why are vials made from valor glass stronger than typical medical vials?
I. They contain aluminum instead of boron.
II. They were submerged in a hot salt bath.
III. They were formed out of long glass tubes

a. I and II
b. I and III
c. I, II, and III

The concentration of the HCl solution is 0.816 M (or 816 mM).

To determine the concentration of the HCl solution, we can use the balanced equation for the neutralization reaction between HCl and NaOH:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

This reaction is a neutralization reaction, which means that the number of moles of H+ ions (from the HCl) is equal to the number of moles of OH- ions (from the NaOH).

Given that 20.4 mL of a 1 M NaOH solution was needed to completely react with 25 ml of HCl, we can use the volume and molarity of the NaOH solution to determine the number of moles of OH- ions:

Molarity = moles of solute/litres of solution

Moles of OH- = (20.4 mL x 1 M) / 1000 mL/L = 0.0204 moles

Since the number of moles of H+ ions is equal to the number of moles of OH- ions, we can use the number of moles of OH- to determine the number of moles of H+ ions:

moles of H+ = moles of OH- = 0.0204 moles

We can then use the number of moles of H+ ions and the volume of the HCl solution to determine the concentration of the HCl solution:

Molarity = moles of solute/litres of solution

concentration of HCl = moles of H+ / (25 mL / 1000 mL/L) = 0.0204 moles / 0.025 L = 0.816 M (or 816 mM)

Therefore, the concentration of the HCl solution is 0.816 M (or 816 mM).

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Answer:

c

c

Explanation:

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Answer:

1.68 atm

Explanation:

Applying

P/T = P'/T'................... Equation 1

Where P = Initial pressure, T = Initial Temperature, P' = Final pressure, T' = Final Temperature.

Make P' The subject of the equation

P' = PT'/T.............. Equation 2

From the question,

Given: P = 2.25 atm, T = 125°C = (125+273) K = 398 K, T' = 25°C = (25+273) K = 298 K.

Substitute these values into equation 2

P' = (2.25×298)/398

P' = 1.68 atm

Answer:

1.) The property by the virtue of which the metals can be beaten into sheets.

Malleability

Click it while asking question it allow u to paste image.

The new volume in litres is 13.29 L.

Given,

Temperature 1 (T₁) = 298 K

Volume 1 (V₁) = 15 L

Temperature 2 (T₂) = 383 K

Pressure 1 (P₁) = 3.36 atm

Pressure 2 (P₂) = 4. atm

By Ideal Gas Equation,

P₁V₁/T₁ = P₂V₂/T₂

⇒ (3.36*15)/298 = (4.90*V₂)/383

⇒50.4/298 = 4.90V₂/383

⇒ (50.4*383) / (298*4.90) = V₂

⇒ 19303.2/1460.2 = V₂

⇒13.29 L = V₂

New Volume will be 13.29 L.

Ideal gas law is a precise approximation of the behavioral of several gases under various situations in thermodynamics. The Ideal Gas Equation is a mathematical formula that combines empirical laws such as Charle's law, Boyle's law, Gay-law, Lussac's and Avogadro's law.

The equation characterizing the states of hypothetical gases stated mathematically by combinations of empirical or physical constants is known as the Ideal Gas Equation. It's also known as the universal gas equation. It is defined as follows:

"The ideal gas law seems to be the equation of state for a fictitious perfect gas. It is a decent approximation of the behavioral of various gases under numerous situations, while it has significant limitations".

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Answer:

Chemical reaction, not physical

Explanation:

completely saturated with salt to freshwater. ... Salt marshes, estuaries, and mangrove forests are each unique ecosystems in ... These areas often serve as nursing grounds where young marine life is ... Several reptiles reside in the salt marsh habitat,

To balance a chemical equation, you can only adjust the coefficients; you cannot change the subscripts. The ratios of the atoms in the molecule and the resulting chemical characteristics are altered by changing subscripts.

Changing the coefficients solely affects the quantity of molecules in that specific chemical. However, modifying the subscripts will change the substance itself, rendering your chemical equation incorrect.

You can determine the substance from the subscript. You can find out how much of each element is present in the molecule. The substance itself would change if it were altered. In contrast, simply altering the coefficient results in a change in the number of molecules.

