A label on an empty sample container reads 12. 00 g. you add in a sample of a compound and mass the sample container obtaining 15. 5465 g. what should the mass of the sample be reported as?

Answer:

K2O2 and KO2 is the answer

Answer:

C.) K₂O

Explanation:

Potassium (K) always forms the cation K⁺.

Oxygen (O) always forms the anion O²⁻.

The ionic compound these two ions form must have an overall charge of 0. However, the dissimilar charges on the ions do not cancel out (+1 + (-2) = -1). Therefore, you need to have two K⁺ ions in the compound to balance the charges (+1 +1 + (-2) = 0).

Answer:

2.3 atm

Explanation:

-7.50°C = 265.5 K

38°C = 315 K

(2.72)(265.5)=(pressure)(315)

pressure = 2.3 atm

The correct answer is nitrates.

The chemical formula for nitrate is [tex]NO_{3}[/tex]. Nitrate is a polyatomic ion. Nitrates are the name for salts that contain this ion. Common nitrate ingredients in explosives and fertilizers. Water is a solvent for almost all inorganic nitrates. Bismuth oxynitrate is one type of insoluble nitrate. Tests of endurance exercise performance show that dietary nitrate supplementation is effective. When young, healthy people consume significant amounts of nitrate, whether in the form of pure sodium nitrate or beet juice, their plasma nitrate concentration is quickly increased by a factor of 2 to 3, and this elevated nitrate concentration can be sustained for at least two weeks. Nitric oxide, or NO, is produced in response to elevated plasma nitrate.

Metal nitrates other than the alkali elements on heating decompose into the corresponding metal __ along with the evolution of nitrogen peroxide and oxygen.

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The combination of a carbonyl group and a hydroxyl group on the same carbon atom is called a Carboxyl group.

The carboxyl group is an organic functional group. Carboxyl group contains two functional groups that is carbonyl group (C = O) and hydroxyl group (OH) which attached with single carbon atom. The structural formula for carboxyl group is RCOOH where R is alkyl or aryl group.

Example of Carboxyl group: HCOOH (Methanoic Acid) , CH₃COOH (Ethanoic acid) etc.

Thus from the above conclusion we can say that The combination of a carbonyl group and a hydroxyl group on the same carbon atom is called a Carboxyl group.

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Reduction is the general term for the process in which a chemical species gains electrons

The number of electrons attached to an atom or group of atoms increases during a reduction, a type of chemical reaction. One material is provided with electrons by another, which is oxidized as a result of the reduction.

One of the atoms participating in the chemical interaction between two chemicals gains electrons during reduction, which is a type of chemical reaction. The term refers to the element that receives electrons since the element that gains electrons have a lower oxidation state. A chemical species lowers its oxidation number during a half-reaction known as reduction, typically by acquiring electrons.

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Natalie is observing electron cloud model of an atom.

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The coefficient that should be placed in front of O₂ to balance the oxygen atoms is 2.

The balanced chemical equation is the equation  in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

The given equation is

CH₃COOH (l) + O₂ → 2CO₂ (g) + 2H₂O(l).

Reactant Side               Product Side

C = 2                               C = 2

H = 4                               H = 4

O = 4                               O = 6

The number of atoms of O in the reactant side is 4 and the number of atoms of O in the product side is 6. So multiply by 2 to balance the oxygen atoms.

CH₃COOH (l) + 2O₂ → 2CO₂ (g) + 2H₂O(l).

Reactant Side               Product Side

C = 2                               C = 2

H = 4                               H = 4

O = 6                               O = 6

Now the given equation is balanced the number of atoms in reactant side is equal to the number of atoms in the product side.

Thus from the above conclusion we can say that The coefficient that should be placed in front of O₂ to balance the oxygen atoms is 2.

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When a 0. 1 m aqueous solution of hydrocyanic acid, HCN, reaches equilibrium, the ka for hydrocyanic acid is  3.969 x 10⁻¹⁰.

It's the value of equilibrium constant for the dissociation of ions into a solution. The more the Ka value the more will be dissociation.

Ka = [H₃O⁺]² / [HCN] [H₃O⁺]

The pH is 5.20

-log [H₃O⁺] = 5.20

Putting antitlog both side.

The value will be 6.30 x 10⁻⁶

Ka = (6.30 x 10⁻⁶)²  / 0.1 - 6.30 x 10⁻⁶

0.1 - 6.30 x 10⁻⁶ = 0.1

Ka = 3.969 x 10⁻¹⁰

Thus, the Ka value for hydrocyanic acid is 3.969 x 10⁻¹⁰.

