1.
A cork has a mass of 30 g and a volume of 160 cm. What is its
density? (4 points)

Answer:

Oxygen has a molar mass of 16.0 g mol−1 , so 1 mole of oxygen atoms has a mass of 16.0 g . Therefore, carbon dioxide has a percent composition of 72.7% oxygen, i.e. for every 100 g of carbon dioxide you get 72.7 g of oxygen

Explanation:

According to the given statement 0.3345 M  is the molarity of the Ba(OH)₂ solution.

The quantity of a substance in a specific solution volume is known as its molarity (M). The amount of moles of a solute every litre of a solution is referred to as molarity. The molecular concentration of a solution is another name for molarity.

Ba(OH)₂+  2HCl ⇒ BaCl₂ + H₂O

Ratio of moles Ba(OH)₂ ,  

HCl  = 1:2

Amount of HCl consumed

= 0.500/1000*66.90

= 0.03345 mol

So Ba(OH)₂ reacted = 1/2 * 0.03345 mol

If  molarity of Ba(OH)₂ is x then

x/1000*50 = 1/2 * 0.03345

x =  0.3345 M.

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Answer:

The correct answer is Water boiling. Examples of physical change are freezing of water, boiling of water, melting of wax, etc. Examples of chemical change are digestion of food, burning of paper, rusting of metal, silver tarnishing, etc.

The question is incomplete, the complete question is;

A student investigated some properties of lead and recorded the findings in the table below. Property Observation Color Dull gray Density 11.3 g/cm3 Volume of sample 3 cm3 Melting point 327°C Reaction with air Lead oxide forms on the surface Electrical conductivity Conducts some electricity Which statement is true with respect to the investigation?

A. The student investigated only the physical properties of lead.  

B. The student investigated only the chemical properties of lead.  

C. The student investigated more physical properties than chemical properties of lead.  

D. The student investigated more chemical properties than physical properties of lead.

Answer:

The student investigated more physical properties than chemical properties of lead.

Explanation:

If we look at the things mentioned in the question as regards the approach of the student towards the study of lead, we will discover that he/studied;

Colour, density, volume, melting point and electrical conductivity which are all physical properties. Physical properties of a substance does not involve chemical reactions of the substance under study.

The only chemical property studied by the student is its formation of an oxide- lead oxide with air as follows;

2Pb(s) + O₂(g) → 2PbO(s)

Thus we can safely conclude that; the student investigated more physical properties than chemical properties of lead.

Answer:

c

Explanation:

edge 2021 :)

Answer:B

Explanation:

It’s pretty obvious lol

Answer:

Option A

250 degrees Celcius

Explanation:

If 1046J of heat energy is added to water, the water will experience a rise in temperature, at a rate that is directly proportional to its specific heat capacity.

Mathematically, this can be seen as \(Q=C\Delta T\)

Where C = specific heat of water = 4.184 J/g • °C.

Q = heat energy = 1046 J

\(\Delta T =1046/4.148=250 degrees Celcius\)

Therefore, the increase in temperature that will be experienced, is for 250 degrees Celcius

The rate constant for this reaction is –0.29 s–1, which represents the rate of change in concentration of no over time.

To find the rate constant, we can use the equation for the first-order rate law, which is:
Rate = k [A]

Where Rate is the change in concentration of the reactant (in this case NO) over time, k is the rate constant, and [A] is the concentration of the reactant.

We are given the initial concentration of NO (2.8 × 10–3 mol/l) and the concentration after a period of time (2.0 × 10–3 mol/l). We can use this information to calculate the change in concentration:
Δ[A] = [A]final – [A]initial
Δ[A] = (2.0 × 10–3 mol/l) – (2.8 × 10–3 mol/l)
Δ[A] = –0.8 × 10–3 mol/l

Note that the negative sign indicates that the concentration of NO is decreasing over time.
We are also given the time period, s, but we don't need it to solve for the rate constant.

