Find the volume, in mL, of an object whose density is 400 g/mL and has a mass of 600
mg.

The concentration of the hydrogen carbonate ion ([HCO₃⁻]) is approximately 3.15 x 10⁷ and the concentration of the carbonic acid ([H₂CO₃]) is approximately 2.52 x 10⁷.

The carbonate buffer system plays a crucial role in maintaining the pH balance of the extracellular environment. In this system, carbonic acid (H₂CO₃) and hydrogen carbonate (HCO₃⁻) act as a conjugate acid-base pair. Given that the ratio of carbonic acid to hydrogen carbonate is 1.25:1, we can assume that the initial concentration of carbonic acid is higher. Let's denote the initial concentration of carbonic acid as [H₂CO₃] and the concentration of hydrogen carbonate as [HCO₃⁻]. The dissociation of carbonic acid can be represented by the equation: H₂CO₃ ⇌ H⁺ + HCO₃⁻. The equilibrium constant (Ka) for this reaction is given as 4.5 x 10⁻⁷. At physiological pH (7.35), the concentration of H⁺ is determined by the dissociation of carbonic acid and is tightly regulated. To calculate the concentration of hydrogen carbonate ion ([HCO₃⁻]), we need to make use of the Henderson-Hasselbalch equation:

pH = pKa + log([HCO₃⁻]/[H₂CO₃])

Substituting the given values, we have:

7.35 = -log(4.5 x 10⁻⁷) + log([HCO₃⁻]/[H₂CO₃])

Rearranging the equation, we find:

log([HCO₃⁻]/[H₂CO₃]) = 7.35 + log(4.5 x 10⁻⁷)

Taking antilog of both sides, we get:

[HCO₃⁻]/[H₂CO₃] = 10^(7.35 + log(4.5 x 10⁻⁷))

Simplifying the right-hand side, we have:

[HCO₃⁻]/[H₂CO₃] ≈ 3.15 x 10⁷

Since the initial ratio of H₂CO₃ to HCO₃⁻ is 1.25:1, we can set up the equation:

[HCO₃⁻] = 1.25 x [H₂CO₃]

Substituting the value of [HCO₃⁻]/[H₂CO₃] from above, we find:

1.25 x [H₂CO₃] = 3.15 x 10⁷

Solving for [H₂CO₃], we get:

[H₂CO₃] ≈ 2.52 x 10⁷

Therefore, the concentration of the hydrogen carbonate ion ([HCO₃⁻]) is approximately 3.15 x 10⁷ and the concentration of the carbonic acid ([H₂CO₃]) is approximately 2.52 x 10⁷.

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Answer:

Wind, ocean currents, waves

Explanation:

The plastic bottle travelled this much distance due to the following factors:

a) Wind

b) Ocean currents

c) Waves

Answer:

An Exothermic Reaction , gives off more heat, and a little energy to its surroundings.

this can helps us figure out that the answer is , C, More heat is given off into its products.

Explanation:

A higher density water will be obtained in lower temperature water.

Density is the fluid property that describes the amount of solutes that dissolves in a given volume of the fluid.

Density of a fluid is calculated by taking the ratio of mass to volume of the fluid.

\(\rho = \frac{m}{v}\)

When the temperature of the water is low, more mass of the water is obtained while a high temperature evaporates the liquid.

Thus, a higher density water will be obtained in lower temperature water.

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1dl/100ml is the missing part of the equation

Equations are not typically measured in the same way that physical quantities are measured.

An equation is a mathematical statement that shows that two expressions are equal. Equations can be used to represent a variety of relationships in mathematics, physics, chemistry, and other fields.

we know that 0.01 dl is the same as 1 ml = 1 / 100

-3.4 x 10^3 ml x 1 dl/100 = -34 dl

hence the answer to the missing part is given as 1dl/100

The solution of the equation is given as -34dl

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Answer:

C. polar nature

Explanation:

The polarity of water and its ability to hydrogen bonding contributes to it's unique properties

Answer: The gas generated by two antacid tablets has a smaller volume.

Explanation:

Since the antiacid is the limiting reagent, we know that the more tablets there are, the more gas there will be.

This means that there will be more gas generated by the four antiacid tablets when compared to the two antiacid tablets, which gives us that the gas generated by the two antiacid tablets has a smaller volume.

Answer:

peace of vassals

Explanation:

I think i spelled it worng

Answer:

the answer is helium but more of then but for now use helium

When hydrogen bromide (HBr) is added to 2-methylbut-2-ene ((CH3)2CCHCH3), an electrophilic addition reaction takes place, where the π bond of the alkene is broken, and the hydrogen and bromine atoms are added to the resulting carbocation.

The reaction proceeds through a Markovnikov addition, where the hydrogen atom attaches to the carbon atom with the greater number of hydrogen atoms.

In this case, the initial addition of HBr to 2-methylbut-2-ene leads to the formation of a primary carbocation, as the positively charged carbon atom only has one alkyl group attached to it. The primary carbocation is relatively unstable, and it can undergo a rearrangement to form a more stable secondary carbocation.

