance can be calculated as follows;

1282 = m × 4.184 × (43.1 - 15)

1282 = 117.57m

m = 10.9grams

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The equilibrium expression can be written as follows:

K = [CaCO₃] × [H₂]⁴

------------------

[CaO] × [CH₄] × [H₂O]²

In this equilibrium expression, the square brackets represent the concentrations of the species involved in the reaction, and the coefficients of the balanced equation indicate the stoichiometric coefficients of the corresponding species.

The concentration of a pure solid (CaCO₃ in this case) is not included in the equilibrium expression, as it remains constant throughout the reaction.

The equilibrium constant (K) represents the ratio of the product concentrations to the reactant concentrations at equilibrium, each raised to the power of their respective stoichiometric coefficients. The specific values of these concentrations depend on the initial conditions, and K remains constant as long as the temperature is unchanged.

It is important to note that the equilibrium constant expression is written based on the balanced chemical equation. The stoichiometric coefficients determine the relationship between the concentrations of reactants and products, allowing us to express the equilibrium state quantitatively using the equilibrium expression.

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Answer: As they develop theories, chemists use models to attempt to explain their findings. Chemists assess the model they are using as new evidence becomes available and, if required, continue to refine it by making modifications.

Explanation:

Answer:

pH = 5.35

Explanation:

Given 1.60 grams sodium acetate (NaOAc(aq))*** added to 50ml of 0.10M acetic acid (HOAc(aq)) solution.

Applying common ion effect keeping in mind that the addition of NaOAc provides the common-ion (OAc⁻).

      HOAc(aq) ⇄          H⁺(aq)            +                 OAc⁻(aq)

 I      0.10m          1.32 x 10⁻³M ≈ ∅M*     (1.6g/82.03g/mol) / 0.050L = 0.39M

 C       -x                         +x                                0.39M + x ≈ 0.39M**

 E    0.10M - x                  x                                    0.39M

       ≈ 0.10M

Ka = [H⁺][OAC⁻]/[HOAC] => [H⁺] = Ka·[HOAc] / [OAc⁻]

[H⁺] = (1.75 X 10⁻⁵)(0.10) / (0.39) = 4.5 x 10⁻⁶M

∴ pH = -log[H⁺] = -log(4.5 x 10⁻⁶) = -(-5.35) = 5.35

_______________________________________________

* [H⁺] before adding NaOAc = SqrRt(Ka · [HOAc]) = SqrRt(1.75 x 10⁻⁵· 0.10) = 1.32 x 10⁻³M. Since this concentration value is so small, the initial [H⁺] is assumed to be zero molar (∅M).

** The added [H⁺] is negligible and dropped in the ICE table. That is, adding ~[H⁺] in the order of 10⁻³M does not change the H⁺ ion concentration sufficiently to affect problem outcome and is therefore dropped in the ICE table.

*** Acetic Acid and Sodium Acetate are frequently written HOAc and NaOAc where the OAc⁻ anion is the acetate ion (CH₃COO⁻) for brevity.

The pH of the solution measures the acid of the liquid throughout the molar concentration of hydrogen ions in a solution.

\(\bold{CH_3COONa \ mass = 1.60\ g}\\\\\)

Solution Volume (V) \(=50.0\ mL =0.05\ L\\\\\)

\(\bold{CH3COONa \ molarity =0.10\ M }\)

\(\bold{\text{Calculating the CH3COONa moles} =\frac{mass}{molar\ mass}}\)

                                                       \(= \frac{1.60\ g}{82.03\frac{g}{mol}} \\\\=0.0195\ mol\\\\\)

\(\bold{M = \frac{ \text{amount of solute moles} } { \text{solution volume in L} }}\)

     \(= \frac{0.0195\ mol}{0.05\ L}\\\\ =0.39\ M\)

Using Henderson Hasselbach equation:

\(\bold{pH = -\log K_{a} + \log{[salt]}{[acid]}}\\\\\)

     \(=\bold{ -\log (1.75\times 10^{-5}) + \log ( \frac{0.39}{0.10}) }\\\\=\bold{ 4.757 + \log (3.9)}\\\\=\bold{ 4.757 + 0.5910}\\\\=\bold{ 5.348}\\\\=\bold{ 5.35}\\\)

So, the final answer is "5.35".

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The concentration of CO2 in club soda is approximately 1.1964 g/L.

To find the concentration of CO2 in club soda, we need to use the stoichiometry of the reaction and the volume and concentration of the NaOH solution used.

The balanced equation for the reaction is:

CO2(aq) + 2NaOH(aq) → Na2CO3(aq)

From the stoichiometry of the equation, we can see that 1 mole of CO2 reacts with 2 moles of NaOH. Therefore, the moles of CO2 can be calculated using the volume and concentration of NaOH solution used.

Given that 32.14 mL of 0.04202 M NaOH solution was used, we can calculate the moles of NaOH:

moles of NaOH = volume (L) × concentration (M)

moles of NaOH = 32.14 mL × 0.04202 mol/L

moles of NaOH = 0.001351 mol

According to the stoichiometry of the equation, 1 mole of CO2 reacts with 2 moles of NaOH. Therefore, the moles of CO2 can be calculated as:

moles of CO2 = (moles of NaOH) / 2

moles of CO2 = 0.001351 mol / 2

moles of CO2 = 0.0006755 mol

Now, we need to convert the moles of CO2 to grams. The molar mass of CO2 is approximately 44.01 g/mol.

mass of CO2 = moles of CO2 × molar mass of CO2

mass of CO2 = 0.0006755 mol × 44.01 g/mol

mass of CO2 = 0.02979 g

Finally, we need to express the concentration of CO2 in club soda in g/L. We are given that the sample of club soda used is 25.00 mL.

concentration of CO2 = (mass of CO2) / (volume of club soda in L)

concentration of CO2 = 0.02979 g / (25.00 mL × 0.001 L/mL)

concentration of CO2 = 1.1964 g/L

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Answer:

2.) C₆H₆O₂

Explanation:

(Step 1)

To find the molecular formula of hydroquinone, you need to first determine the molar mass of the empirical formula. The molar mass is a sum of the individual atomic weights of each element multiplied by their quantities (denoted by the subscripts).