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The statement is False.

No, whole cloves are not broken up to help release the eugenol during distillation.

Steam distillation:

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A monosaccharide that consists of 5 carbon atoms, one of which is in a ketone group, is classified as a(n) ketopentose.

Ketone family

The suffix "-one" is added after locating the carbonyl group and, if necessary, designating it with a location number. Ketones are given their common names by first identifying the alkyl groups linked to the carbonyl (in alphabetical order), followed by the prefix "ketone." Several Regular Ketones. The three most significant ketones in terms of scale are acetone, methyl ethyl ketone, and cyclohexanone. Though less frequently than in general organic chemistry, they are still widespread in biochemistry. The simplest ketone is dimethyl ketone, or CH3COCH3, often known as acetone.

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(a) 0.807 fraction of the sample will remain 3. 80 yr after its preparation of tritium.

Half life is the time taken for the radioactivity of a substance to fall to half its original value whereas mean life is average lifetime of all the nuclei of a particular unstable atomic species.

So if a radioactive element has a half life of one hour, this means that half of it will decay in one hour. After another hour, half of the remaining material will decay.

Example:

The radioactive isotope cobalt-60, which is used for radiotherapy, has, for example, a half-life of 5.26 years.

Tritium half life = 12.32 (T half)

Formula :

(a) 0.807 fraction of the sample will remain 3. 80 yr after its preparation of tritium.

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An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species.In oxidation loss of electrons takes place and in reduction gain of electrons is takes place whereas nuetralization reaction is a reaction in which an acid and a base react to form water and a salt and involves the combination of [tex]H^{+}[/tex] (proton) and [tex]OH^{-}[/tex] to generate water.

The neutralization reaction is not a redox reaction is because in nuetralization reaction there is no transfer of electron takes place where as in redox reaction there is no exchange of electrons between atoms. Moat bond forming and breaking redox reaction requires the concomitant transfer of protons.

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Answer:

6.34917360^25g

Explanation:

It's been a while since I've done this type of problem so I'm not making any promises that its right hahaha, but I hope it helps anyway. Please let me know whether I'm right or not!

There are one antibonding molecular orbitals present in molecular orbital model of c.

The cyclobutadiene has a pi system comprised of four  individual atomic p - orbital and thus should have a four pi   molecular orbitals. The compound is the prototypical antiaromatic hydrocarbon with 4 [tex]\pi[/tex] - electrons .  Its rectangular structure is the result of jahn teller reaction which disorder the molecule and lowers its symmetry , converting the triplet to a singlet ground state. It is a small annulene . The  delocalisation energy of the  [tex]\pi[/tex]   electrons of the cyclobutene is predicted to be zero .

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(a) The temperature goes up by 3⁰C.

(b) There is increase in temperature, so the reaction is exothermic.

(c) Examples of exothermic reaction include, rusting of Iron, burning of coal, reaction of strong acid and water.

(d) Examples of endothermic reaction include, melting solid salts and evaporating liquid water.

Exothermic reaction is a type of chemical reaction in which energy is transferred to or from the surroundings.

Examples of exothermic reactions include the following;

Endothermic reaction is a type of chemical reaction in which energy is absorbed from the surroundings.

Examples of endothermic reactions include the following;

From the increase in temperature from 15⁰C to 23⁰C, we can conclude that the temperature goes up by 3⁰C.

20⁰C - 17⁰C = 23⁰ - 20⁰C = 3⁰C.

Since there is increase in temperature, the reaction is exothermic.

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Answer:

D

Explanation:

Gamma radiation penetrates the cell wall of prokaryotic organisms such as bacteria and can inhibit their metabolic functions as well as destroy their DNA.

Debunking the other answers:

A is incorrect as Gamma radiation is used in the treatment of cancer via radiotherapy.

B is incorrect as Gamma rays are too small and would just penetrate any smoke particles.

C is incorrect because Gamma rays are used to disinfect food products to prevent food borne illness. Irradiation is safe to use on food and does not make it radioactive.

Thus, D is correct.

Magnesium (Mg) metal is the substance produced at the cathode.

As molten calcium bromide is electrolyzed, calcium (Ca) metal is the substance produced at the cathode.