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Coca-Cola is making a version of Coke that uses sucrose (C12H22O11) instead of high fructose corn syrup. The molarity of sucrose in the Coke is 0.32 mol/L.

Coca-Cola is making a version of Coke that uses sucrose (C12H22O11) instead of high fructose corn syrup. If there are 39.24 grams of sucrose in a 12.0 fluid oz. can of this Coke, what is the molarity of sucrose in the Coke.

To find the molarity of sucrose in the Coke, we first need to find the number of moles of sucrose present in the can. We can use the mass of sucrose (39.24 g) and the molar mass of sucrose (342.3 g/mol) to find the number of moles of sucrose:

= (39.24 g) / (342.3 g/mol)

= 0.114 mol

Next, we need to find the volume of the can in liters. We can convert the volume in fluid oz. to liters using the conversion factor 1 fluid oz.

= 0.0295735 L:

= (12.0 fluid oz.) x (0.0295735 L/fluid oz.)

= 0.354882 L

Finally, we can use the formula for molarity (moles of solute / liters of solution) to find the molarity of sucrose in the Coke:

= molarity

= (0.114 mol) / (0.354882 L)

= 0.32 mol/L

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The syringes' circle measures area 10.17 square inches.

The area of a circle is the area that a circle occupies on a two-dimensional plane. The region contained by the circle's radius or perimeter is another way to define the circle's area. The circle's radius, or r, is equal to its diameter in half.

We need to calculate the area of the circle

You may determine the circle's surface area via

Change the formula's value for r.

Area of circle =π[tex]r^{2}[/tex]

Area of circle =3.14[tex]\times[/tex]1.80[tex]\times[/tex]1.80

Area of circle =10.17 [tex]cm^{2}[/tex]

Hence the area of the Circle is 10.17[tex]cm^{2}[/tex]

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NaN3→ Na + 32N2 1 mole of sodium azide gives 32 moles of N2

37 gallons

The amount of nitrogen gas produced by a handful (130 grammes) of sodium azide is 67 litres, which is sufficient to inflate a typical air bag. An air bag can be inflated in as little as 0.03 SECOND. There is another chemistry at play.

Even though we don't typically link chemistry with cars, a lot of chemistry happens in a functioning car, such as the chemical processes that take place in the battery to produce electricity and the burning of gasoline to power the engine. Another reaction includes the air bag, which is one that most drivers would much rather not experience firsthand.

Instead of using compressed gas, air bags are inflated using the byproducts of a chemical reaction. Sodium azide, often known as NaN3, is the principal chemical involved in the air bag reaction

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a. Acetylacetone(acacH) can form up to 4 bonds with a metal ion.

b. Dirthylenetriamine can form up to 4 bonds with a metal ion.

c. Salen can form up to 6 bonds with a metal ion.

d. Porphine can form up to 8 bonds with a metal ion.

Answer:

A battery contains electrochemical cells that can store chemical energy to be converted to electrical energy. A dry-cell battery stores energy in an immobilized electrolyte paste, which minimizes the need for water. Common examples of dry-cell batteries include zinc-carbon batteries and alkaline batteries.

Explanation:  i hope this helps sorry if it didnt

Answer:

Your answer would Graphite Rod I got this correct on my quiz

that is of course if you had the multiple choice answer like how I did on my quiz.

Sorry if its late, Hope this helps!

A 16. 81 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21. 33 grams of CO₂ and 5. 822 grams of H₂O are produced. In a separate experiment, the molar mass is found to be 104. 1 g/mol.  The empirical formula is CH₃   and molecular formula is C₇H₇O₂₁

The empirical formula can be calculate as follows:

The reaction can be write

CxHyOz  → CO₂ + H₂O

As a result, we may state that there are two O-atoms for every one C-atom (12g) in CO2 (32g).

This indicates that the ratio is 12g C/32g O.

The molar mass of CO₂ is 44 g/mole.

(12/44) x 33 = 9 g

There are 9 grams of carbon in every 33 grams of CO₂

The same can be done for water:

In H₂O, each element is represented by two H atoms and one O atom. 2 g H and 16 g O.

[2 g H / 18 g H2O] are present in 5. 822 g of water

(2/18)x 5.822= 0.6468 g

The initial sample had 14.16 g of sample, 9 g of C, and 0.6468 g of H, which resulted in 34.3532g of oxygen.

Determine the amount of moles.

