How would each of the following procedural errors affect the value obtained for the
molar volume of hydrogen gas? Explain your reasoning.
(a) Some bubbles of hydrogen gas remained clinging to the sides of the tube.
(b) Some magnesium metal was left un-reacted at the end of the experiment.
(c) 7 mL of the HC1 were used instead of 5 mL.

sodium hydroxide (NAOH) a.k.a causic soda commonly used in labs​.

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Answer: Sodium hydroxide (NaOH), also known as caustic soda or lye, is a highly versatile substance used in a variety of manufacturing processes.

Explanation: For Example: Sodium hydroxide is also the most common base used in chemical laboratories, being able to test for quite a number of cations (this is called Qualitative Inorganic Analysis), as well as to provide alkaline mediums for some reactions that need it, such as the Biuret test.

The calculated answer is 1.99 x 10/21 x Au

Gold has a molar mass of 196.96657 g/m. The mass of 1 mole of gold atoms, which is commonly given in grams, is referred to as the molar mass of gold.

Divide the specified mass by the molar mass of gold, which is 196.966569 g/mol, to determine the moles.

The term "atomic weight" refers to an atom's overall weight. With the addition of a small amount from the electrons, it is roughly equal to the sum of the protons and neutrons. The number of neutrons an atom has has a significant impact on the stability of the nucleus and, consequently, the radioactivity of the atom. The average mass of an element's atoms expressed in atomic mass units is known as its atomic mass (amu, also known as daltons, D). Gold has a molar mass of 196.96657 g/m. The mass of 1 mole of gold atoms, which is commonly given in grams, is referred to as the molar mass of gold.

Divide the specified mass by the molar mass of gold, which is 196.966569 g/mol, to determine the moles.

(Atomic weight in g/mol on the periodic table).

0.00330 mol = 0.650 g Au = 196.966569 gmol Au Gold atoms 1 mole of atoms is equal to 6.02210/23 atoms.

double the estimated mol Au by several.

6.022×10/23 atoms1mol.

0.00330mol Au = 6.022 x 10/23 x 1 mol

= 1.99 x 10/21 x Au

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Answer:

The structure has a six-carbon ring which is represented by a hexagon and it includes 3-double bonds.

Comparing consumer products on the basis of their components like use, safety, quality, and cost is necessary as it creates awareness about the product among consumers.

Consumer Products - It is a kind of product that is bought by individuals for personal use or consumption. These are the end products or final goods reached to the end-user i.e consumer by retailer via various distribution channels. Ex- grocery, clothing, etc.

Marketing strategies are built based on different market segmentation and types of consumer products available to the target market.

Types of Consumer Products

There are four types of consumer products, they are:-

Convenience Products

Shopping Products

Specialty Products

Unsought Products.

Hence, it is necessary to compare consumer products on basis of the components like use, safety, quality, and cost. It also increases the competition as well as creates awareness among consumers.

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After 90 years amount of  Cs-137 isotopes has a half-life of 30 years. would remain is equal to = 0.125 grams.

An important and useful factor of radioactive decay is half of life, which is the amount of time it takes for one half of a radioactive isotope to decay. The half of-life of a selected radioactive isotope is constant; it is unaffected by way of situations and is independent of the initial quantity of that isotope.

The half of life of a chemical reaction can be defined because the time is taken for the awareness of a given reactant to attain 50% of its preliminary awareness. The half of-life of a response is the time required for the reactant's attention to decrease to at least one-half of its initial price. The half of-life of a first-order response is consistent this is associated with the charge consistent for the reaction. t1/2 = 0.693/k

Since all radioactive reaction follows first-order reaction, The rate of reaction does not depend on initial concentration.

calculation:-

1 gram Cs-137 at T = 0

after the first half,

⇒0.5 gram Cs-137 at T = 30

After the second half,

⇒0.25 gram Cs-137 at T = 30.

After the third half,

⇒0.125 gram Cs-137 at T = 90 years.

Hence, after 90 years amount of  Cs-137 would remain is equal to = 0.125 grams.

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Answer:

250 cm ^3

Explanation:

Explanation:

When magnesium reacts with oxygen, it produces light bright enough to blind you temporarily. Magnesium burns so bright because the reaction releases a lot of heat. As a result of this exothermic reaction, magnesium gives two electrons to oxygen, forming powdery magnesium oxide (MgO).