Now we can plug in the values we have into the rate law equation:
Rate = k [A]
Rate = (–0.8 × 10–3 mol/l) / s
k = Rate / [A]
k = (–0.8 × 10–3 mol/l) / (2.8 × 10–3 mol/l)
k = –0.29 s–1

Note that the rate constant is negative, which is expected for a decreasing concentration of a reactant. The units of the rate constant are s–1, which means that the concentration of NO decreases by 0.29 mol/l per second.

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Answer:

Kinetic Energy

Explanation:

Answer:

Explanation:

Burning a magnesium ribbon in the air is an addition reaction while heating potassium manganate 7 is a decomposition reaction.

Magnesium burns in air to produce magnesium oxide as follows:

\(2Mg + O_2 --- > 2MgO\)

Potassium manganate 7 burns to produce multiple products as follows:

\(2 KMnO_4 --- > K_2MnO_4 + MnO_2(s) + O_2\)

Thus, the MgO will be heavier than Mg. On the other hand, \(MnO_2\) will be less heavy than \(KMnO_4\).

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Answer:

B)−6,942 J /mol

Explanation:

At constant temperature and pressure, you cand define the change in Gibbs free energy, ΔG, as:

ΔG = ΔH - TΔS

Where ΔH is enthalpy, T absolute temperature and ΔS change in entropy.

Replacing (25°C = 273 + 25 = 298K; 25.45kJ/mol = 25450J/mol):

ΔG = ΔH - TΔS

ΔG = 25450J/mol - 298K×108.7J/molK

ΔG = -6942.6J/mol

Right solution is:

Toner particles are composed mostly of plastic, which allows them to be melted to the page. The option c is coorect.

The plastic used in toner particles is typically made from a type of polyester called styrene-acrylic. This plastic has a low melting point, which allows it to be melted by the heat of the printer and fused to the paper.

In addition to plastic, toner particles may also contain carbon black, which is used to give the toner its black color.

Other colors of toner may contain different pigments to produce the desired color. Overall, the plastic composition of toner particles plays a crucial role in the printing process, allowing the toner to adhere to the page and produce high-quality prints.

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The atomic mass of oxygen is 16 amu. To find the number of grams of oxygen that contain one mole of oxygen atoms, we need to convert from amu to grams.

The conversion factor between amu and grams is the molar mass of an element. The molar mass of oxygen is 16 g/mol. To calculate the number of grams of oxygen in one mole of oxygen atoms, we can use the equation: grams = number of moles * molar mass

Since we are looking for the number of grams of oxygen in one mole of oxygen atoms, the number of moles is 1 mole. Therefore, the number of grams of oxygen in one mole of oxygen atoms is 1 mole * 16 g/mol = 16 grams. So, one mole of oxygen atoms contains 16 grams of oxygen.

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Answer:

Glucose. C6H12O6

Explanation:

In aerobic respiration glucose generate ATP. Which are the source of energy.

The solution has a boiling point of 100.354 °C and a freezing point of -1.2927 °C.

The boiling and melting points of a solution are affected by the number of solute particles present in the solvent. To determine these points, we need to calculate the molar concentration of glucose (C₆H₁₂O₆) in the solution.

First, we need to convert 25 g of glucose to moles. The molar mass of glucose is 180 g/mol, so 25 g of glucose is 0.139 moles.

Next, we need to calculate the molarity of the solution. Since we have 0.20 L of water, the total volume of the solution is 0.20 L. Therefore, the molarity of the solution is 0.139 moles / 0.20 L = 0.695 M.

Using the molal boiling point elevation constant for water of 0.51 °C/m, we can calculate the boiling point elevation of the solution. ΔTb = Kbm, where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant, and m is the molality of the solution. For a 1.00 m aqueous solution, ΔTb is 0.51°C. So for our 0.695 m solution, ΔTb is (0.51°C/m) x (0.695 m) = 0.354°C. Therefore, the boiling point of the solution is slightly elevated from the boiling point of pure water (100°C) to 100.354°C.