Atomic Weights:

Carbon (C): 12.011 g/mol

Hydrogen (H): 1.008 g/mol

Oxygen (O): 15.998 g/mol

Molar Mass (C₃H₃O): 3(12.011 g/mol) + 3(1.008 g/mol) + 15.998 g/mol

Molar Mass (C₃H₃O): 55.055 g/mol

(Step 2)

As you can see, the molar mass of the empirical formula is not the same as the molar mass of the molecular formula. Rather, the empirical molar mass is approximately half of the molecular molar mass. Therefore, to make these masses equal, we need to double the empirical formula. This can be done by multiplying all of the subscripts by 2.

55.055 g/mol x 2 = 110.11 g/mol

110.11 g/mol = 110.108 g/mol

C₃ x 2 = C₆

H₃ x 2 = H₆

O x 2 = O₂

Therefore, the correct molecular formula of hydroquinone is C₆H₆O₂.

The three main components of electrolytic cells are cathode, anode and electrolyte. The negatively charged electrolytic cells are cathode and the positively charged electrolytic cells are anode.

An electrolytic cell can be defined as the electrochemical device which uses the electrical energy to perform a non-spontaneous redox reaction. They are mainly used for the electrolysis of certain compounds.

Here the anode cell is:

3H₂O (l) + I⁻ (aq) → IO⁻₃ (aq) + 6e⁻ + 6H⁺ (aq)

cathode cell is:

14H⁺ (aq) + Cr₂O₇²⁻ (aq) + 6e⁻ → 2Cr³⁺ (aq) + 7H₂O (l)

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Answer:

Option A, The pH of the solution is 2.7 and the solution is acidic

Explanation:

As we know,

pH =  - log [H3O+]

Substituting the given values, we get

pH =  - log [2.1 x 10-3 M]

pH =  - (-2.678)

pH = 2.678 or 2.7

The solution is acidic as pH value is less than 7

Hence, option A is correct

Explanation:

if they aren't in a solvent, then you magnetize first, to remove the iron fillings, secondly, you can sieve to separate the sand and pebbles

Separate the pebbles by either straining them out or pulling each one out separately with tweezers or your fingertips.

To separate their soil mixes using the tools you have supplied, ask the groups to develop a plan. They might consider using forceps to separate the soil, shaking it between two tumblers, placing it in a water-filled tumbler, or pouring it through air. Invite student groups to present their ideas.

Pbbles  - Separate the pebbles by either straining them out or pulling each one out separately with tweezers or your fingertips. As a strainer, a colander or a section of window screen works well. Iron fillings Using a magnet, it is simple to remove the tiny iron filings from the mixture.

Therefore, separate the pebbles by either straining them out or pulling each one out separately with tweezers or your fingertips.

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Answer:

I do because there are certain people that are right for you. If there was no such thing as soulmates, you could just get with anyone and it wouldnt matter. You dont just fall in love with anyone. You may mistake love for something else. But when you're in love you'll know. Twin flames are not the same. Some people mistake twin flames with soulmates. They are close but not so much in a way.

Explanation:

Answer:

Ye

Explanation:

Answer:

Electron cloud

Explanation:

If you were to scale the size of the atom up to have the nucleus be about the size of a basket ball, then the placement of the actual electrons (about the size of a large marble at this scale) throughout the cloud would be many yards away from the nucleus.

Answer:

\(V_{average}=15.34L\)

Explanation:

Hello,

In this case, given the volumes of pollutants:

Macon 0.519 L

Ogiala 44.67 L

Pitt 0.826 L

The average he ought report is computed by adding them and subsequently dividing by three since there are three pollutants as shown below:

\(V_{average}=\frac{0.519L+44.67L+0.826L}{3} \\\\V_{average}=15.34L\)

Regards.

When ammonia is produced at a rate of 0.397 M/s, nitrogen is consumed at a rate of 0.199 M/s, and hydrogen is consumed at a rate of 0.596 M/s.

The rate of reaction is the speed at which a chemical reaction takes place, defined as proportional to the increase in the concentration of a product per unit time and to the decrease in the concentration of a reactant per unit time.

N₂ + 3 H₂ ⇄ 2 NH₃

The molar ratio of N₂ to NH₃ is 1:2.

0.397 mol NH₃/L.s × (1 mol N₂/2 mol NH₃) = 0.199 mol N₂/L.s

The molar ratio of H₂ to NH₃ is 3:2.

0.397 mol NH₃/L.s × (3 mol H₂/2 mol NH₃) = 0.596 mol H₂/L.s

When ammonia is produced at a rate of 0.397 M/s, nitrogen is consumed at a rate of 0.199 M/s, and hydrogen is consumed at a rate of 0.596 M/s.

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The equilibrium constant for the reaction given that the equilibrium concentration of [PH₃] = 0.250 M, [H₂] = 0.580 M, and [P₄] = 0.750 M is 7.3

From the question given above, the following data were obtained:

The equilibrium constant for the reaction can be obtained as shown below:

Equilibrium constant = [Product]ᵐ / [Reactant]ⁿ

Where

Equilibrium constant = [H₂]⁶[P₄] / [PH₃]⁴

Equilibrium constant = [(0.580)⁶ × 0.750] / (0.250)⁴

Equilibrium constant = 7.3

Thus, the equilibrium constant for the reaction is 7.3

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