The mass of carbon dioxide is obtained as 236 g of carbon dioxide.

We know that the first step when we are trying to solve any type of question that involves stoichiometry is that we have to obtain the balanced reaction equation. This would show us the kind of interaction that is able to take place in order to obtain the results that we seek.

The reaction equation is given as;

C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O

If 1 mole of propane occupies 22.4 L

x mole of propane occupies 40.2 L

x = 1 mole  * 40.2 L /22.4 L

x = 1.79 moles

Now;

1 mole of propane produced 3 moles of carbon dioxide

1.79 moles of propane produced 1.79 moles * 3 moles/1 mole

= 5.37 moles of carbon dioxide

Mass of carbon dioxide =5.37 moles * 44 g/mol = 236 g of carbon dioxide.

The mass of carbon dioxide is obtained as 236 g of carbon dioxide.

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The mass of chlorine is 9.9 * 10^6 kg.

Total mass of CFC = M(CFC) * number of cars = 1.0 kg * 1.00 * 10^8

= 1.00 * 10^8 kg

Mass of the leaked CFC ;

% leaked = m(leaked) *100 / m(total)

m(leaked) = o.24 * 1.00 * 10^8 kg

m(leaked) = 2.4 * 10^6 kg

Now no. of moles of CFC:

n(CFC) = m(CFC) / MM(CFC) = 2.4 * 10^9 g / 170.92 g/mol = 0.014 * 10^9 mol

= 1.4 * 10^7 mol

No. of moles of chlorine ;

n(Cl) = n(CFC)*2

= 1.4  * 10^7 mol *2

= 2.8 * 10^7 mol

Now at last the mass of the chlorine :

m(Cl) = n(Cl) * MM(Cl)

= 2.8 * 10^7 mol * 35.453 g/mol

= 99.2 * 10^10 g

Mass of Cl = 9.9 * 10^6 Kg

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The orbitals where you can find each of the bonds;

a - sp3

b - sp3

c - sp2

d- sp3

The term hybridization is the term that is used to describe the mixture of atomic orbitals which can be used to combine with other orbitals of relevant energy.

Let us recall that in some cases, the energy that is required in bonding could only be obtained by the hybridization of orbitals in which orbitals are mixed. The mixed orbitals are orbitals that are very close to each other in energy and help in the formation of bond.

Now let us try to write the orbitals where you can find each of the bonds;

a - sp3

b - sp3

c - sp2

d- sp3

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The total number of atoms in 0.160g of P2O5 is 3 x 10^21 atoms.

Atoms are defined as a smallest unit into which a matter can be divided without the release of electrically charged particles.

First calculate the number of mole of P2O5

Mass of P2O5 = 0.160 g

Molar mass of P2O5 = 142g/mole

Number of mole of P2O5 = mass/ molar mass

n = 0.160/142 g/mole

n = 0.0011 mole of P2O5

One mole of P2O5 has five mole of oxygen.

Using conversion factor

= 0.0011 mole of P2O5 x [ 5 mole of oxygen/ 1 mole of P2O5]

= 0.0055 mole of oxygen

Total number of atoms = 0.0055 x 6.02 x 10^23 atom/mole

                                      = 3 x 10^21 atoms

Thus, the total number of atoms in 0.160g of P2O5 is 3 x 10^21 atoms.

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Taking into account the reaction stoichiometry, 633.75 grams of nitrogen monoxide can form by the reaction of 507 g of ammonia and 845 g of oxygen

In first place, the balanced reaction is:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. This means the when the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 68 grams of NH₃ reacts with 160 grams of O₂, 507 grams of NH₃ reacts with how much mass of O₂?

mass of O₂= (507 grams of NH₃×160 grams of O₂)÷ 68 grams of NH₃

mass of O₂= 1,192.94 grams

But 1,192.94 grams of O₂ are not available, 845 grams are available. Since you have less mass than you need to react with 507 grams of NH₃, O₂ will be the limiting reagent.

The following rule of three can be applied: if by reaction stoichiometry 160 grams of O₂ form 120 grams of NO, 845 grams of O₂ form how much mass of NO?

mass of NO= (845 grams of O₂×120 grams of NO)÷ 160 grams of O₂

mass of NO= 633.75 grams

Finally, 633.75 grams of nitrogen monoxide can form by the reaction of 507 g of ammonia and 845 g of oxygen.