Due to the fact that oxidation always takes place at the anode and reduction always occurs at the cathode,

Consider the two cations that were produced in the two ionisation processes of the two substances: mathrm Ca +2 and mathrm Mg +2. The following is how to express two cations' potential cathode reactions:

[tex]ca { }^{ + 2} + (aq) + 2e {}^{ - } - > \\ ca(s)E {}^{0} _{ca {}^{ + 2} \div ca } = - 2.87v[/tex]

[tex] Mg{ }^{ + 2} + (aq) + 2e {}^{ - } - > \\Mg(s)E {}^{0} _{Mg {}^{ + 2} \div Mg} = - 2.37v[/tex]

[tex] Mg^{ + 2} [/tex]

will be reduced quickly since it has a low electronegative electrode potential.

Thus, magnesium (Mg) metal is the substance produced at the cathode.

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It is Cobalt(II)chloride dihydrate. The value of x is 2.

First, we have to determine the mass of H₂O that was evaporated upon heating which will be = 2.00g - 1.56 g

                                 = 0.435 g

So,

CoCl₂ + x.H₂O → CoCl₂.xH₂O

Molecular mass of CoCl₂ = 129.839 g/mol

Molecular mass of H₂O = 18 g/mol

Weight of sample after heating = 1.565 g

Weight of H₂O evaporated = 0.435 g

Hence,

           [tex]18x = \frac{129.839X0.435}{1.565} \\x = \frac{55.48}{28.17} \\x = 1.9[/tex]

           x ≈ 2

So, it is Cobalt(II)chloride dihydrate.

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The lipids are mostly made up of non-polar molecule type .

So , first option is correct one.

Lipids contains same elements as the carbohydrates like carbon , hydrogen , and oxygen ( C , H and O) . However lipids are mainly made up of hydrocarbon chain  ( or  rings ) and contain few polar hydroxyl groups ( -OH ) . This makes the most lipids non - polar hydrobhobic molecules and hey are not dissolve well in water. The lipids are mainly made up of glycerol and fatty acids .

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An atom which an electron moves from the first to the third shell atom absorbs more energy.

An atoms may occupy different energy states . The energy states are discrete , that means they occur at specific values only. Therefore an atom can only move to a new energy level if it absorbs or emits an amount of energy that exactly corresponds to the difference between two energy levels.

The lowest possible energy levels that the atom can occupy is called ground state.The energy levels which is higher to the ground state is called excited state.

The more energy absorb when electron move from first to third because in this second energy level have to pass to reach at third energy level.

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Answer:

Mixing chamber

Answer:

1.1 atm

Explanation:

moles of neon =  10.0g /20.2g/mole =.495 mole

moles of ar      =  10.0 g /39.9 g/mole= .251 mole

the partial pressure of each is determined  by their mole fractions of the entire mix

Ne   .495 mole / (.495 + .251 mole)   * 1.6 atm =

      1.1 atm    ( to two significant figures)

Answer:

25.6 grams of ethanol in 32.5 ml of the substance.

Explanation:

The density of ethanol can be used as a conversion factor.  We know that:

Ethanol is 0.789 g/ml.  That becomes a conversion factor that we may write as (0.789 g/ml) and then convert either grams ethaanol or ml ethanol into the other unit.  We are given volume of ethanol:  32.5 ml.

Note that when we multiply (0.789 g/ml) by (32.5 ml) the ml unit cancels, leaving just grams.  That's what we want, so:

(0.789 g/ml)(32.5 ml) = 25.6 grams of ethanol in 32.5 ml of the substance.

Agar is used to assist establish an anaerobic environment that promotes nitrate reduction.

Nitrate Reduction test:

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Answer:

21.92 gm

Explanation:

A 1.5 M  solution would have 1.5 * 58.44 = 87.66 gm per liter

   you only need 1/4 of this (250 ml)    =  21.92 gm

The pH after 10. 0 ml of 0. 40 M NaOH is added to 20. 0 ml of 0. 50 M HCl is 13.3 .

Calculation ,

[ [tex]OH^{-}[/tex] ] = [tex]M_{1} V_{1} -M_{2} V_{2} /V_{1} +V_{2}[/tex]

where M is molarity and V is the volume .