C: 9.0 g / 12.0 g/mol = 0.75 mol

H: 0.6468 g / 1.0 g/mol = 0.6468 mol

O: 34.3532g / 16.0 g/mol =  2.14 mol

Then we divide by the smallest number

C: 0.75 mol / 0.6468 mol = 1

H: 0.6468 mol / 0.6468 mol= 1

O:  2.14 mol/ 0.6468 mol =  3

so the empirical formula will be CH₃

Determine the molecular formula

mass molecular formula / mass of empirical formula = n

We have to multiply the empirical formula by n to get the molecular formula.

104.1 /15 = 7= n

This means the  molecular formula = C₇H₇O₂₁

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The Lucas Test is used to detect the presence of a primary alcohol.

When the test is positive, the mixture of the primary alcohol with the Lucas reagent changes from a clear and colourless characteristic to a turbid, cloudy, and hazy characteristic. This visual change is due to the reaction of the primary alcohol with the Lucas reagent which is a mixture of concentrated hydrochloric acid and zinc chloride.

The Lucas Test is used to detect the presence of a primary alcohol, and when a positive result is obtained, the reaction of the primary alcohol with the Lucas reagent, a mixture of hydrochloric acid and zinc chloride, causes the visual change from a clear and colourless characteristic to a turbid, cloudy, and hazy one.

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A flask contains a mixture of N₂ and O₂ at STP. If the partial pressure of N₂ is 40 kPa, the partial pressure of O₂ is 61.325 kPa.

STP is the abbreviation which is used for a Standard Temperature and Pressure. The standard temperature is 273 K as well as the standard pressure is 1 atm pressure.

1 atm = 101.325 kPa

According to Dalton's law, state's that the sum of the individual pressures exerted by the gases in a mixture is the total pressure.

[tex]P_{total}[/tex]= [tex]P_{A}[/tex] + [tex]P_{B}[/tex]

Therefore, [tex]P_{total}[/tex]= [tex]p_{n2}[/tex] + [tex]P_{O2}[/tex]  

[tex]P_{N2}[/tex] = 40kPa

[tex]P_{total}[/tex] = 101.325kPa

101.325kPa = 40kpa + [tex]P_{O2}[/tex]    

[tex]P_{O2}[/tex] = 61.325kpa

--The given question is incorrect, the correct question is

--"A flask contains a mixture of N₂ and O₂ at STP. If the partial pressure exerted by the N₂ is 40.0 kPa, the partial pressure of O₂ is?"--

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Answer:

b is your answer

Explanation:

Answer:

here is ur ans Lilly

Explanation:

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Answer:

The hydroxide [OH-] concentration of the solution is 1.26*10⁻⁵ M.

Explanation:

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution.

POH indicates the concentration of hydroxyl ions [OH-] present in a solution and is defined as the negative logarithm of the activity of hydroxide ions (that is, the concentration of OH- ions):

pOH= -log [OH-]

A solution has a pOH of 4.90. Replacing in the definition of pOH:

4.90= -log [OH-]

Solving:

-4.90= log [OH-]

1.26*10⁻⁵ M= [OH-]

The hydroxide [OH-] concentration of the solution is 1.26*10⁻⁵ M.

Answer:

La definición de un elemento es una parte o aspecto de algo abstracto, especialmente uno que es esencial o característico.

The ratio of effusion rates of Sulphur dioxide and methane is 0.5

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

Rate of Diffusion ∝ 1 ÷ √molar mass of the gas

We are given:

Molar mass of Sulphur dioxide = 64.066 g/mol

Molar mass of methane = 16.04 g/mol

By taking their ratio, we get:

Rate of SO₂ ÷ Rate of CH₄ = √MCH₄ ÷ √MSO₂

Rate of SO₂ ÷ Rate of CH₄ = √16.04 ÷ √64.066

Rate of SO₂ ÷ Rate of CH₄ = 0.5

Hence, the ratio of rate of effusion of Sulphur dioxide and methane will be 0.5

The isostasy is relates to  the buoyancy , the deepest oceans depths relates to the trench, the theory of the crustal structure relates to the plate tectonics.

The match  the following for the given options are given below as follows :

1 ) isostasy                               buoyancy

2) deepest ocean depths        trench

3) drowned glaciated valley     fjord

4) theory of crustal structure   tectonic plates

5)  oceanic plate going down   subduction

The above are the correct order for the given following relations. these given options are related to the earth movement. like isostasy is the rising or the settling of the part of the earth's lithosphere.