The major product that is typically obtained is the 2-bromo-2-methylbutane. The hydrogen atom from HBr adds to the carbon with three hydrogen atoms (the more substituted carbon), resulting in the formation of a secondary carbocation.

On the other hand, a minor product is also formed, which is 3-bromo-2-methylbutane. This product arises from the addition of HBr to the primary carbocation, which is less stable. Although the primary carbocation is less favored, it can still be formed and lead to the formation of the minor product.

In summary, the addition of HBr to 2-methylbut-2-ene yields two products: the major product is 2-bromo-2-methylbutane, resulting from the addition of HBr to the more stable secondary carbocation, and the minor product is 3-bromo-2-methylbutane, originating from the less stable primary carbocation.

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Answer

ΔH for the reaction is +12.2 kJ

Explanation

Given:

The given reaction is:

The following are the given data:

What to find:

To calculate the ΔH for the given reaction.

Step-by-step solution:

Step 1: Multiply (a) through by 2

Step 2: Reverse (b)

Step 3: (c)

Step 4: Combine (a), (b), and (c) and simplify.

Since the given equation is in 1 mole, then divide through by 2

Therefore, ΔH for the given reaction is +12.2 kJ

J.J. Thomson sent a cathode ray between electrically charged metal plate. The aim of the experiment was to find the nature of the particles. The correct option is option D

He lighted the metal plates. A gold foil was subjected to an alpha radiation beam by B. C. In a chamber, he suspended charged oil drops. OD. He produced a cathode ray discharge between two electrically charged metal plates. All atoms include tiny, negatively charged subatomic particles that mimic electrons, as demonstrated by J.J. Thomson's cathode ray tube studies. In Thomson's "plum pudding" atom model, a negatively charged "electon" encircled a positively charged "soup."

He made the decision to attempt to identify the particle's makeup. He sought to calculate their mass and charge by measuring how much they were bent by electrical currents of various intensities because they were too small to have their mass or charge directly ascertained.

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One piece of evidence may occasionally be helpful in both investigations. A ballistic comparison, for instance, can prove that a gun was used in a murder case as well as other crimes that were committed in another nation, which already indicates the routing of that gun.

Examining the facts pertaining to weapons and ammunition found at crime scenes is the focus of forensic ballistics. Analyses of the outcomes and behaviours of projectiles and explosives are also included.

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Answer:

There can be numerous derived quantities all branching from the basic 7 base quantities.

Explanation:

The  derived SI units are

m/s² that is of acceleration that is rate of change of velocity

m/s is the SI unit of velocity that is rate of change of displacement

Nm is the SI unit of Torque i.e T= rxF

and so on

Answer:

\(\fbox{c = - 4.01 \: joule/g°C}\)

Step by step explanation:

Given:

Mass of given sample (m) = 2.50 g

Initial temperature (T1) = 25°C

Final temperature (T2) = 20°C

Heat Energy Q = 12 cal

To find:

\(Specific \: Heat \: c = \: ?\)

Solution:

We know that,

Specific heat of any substance is directly proportional to the mass and change in temperature.

Represented by equation,

\(Q = mc \triangle T\)

Where,

Q = Heat Energy

m = mass of given sample

c = specific heat

∆T = change in temperature

Substituting corresponding values,

\(Q = mc \triangle T \\ 12 = 2.5\times c \times (20-25) \\ c = \frac{12}{2.5 \times ( - 5)} \\ c = - 0.96 \: cal/g°C \\ \)

We also know that,

\(1 \: cal = 4.184 \: joules\)

multiplying above answer by 4.184,

\(c = - 0.96 \times 4.184 \\ \fbox{c = - 4.01 \: joule/g°C}\)

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Answer:

I don't know the second one but one of them is Aristotle

Answer:

Aristrotle and John Dalton

Explanation:

Answer:

Explanation:

20mL = 0.020L

0.115M = mol/0.02L

mol=0.0023

equivalence point → mols of weak acid = mols of strong base

0.1014 = 0.0023/volume

volume = 0.02268 L → 22.68 mL

To know the end point we use phenolphthalein as indicator. End point is a point where completion of reaction happen. Therefore, volume of base is 22.6ml.

Titration is a technique by which we know the concentration of unknown solution using titration of this solution with solution whose concentration is known.

The balanced equation between sulfurous acid and potassium hydroxide.

KOH + H₂SO\(_3\) →K\(_2\)SO\(_4\) +H\(_2\)O

concentration of sulfurous acid= 0.115 M

Volume of  sulfurous acid= 20.0-mL

Concentration of potassium hydroxide=0.1014 M

Volume of potassium hydroxide= ?

At equivalent point,

moles of  H₂SO\(_3\)      moles of KOH

                      M₁×V₁=M₂×V₂

    0.115 M×20.0-mL = 0.1014 M×V₂

                              V₂=22.6ml

Therefore, volume of base is 22.6ml.