To calculate the freezing point depression of the solution, we can use the formula ΔTf = Kfm, where ΔTf is the freezing point depression, Kf is the molal freezing point depression constant, and m is the molality of the solution. The molal freezing point depression constant for water is 1.86°C/m [3]. Thus, ΔTf = (1.86°C/m) x (0.695 m) = 1.2927°C. Therefore, the freezing point of the solution is depressed from the freezing point of pure water (0°C) to -1.2927°C.

In summary, the boiling point of the solution is 100.354°C, and the freezing point of the solution is -1.2927°C.

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Answer:

Its when like two pure substances are like combined into one.

Explanation:

The compound form with a sodium cation and a carbonate anion has the following molecular formula: Na2CO3, and the name is the same as the individual names but now together, Sodium carbonate, letter B

Answer:

Protons = 89

Neutrons = 142

Electrons = 89

Option C

Explanation:

Hello,

Were required to find the proton, neutron and electron of an element Y

The atomic mass of element Y is composed of the protons and neutrons while the atomic number is the exact amount of protons present in the element.

Element = ₈₉²³¹Y

Atomic number = 89

Number of protons = 89

Number of electrons = 89

Atomic mass = 231

Atomic mass = protons + neutrons

231 = 89 + neutrons

Neutrons = 231 - 89

Neutrons = 142

Therefore the number of protons is 89, number of neutrons is 142 and number of electrons is 89.

Note: for an electrically neutral atom, the number of protons must be equal to the number of electrons

Answer: Time needed: 2.5 s

Distance covered: 31.3 m

Explanation:

I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by

on one side of the equation and solve by plugging your values

To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation

Explanation:

Answer:

Option d is correct

Explanation:

The basis of the VSEPR model of molecular bonding is that to minimize repulsions, electron domains in the valence shell of an atom arrange on their own.

According to VSEPR theory, if there are two electron domains on a central atom, they will be arranged such that the angles between the domains are 180°

Answer:

3.329 g

Explanation:

First you need to determine the molar mass of H2S which is 34.1 g/mol.

With that we know that to find the moles of H2S we just divide the mass of sample with the molar mass.

3.54 g / 34.1 g/mol = 0.103812317  mol of H2S

This means that there is also 0.103812317  mol of sulfur since there is 1 mole of sulfur per 1 mole of H2S.

The molar mass of sulfur is 32.065 g/mol and to find the mass of sulfur you need to multiply the molar mass with the moles of the compound.  

0.103812317  mol * 32.065 g/mol = 3.329 g of sulfur

Let me know if you get something else or if something is unclear in the comments so that we can figure it out.

An electron will emit energy in quanta when its energy state changes from 4p to 3s.

When a quantum mechanical system or particle is bound, or spatially constrained, it can only take on specific discrete energy values, or energy levels. As opposed to classical particles, which can have any energy level, this. The phrase can also refer to the energy levels of nuclei or the vibrational or rotational energy levels in molecules. The term is most frequently used to describe the energy levels of the electrons in atoms, ions, or molecules that are confined by the electric field of the nucleus. A system with such distinct energy levels is said to have a quantized energy spectrum.

An electron shell, also known as a primary energy level, is the orbit of one or more electrons around the nucleus of an atom in chemistry and atomic physics. The "1 shell" (also known as the "K shell") is the shell that is closest to the nucleus. The "2 shell" (also known as the "L shell"), "3 shell" (also known as the "M shell"), and so forth are the shells that are further and more away from the nucleus. The shells are either labelled alphabetically with X-ray notation letters or with the primary quantum numbers (n = 1, 2, 3, 4...) (K, L, M, N...).

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3. (a) Ni electrode can serve as the standard electrode as its reduction potential is zero. It can act a reference standard electrode as the species with reduction potential values more positive to it will get reduced and species with reduction potential values more negative to it will get oxidized. Thus Ni electrode can serve as reference standard electrode.

(b) Cu2+ ion is most easily reduced as it has positive reduction value thus it act as good oxidizing agent by oxidizing other an itself get reduced.

(c) Al metal is most easily oxidized as it has most negative reduction potential value thus it act as good reducing agent by reducing others and itself get oxidized.