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Explanation:

Cs(s)+HBr(aq)=H(g)+CsBr(aq)

It is important to know the different apparatus in the laboratory and how to do it because these apparatus are handled carefully, and the experiment is done very carefully to avoid any mistakes.

A laboratory is a place where all scientific apparatus and instruments are put and experiments are done. These places are kept safe and hygienic and all experiment is done precisely to get results.

Thus, knowing the various lab equipment and how to utilize it is crucial because errors might happen during experiments because of how carefully this equipment must be handled.

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The electron is in the 5s orbital.

The term orbital is the region in space where there is a high probability of finding an electron. There are four quantum numbers that define an orbital and these are;

Principal quantum number

Orbital quantum number

Spin quantum number

Magnetic quantum number

Let us now look at the sets of quantum numbers as given, a look at the principal and the orbital quantum numbers (m and l) tell us that the electron is in the 5s orbital.

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A chemist is called by the highway patrol to identify a clear colorless liquid that was spilled on the freeway. The test chemist could do to identify whether the liquid is a mixture or a substance is melting point.

If it is to be identified that whether a liquid is a mixture or a pure substance, testing the melting point is preferably considered.

Melting point is a a range of temperature of a substance from which the first crystal starts melting.

Pure substances have sharp melting point but an impure substance or a mixture melts over a given range of temperature.

By looking at the overall composition, pure substance and mixture can be identified visually but it is quite difficult in case of a clear, colorless liquid.

Therefore, melting points can be used to find out whether a liquid is a mixture or a substance.

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Chromium oxide's empirical formula is Cr₂O₃

The simplest whole-number ratio of atoms in a compound is the empirical formula. The proportion of atoms to moles is the same. Our task then is to determine the molar ratio of Cr to O.

Cr's mass is 7.35 g.

10.74 g of chromium oxide are equal to 10.74 g of chromium and 10.74 g of oxygen.

O's mass is 10.74 g - 7.35 g, or 3.39 g.

Moles of Mg = 52.00 g Mg = 0.1413 mol Mg/7.35 g Mg = 1 mol Mg

Moles of O equal 3.39 g O to 1 mol O and 16.00 g O to 0.2119 mol O

For every 3 mol of O, there are 2 mol of Cr.

Chromium oxide's empirical formula is Cr₂O₃.

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Major periodic trends include atomic radius, ionization energy, electron affinity, electronegativity, valency and metallic character.

Answer:

A sentence fragment is a sentence that is missing either its subject or its main verb.

Explanation:

Incorrect: Went to the store yesterday.

Incorrect: After the classes, the library. My life nowadays

Ice forms as water is poured onto dry ice is a physical change .

So , option D is correct one .

The physical change is explained in following points ,

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Answer:

795.702

Explanation:

Answer:

the answer is A. Chlorine combines with sodium to form table salt

The pH of a 0.04 M Calcium hydroxide solution is 1.39.

pH refers to the degree of alkalinity and acidity of a solution.

It is the quantitative measure of the acidity or basicity of a solution. The term "pH" is used in chemistry to denote the values of the concentration of the hydrogen ion {H+}.

The pH scale runs from 0 to 14 where; a value of seven is considered neutral, less than seven acidic, and greater than seven basic.

To calculate the pH of an aqueous solution, one needs to know the concentration of the hydronium ion in moles per liter (molarity).

The pH of a solution can be calculated as the negative logarithm of a concentration or molarity as follows:

pH = - log {H+}

According to this question, a 0.04 M calcium hydroxide is given. The pH can be calculated as follows:

pH = - log {0.04}

pH = 1.39

Therefore, the pH of a 0.04 M calcium hydroxide solution is 1.39.

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To tolerate biting forces that average 28,000 psi, modern dental restorative materials have strengths in excess of dna psi.

Deoxyribonucleic acid is a polymer made of two polynucleotide chains that coil around one another to form a double helix and which contains the genetic material necessary for the growth, development, and reproduction of all known organisms as well as a large number of viruses. The two types of nucleic acids are DNA and RNA. One of the four major types of macromolecules that are necessary for all known forms of life is nucleic acids, along with proteins, lipids, and complex carbohydrates (polysaccharides).

Due to the fact that the two DNA strands are made up of more straightforward monomeric units known as nucleotides, they are referred to as polynucleotides. Each nucleotide consists of a phosphate group, a sugar called deoxyribose, and one of the four nucleobases that contain nitrogen: cytosine (C), guanine (G), adenine (A), or thymine (T).