[ [tex]OH^{-}[/tex] ] = 0. 50 M × 20. 0 ml - 0. 40 M × 10. 0 ml / 10. 0 ml +20. 0 ml

[ [tex]OH^{-}[/tex] ] = 6/30 = 0.2

pOH = -㏒[ [tex]OH^{-}[/tex] ]  =  -㏒ 0.2 = 0.7

pH = 14 - pOH =  14 - 0.7 = 13.3

Therefore , the pH after 10. 0 ml of 0. 40 M NaOH is added to 20. 0 ml of 0. 50 M HCl is 13.3 .

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Answer:

44.1 L

Explanation:

Since volume is being held constant, we can use the following variation of the Ideal Gas Law to find the new pressure.

[tex]\frac{P_1}{T_1N_1}=\frac{P_2}{T_2N_2}[/tex]

In the equation, "P₁", "T₁", and "N₁" represent the initial pressure, temperature, and moles. "P₂", "T₂", and "N₂" represent the final pressure, temperature, and moles. Your answer should have 3 sig figs to match the sig figs of the given values.

P₁ = 234 atm                                 P₂ = 68.0 atm

T₁ = 295 K                                     T₂ = 280 K

N₁ = 144 moles                              N₂ = ? moles

[tex]\frac{P_1}{T_1N_1}=\frac{P_2}{T_2N_2}[/tex]                                              <----- Equation

[tex]\frac{234 atm}{(295 K)(144 moles)}=\frac{68.0 atm}{(280 K)N_2}[/tex]                        <----- Insert values

[tex]\frac{234 atm}{42480}=\frac{68.0 atm}{(280 K)N_2}[/tex]                                      <----- Multiply 295 and 144

[tex]0.00551=\frac{68.0 atm}{(280 K)N_2}[/tex]                                    <----- Simplify left side

[tex]1.54=\frac{68.0 atm}{N_2}[/tex]                                            <----- Multiply both sides by 280

[tex](1.54)N_2={68.0 atm}[/tex]                                   <----- Multiply both sides by N₂

[tex]N_2 = 44.1L[/tex]                                                <----- Divide both sides by 1.54

There will be approximately 49.7 moles of air in the tank at the end of the dive.

The Ideal Gas Law, which asserts the following, can be used to solve this problem:

PV = nRT

Where:

P = pressure

V = volume

n = moles of gas

R = ideal gas constant

T = temperature

Let us first use the given data to obtain the initial amount of moles of air in the tank:

[tex]P_1[/tex]= 234 atm

V = 15.0 L

[tex]T_1[/tex] = 295 K, and

R = 0.0821 L/(Kmol) atm.

Using the ideal gas law equation, we can solve for [tex]n_1[/tex]:

[tex]n_1 = (P_1 * V) / (R * T_1)[/tex]

= (234 atm * 15.0 L) / (0.0821 L·atm/(K·mol) * 295 K)

≈ 189.6 moles

Using the new pressure and temperature, we can determine the exact amount of air in the tank:

P2=68.0 atm and T2=280 K

We can determine [tex]n_2[/tex] by applying the same formula:

[tex]n_2 = (P_2 * V) / (R * T_2)[/tex]

= (68.0 atm * 15.0 L) / (0.0821 L·atm/(K·mol) * 280 K)

≈ 49.7 moles

As a result, there will be approximately 49.7 moles of air in the tank at the end of the dive.

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The mass or the weight of the substance is given as the product of moles and molar mass. The weight of CO as the same oxygen atom in carbon dioxide is 28 gm.

Mass of the substance has been given by the moles and the molar mass of the substance and also by the Avogadro's number (Na).

It is known that the molar mass of carbon dioxide is 44, and 1 mole of it will contain 44 gm or 6.022 × 10²³ molecules.

In a carbon dioxide molecule oxygen has 2 atoms so,

6.022 x 10²³ molecules of CO₂ = 2 × 6.022 x 10²³ atoms of Oxygen

so, 44g of carbon dioxide = 2 × 6.022 × 10²³ atoms of Oxygen

22 gm of carbon dioxide will also have 6.022 × 10²³ atoms of Oxygen

1 mole of carbon monoxide = 28 gm

So, 28 gm will have 6.022 × 10²³ atoms of Oxygen and will weigh 28 gm.

Therefore, the mass of carbon monoxide is 28 gm.

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Explanation:

1 mol of CO2=12+32=44 g

Number of oxygen in 1 molecule of CO2=2

Therefore, 44 g CO2 contains 2 moles of oxygen

22 g CO2 contains =442×22=1 mole oxygen atom

1 mole of CO=12+16=28 g

∴ 28 g CO contains 1 mol oxygen atom.