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Answer: Solid ammonium chloride can be separated from a solid mixture of ammonium chloride and anhydrous chloride through the process of sublimation.

Explanation:

A process in which solid substance converts directly into gas without undergoing liquid state is called sublimation.

For example, when solid ammonium chloride is heat in the apparatus then it will convert directly into vapors without undergoing liquid state. During this process it leaves behind anhydrous chloride into the mixture itself.

When these vapors are cooled down at the surface of funnel then it gives ammonium chloride in the solid state.

Thus, we can conclude that solid ammonium chloride can be separated from a solid mixture of ammonium chloride and anhydrous chloride through the process of sublimation.

Normal

BeI2 + SrSO4 → SrI2 + BeSO4

Complete ionic:

Be2+ + 2I- + Sr2+ + SO42- → Sr2+ + 2I- + Be2+ + SO42-

Net ionic:

Be2+ + 2I- → Sr2+ + 2I- + Be2+

The complete ionic equation for this reaction is the same as the normal equation, except that the reactants and products are written as their respective ions.

The net ionic equation is the same as the complete ionic equation, except that the spectator ions, which are ions that appear on both the reactant and product sides, are omitted.

In this reaction, strontium sulfate and beryllium iodide react to form strontium iodide and beryllium sulfate. The spectator ions are Sr2+ and 2I-, which appear on both sides of the equation and cancel out, leaving the net ionic equation.

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Answer: Memory B and memory T cells are types of lymphocytes.

Explanation:

Memory B cells forms the part of adaptive immune system. These cells develop inside the germinal centers of the secondary lymphoid organs. These cells circulate in the blood. These are produced by the bone marrow and can be found in tonsil and spleen. Memory B cells are responsible for recognizing previous infections with a foreign antigen and prepare antibody against the new infection. Memory T cells recognize the antigen lived in the body for longer and produce quick response against it on second exposure. These are one of the types of T lymphocytes.

Answer:

pH= 0.8

pOH=  13.2

Explanation:

Number of moles = mass/molar mass

mass= 15 g

molar mass =36.5 g/mol

Number of moles = 15 g/36.5 g/mol = 0.4 moles

From;

n= CV

Where;

C= concentration of solution

V= volume of solution

n= Number of moles

C = n/V

n= 0.4 moles

V= 2500mL or 2.5 L

C= 0.4/2.5

C= 0.16 M

Hence [H^+] = 0.16 M

pH= - log [H^+]

pH= 0.8

pOH= 14 - pH

pOH= 14 - 0.8

pOH=  13.2

The energy of a photon can be calculated using the formula: E = hν = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 Js), ν is the frequency, c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength which is 6.53 x 10^-38 kJ/mol

Part A: ν = 5.95 x 10^19 s^-1

E = hν = (6.626 x 10^-34 Js) x (5.95 x 10^19 s^-1) = 3.948 x 10^-14 J/photon

To convert to kilojoules per mole, we need to divide by Avogadro's number (6.022 x 10^23 photons/mol)E = (3.948 x 10^-14 J/photon) / (6.022 x 10^23 photons/mol) = 6.53 x 10^-38 kJ/mol

Part B: ν = 20.00 x 10^3 s^-1

E = hν = (6.626 x 10^-34 Js) x (20.00 x 10^3 s^-1) = 1.325 x 10^-31 J/photon

To convert to kilojoules per mole, we need to divide by Avogadro's number (6.022 x 10^23 photons/mol). E = (1.325 x 10^-31 J/photon) / (6.022 x 10^23 photons/mol) = 2.20 x 10^-55 kJ/mol. Part C: λ = 2.57 x 10^2 m. E = hc/λ = (6.626 x 10^-34 Js) x (2.998 x 10^8 m/s) / (2.57 x 10^2 m) = 3.948 x 10^-14 J/photon. To convert to kilojoules per mole, we need to divide by Avogadro's number (6.022 x 10^23 photons/mol) .E = (3.948 x 10^-14 J/photon) / (6.022 x 10^23 photons/mol) = 6.53 x 10^-38 kJ/mol. In all the parts, the energy is expressed in kilojoules per mole to three significant digits.

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