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When we assume the specific heat of the solution is equal to the specific heat of water ΔHdissolution is 27×10⁻⁵

A calorimeter is a device used for heat measurements necessary for calorimetry

Here given data is 4.79g = KBr and are dissolved in 109.00g water and temperature is 21.08°C and according to calorimeter heat of capacity = 1.84 J/°C

We have to calculate enthalpy of dissolution of KBr = ?

ΔHdissolution = 27×10⁻⁵

KBr+H₂0→KBr ΔH=?

qreaction = -qcal

1.79×1.84(-2.4) = 0.8936 J

1000 = -0.0008936

4.79+109.00 = 522.11 g

qcal = 522.11 g×1.84(-2.4) = -2305 J

qreaction = -qcal = -2305 J

n(KBr) = 4.79/74.55gmol⁻¹ = 0.0642mol

2305/ 0.0642mol = 0.00002787mol⁻¹

= 27×10⁻⁵

ΔHdissolution = 27×10⁻⁵

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The vapour pressure of methanol at 57.2°C is 66.3 kpa.

The enthalpy of vaporization for methanol,

              =35.2 kj/mol = 35.2×1000 j/mol                  

at temperature T2=  64.7°C = 64.7 + 273 = 337.4 K

temperature T1 = 57.2 °C = 57.2 + 273 = 330.2 K

vapour pressure P2 = 101.3 kpa

vapor pressure P1 = ?

Clausius - Clapeyron equation is geven as :

ln ( P1 / P2 ) =  ΔH / R ( 1 / T2 - 1 / T1 )

ln ( P1 / 101.3 kpa ) = 35200 j/mol / 8.314 j/Kmol × ( 1/337.4 K - 1/330.2 K)

ln ( P1 / 101.3 kpa)  =  [tex]e^{-0.423}[/tex]

P1 / 101.3 kpa = 0.655

P1 = 0.655 × 101.3 kpa

P1 = 66.3 kpa

Hence, The vapor pressure at 57.2°C is 66.3 kpa.

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The equation of the double displacement reaction is:

The reaction between solid zinc(II) sulfide reacts with aqueous hydrobromic acid (HBr) is an example of a double displacement reaction.

In the reaction, bromide ion from the hydrobromic acid and the sulfide ion from the zinc sulfide both displace each other from their respective positive ions to form new products. The products formed are aqueous zinc(II) bromide and dihydrogen sulfide gas.

The equation of the double displacement reaction is given below:

ZnS (s) + 2 HBr (aq) → ZnBr₂ (aq) + H₂S (g)

The zinc sulfide exists as a solid whereas the hydrobromic acid is an aqueous solution. The zinc (ii) bromide formed exists as an aqueous solution whereas dihydrogen sulfide is a gas.

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Note that the complete question is given below:

Solid zinc(II) sulfide reacts with aqueous hydrobromic acid (HBr) to form aqueous zinc(II) bromide and dihydrogen sulfide gas.

Express your answer as a chemical equation. Identify all of the phases in your answer.

The pH of the solution prepared by dissolving sodium acetate and acetic acid is 5.05.

Given -

CH3COONa = 2.05g

So, Number of moles of CH3COONa = 2.05 / 82.03 = 0.024

∴ Molarity = 0.024  / 0.05

                = 0.48 M

and CH3COOH =  0.25 M  

pH of buffer - The equilibrium constant (Ka) of the weak acid and the ratio of weak base (A-) to weak acid (HA) in solution are the two variables that determine the pH of a buffer.

1) Various weak acids have different equilibrium constants (Ka).

2) Ka tells us how much of the HA in the solution will be dissolved into H+ and A-.

According to the question -

CH3COOH ⇄ CH3COO⁻ (aq) + H⁻ (aq)

Ka = [CH3COO⁻] [H⁺] /  {CH3COOH]

 [H⁺] = Ka x [CH3COOH] / [CH3COO⁻]

[H⁺] = 1.75 x 10⁻⁵ [0.25] / [ 0.48]

[H⁺] = 0.43 x 10⁻⁵ / 0.48

      = 0.89 x 10⁻⁵

pH =  -log [H⁺]

∴ pH = - log [0.89 x 10⁻⁵]

pH =  5.05

Hence, the pH of the solution will be 5.05.

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