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A hypothesis is a proposed explanation or prediction based on limited evidence or observations, which can be tested through further investigation or experimentation. It should be specific, testable, and based on existing knowledge.

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Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2

use this reaction to look for a limiting reagent

Mg

n=2.37/24.305=0.098 mol

use the ratio to find the number of moles for H2

1:1

0.098:x

no of moles for H2=0.098mol

use the same procedure

HCl

n=4.56/(1.0079+35.453)=0.125mol

H:HCl

1:2

x:0.125mol

H2 moles =0.0625mol

therefore HCl is a limiting reagent because it produces less number of moles

look fo H2 mass

m=nMr

m=(0.0625)(2.0158)=0.126g

percentage yield =0.11/0.126×100=87.31%

Answer:

The answer is option A.

Na4C

Hope this helps you

Answer:

the correct answer is option A

Approximately 4.37 x 10^23 molecules of HCl would be required to form 34.52 g of MgCl₂.

To determine the number of molecules of HCl required to form 34.52 g of MgCl₂, we need to use the molar mass and stoichiometry of the balanced equation:

Mg(OH)₂(s) + 2 HCl(aq) → 2 H₂O(l) + MgCl₂(aq)

The molar mass of MgCl₂ is 95.21 g/mol.

First, we need to calculate the number of moles of MgCl₂ formed:

Moles of MgCl₂ = mass of MgCl₂ / molar mass of MgCl₂

Moles of MgCl₂ = 34.52 g / 95.21 g/mol

Moles of MgCl₂ = 0.363 mol

According to the balanced equation, the stoichiometric ratio between HCl and MgCl₂ is 2:1. Therefore, the moles of HCl required can be calculated as follows:

Moles of HCl = 2 * Moles of MgCl₂

Moles of HCl = 2 * 0.363 mol

Moles of HCl = 0.726 mol

To calculate the number of molecules, we need to use Avogadro's number, which is approximately 6.022 x 10^23 molecules/mol.

Number of molecules of HCl = Moles of HCl * Avogadro's number

Number of molecules of HCl = 0.726 mol * 6.022 x 10^23 molecules/mol

Number of molecules of HCl = 4.37 x 10^23 molecules

Therefore, approximately 4.37 x 10^23 molecules of HCl would be required to form 34.52 g of MgCl₂.

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There are 5.79 x 10²¹ representative particles in 2.62g of the molecular compound.

Determine the number of moles of the molecular compound.

We can use the formula:

moles = mass / molar mass

where mass is 2.62g and molar mass is 273g/mol.

moles = 2.62g / 273g/mol

moles = 0.00961 mol

Use Avogadro's number to convert from moles to representative particles.

We can use the formula:

representative particles = moles x Avogadro's number

where Avogadro's number is 6.022 x 10²³.

representative particles = 0.00961 mol x 6.022 x 10²³

representative particles = 5.79 x 10²¹

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In general, the diffusion coefficient is inversely proportional to pressure. This is also an observed fact: gas production rates from coal seams tend to increase as the reservoir pressure goes down.

Answer:

See below

Explanation:

Conversion factor :  1 ft^3  = 28 .317  liters

$ 1.30 /( 17 ft^3 *  28.317 liters/ft^3)    =   $.0027  per liter  

Answer:

to answer your question "which state of matter have most space between particles"

gas is your best answer

Explanation:

when in gas form..particles gain energy and move further apart

for solids particles are fixed (stays in one spot) and vibrates and for liquids particles gain energy but doesn't have enough space to move around freely..so gas is you best answer

In the given reaction, the limiting reactant is C₆H₆ (benzene).

To determine the limiting reactant as well as calculate the mass of HCl produced, compare the moles of each reactant.

The number of moles for each reactant:

Molar mass of Cl₂ = 35.5 g/mol + 35.5 g/mol = 71 g/mol

Moles of Cl₂ = mass of Cl₂ / molar mass of Cl₂

                     = 45 g / 71 g/mol

                     = 0.634 moles of Cl₂

Molar mass of C₆H₆ (benzene) = 12 g/mol + 6(1 g/mol) = 78 g/mol

Moles of C₆H₆ = mass of C₆H₆ / molar mass of C₆H₆

                        = 45 g / 78 g/mol = 0.577 moles of C₆H₆

Determine the stoichiometry between Cl₂ and HCl:

Cl₂ + C₆H₆ → C₆H₅Cl + HCl

Here, we can see that 1 mole of Cl₂ produces 1 mole of HCl.

Thus, the limiting reactant is C₆H₆ (benzene).

Calculate the mass of HCl produced:

Molar mass of HCl = 1 g/mol + 35.5 g/mol = 36.5 g/mol

Moles of HCl produced = moles of C₆H₆ = 0.577 moles

Mass of HCl produced = moles of HCl produced × molar mass of HCl

Mass of HCl produced = 0.577 moles × 36.5 g/mol

                                     ≈ 21.04 g

Therefore, approximately 21.04 grams of HCl will be produced.

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