(d) Copper and aluminium electrodes are connected to battery to form a spontaneous cell

(i) Al is oxidized as it has negative reduction potential value, thus it loses 3 electrons and form Al3+ ion.

(ii) Al metal act as anode. Because at anode always oxidation takes place. So Al is oxidized to Al3+ at anode.

(iii) At anode: 2 Al (s) 1  2 Al3+ (aq) + 6e-

At cathode: 3 Cu2+ (aq) + 6e- 1  3 Cu (s)  

Net ionic equation: 2 Al (s) +  3 Cu2+ (aq) 1  2 Al3+ (aq) + 3 Cu (s) 

(iv) E0cell = E0cathode - E0anode 

E0cathode = E0Cu2+/ Cu  = 0.62 V                

E0anode = E0Al3+ / Al = -1.38 V

E0cell  = 0.62 V - (-1.38) V = 2 V

As the standard electrode potential of cell is positive, thus the cell reaction is spontanoeus and will proceed in forward direction.

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Under acidic conditions, the first step of nucleophilic addition to an aldehyde is proton transfer to carbonyl oxygen. This results in the formation of a resonance-stabilized intermediate known as the protonated hemiacetal.

Subsequently, the nucleophile can attack the carbonyl carbon, leading to the formation of a new carbon-oxygen bond and the elimination of the protonated leaving group. Option b, nucleophilic attack of the carbonyl carbon, is the second step of the reaction. Option c, formation of an enolate ion, occurs under basic conditions, while option d, formation of a hydrazone, involves the reaction of the aldehyde with hydrazine and is not typically the first step in a nucleophilic addition reaction.
Under acidic conditions, the first step of nucleophilic addition to an aldehyde is: a. Proton transfer to carbonyl oxygen.

In this step, the acidic conditions provide a proton (H+) that is transferred to the carbonyl oxygen, which has a partial negative charge due to its electronegativity. This protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic, allowing the subsequent nucleophilic attack to occur more easily.

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Answer:

p=2.74 atm

Explanation:

According to Boyle's Law, at a constant temperature, a gas' pressure is inversely proportional to volume.  This relationship is used to establish the following equation:

\(p_1v_1=p_2v_2\)

For this problem, let

\(v_1=11.24l\\v_2=4.63l\\p_2=6.64atm\)

So,

\(p_1(11.24l)=(4.63l)(6.64atm)\\p_1(11.24l)=30.74l*atm\\\frac{p_1(11.24l)}{11.24l}=\frac{30.74l*atm}{11.24l}\\p_1=2.74atm\)

This answer makes sense according to Boyle's Law.  As the volume decreases, pressure must increase.  Since the initial volume is higher, the initial pressure should be lower.

Answer:

238 amu

Explanation:

(234 * 0.0001) + (235 * 0.0071) + (238 * 0.9928) = 238

The average atomic mass of uranium is 237.98 amu.

Given:

Three isotopes of uranium element.

To find:

The average atomic mass of uranium.

Solution:

Mass of uranium-234 isotope = 234 amu

Percentage of the abundance of uranium-234 isotope = 0.01%

Fractional abundance of uranium-234 isotope = 0.0001

Mass of uranium-235 isotope = 235 amu

Percentage of the abundance of uranium-235 isotope = 0.71%

Fractional abundance of uranium-235 isotope = 0.0071

Mass of uranium-238 isotope = 238 amu

Percentage of the abundance of uranium-238 isotope = 99.28%

Fractional abundance of uranium-238 isotope = 0.9928

The average atomic mass of uranium will be given by:

\(M=\sum [\text{Mass of an that isotope}\times \text{Fractional abundance of isotope}]\\\\=234amu\times 0.0001+235amu\times 0.0071+238amu\times 0.9928\\\\=237.9783 amu\approx 237.98 amu\)

The average atomic mass of uranium is 237.98 amu.

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Answer:

No

Explanation:

ecause it is an amino acid and does not contain peptide bonds