g An example of an element is ____.
a) chicken noodle soup
b) powerade
c) air inside a balloon
d) lead pipe
e) baking soda (NaHCO)

Answer:

The answer to your question is below

Explanation:

a) Acid dissociation reaction for HCl

                  HCl(g)  +   H₂O(l)    ⇒    H⁺¹  (aq)  +  Cl⁻¹ (aq)

b) pH = -log [H⁺¹]

Concentration = 5 x 10⁻⁴

                   pH = -log [5 x 10⁻⁴]

                    pH = 3.3

c) Acid dissociation reaction

                       NaOH(s)  + H₂O   ⇒   Na⁺¹(aq)   +   OH⁻¹(aq)

d) pH

                      pOH = -log [7 x 10⁻⁵]

                      pOH = 4.2

                      pH = 14 - 4.2

                     pH = 9.8

The acid dissociation and pH for HCl and NaOH were detailed. For HCl, the reaction and solution pH were HCl -> H+ + Cl- and 3.3 respectively. For NaOH, the dissociation and solution pH were NaOH -> Na+ + OH- and 9.8 respectively.

a) The acid dissociation reaction for hydrochloric acid (HCl) is as follows:

HCl → H+ + Cl-

b) The pH of a solution of hydrochloric acid can be calculated using the formula:

pH = -log[H+].

As HCl is a strong acid, it will completely ionize in water, meaning the concentration of H+ ions equals the concentration HCl. Therefore, the pH for a 5.0 x 10^-4 M HCl solution is -log(5.0 x 10^-4) = 3.3.

c) Sodium hydroxide is a base, not an acid, so its ionization does not involve releasing protons into the solution, but rather attracting them. It has the following dissociation reaction:

NaOH → Na+ + OH-.

d) As for the pH of a solution of 7.0 x 10^-5 M NaOH, we first need to find the pOH using the formula pOH = -log[OH-], since NaOH being a strong base completely ionizes. This equals -log(7.0 x 10^-5) = 4.2. Given that pH and pOH are related by the expression pH + pOH = 14 at 25°C, we can solve for the pH of the NaOH solution = 14 - pOH = 14 - 4.2 = 9.8.

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The question is incomplete, here is the complete question:

What is the mass % of ammonium chloride in a 1.73 M ammonium chloride aqueous solution at 20 °C? The density of the solution is 1.0257 g/mL

Answer: The mass percent of ammonium chloride in solution is 9.03 %

Explanation:

We are given:

Molarity of ammonium chloride solution = 1.73 M

This means that 1.73 moles of ammonium chloride is present in 1 L or 1000 mL of solution.

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of solution = 1.0257 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

[tex]1.0257g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.0257g/mL\times 1000mL)=1025.7g[/tex]

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of ammonium chloride = 1.73 moles

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in above equation, we get:

[tex]1.73mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.73mol\times 53.5g/mol)=92.6g[/tex]

[tex]\text{Mass percent of ammonium chloride}=\frac{\text{Mass of ammonium chloride}}{\text{Mass of solution}}\times 100[/tex]

Mass of solution = 1025.7 g

Mass of ammonium chloride = 92.6 g

Putting values in above equation, we get:

[tex]\text{Mass percent of ammonium chloride}=\frac{92.6g}{1025.7g}\times 100=9.03\%[/tex]

Hence, the mass percent of ammonium chloride in solution is 9.03 %

Answer: False

Explanation:

Strong acid is defined as the acid which completely dissociates when dissolved in water. They have low pH. These releases  ions in their aqueous states.

[tex]HNO_3(aq.)\rightarrow H^+(aq.)+NO_3^-(aq.)[/tex]

Weak acid is defined as the acid which does not completely dissociates when dissolved in water. They have high pH. These releases  ions in their aqueous states.

[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]

Thus as the given acid is only 92% dissociated and cannot be completely dissociated, it is termed as a weak acid.

Final answer:

The given statement," An acid that dissociates to the extent of 92% in water would be termed a strong acid." is false.

Explanation:

In chemistry, the degree to which an acid dissociates in water classifies it as either strong or weak. An acid that dissociates 100% into ions when dissolved in water is deemed a strong acid. In contrast, acids that do not reach 100% dissociation are termed weak acids. An example of a weak acid is acetic acid (CH3COOH), which has a dissociation level significantly lower than 100% when dissolved in water.

Considering the given information, an acid that dissociates to the extent of 92% in water would not meet the criteria for a strong acid, as it does not fully dissociate. Thus, even with a high dissociation percentage such as 92%, this acid would still be classified as a weak acid. It is important to note that in practical terms a 92% dissociation indicates a relatively strong acid amongst weak acids, but it is not categorized as a strong acid in the technical sense.

Answer: The rate of disappearance of [tex]N_2[/tex] is 0.172 M/s

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of [tex]N_2[/tex] = [tex]-\frac{1d[N_2]}{dt}[/tex]

Rate in terms of disappearance of [tex]H_2[/tex] = [tex]-\frac{1d[H_2]}{3dt}[/tex]

Rate in terms of appearance of [tex]NH_3[/tex] = [tex]\frac{1d[NH_3]}{2dt}[/tex]

The rate of formation of ammonia is = 0.345 M/s

[tex]\frac{1d[NH_3]}{dt}=0.345 M/s[/tex]

The rate of disappearance of [tex]N_2[/tex] is one by three times the rate of appearance of [tex]NH_3[/tex]

[tex]-\frac{1d[N_2]}{dt}=\frac{d[NH_3]}{2dt}[/tex]

[tex]-\frac{1d[N_2]}{dt}=\frac{0.345M/s}{2}=0.172M/s[/tex]

Thus the rate of disappearance of [tex]N_2[/tex] is 0.172 M/s

If the rate of formation of ammonia is 0.345 M/s, the rate of disappearance of nitrogen is 0.173 M/s.

Let's consider the balanced equation for the formation of ammonia.

N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)

The molar ratio of N₂ to NH₃ is 1:2. If the rate of formation of ammonia is 0.345 M/s, the rate of disappearance of nitrogen is:

[tex]\frac{0.345molNH_3}{L.s} \times \frac{1molN_2}{2molNH_3} = \frac{0.173molN_2}{L.s} = 0.173 M/s[/tex]

If the rate of formation of ammonia is 0.345 M/s, the rate of disappearance of nitrogen is 0.173 M/s.

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Answer:

Radius of platinum atom is [tex]1,38\times10^{-8} \ cm.[/tex]

Explanation:

We know, Density in solid state is also given by , [tex]\rho=\dfrac{Z\times M}{N_A\times a^3}[/tex]

Here, Z= total no of atom or molecule per unit cell. ( Z = 4 for FCC structure )

M = atomic weight = 195.08 gm/mol.

N_A= Avogadro's number = [tex]6.023\times 10^{23} .[/tex]

a= size of side of cube.

Also for, FCC crystal [tex]a=\sqrt2\times 2r[/tex]    ....1

Since,[tex]\rho=\dfrac{Z\times M}{N_A\times a^3}\\[/tex]

Therefore, [tex]a^3=\dfrac{Z\times M}{N_A\times \rho}[/tex]

Putting all values , [tex]a=3.9\times 10^{-8} \ cm.[/tex]

Putting value of a in equation 1 we get,

[tex]r=\dfrac{a}{\sqrt2\times2}=1.38\times 10^{-8} \ cm.[/tex]

Hence, this is the required solution.

The radius of a platinum atom can be found using the density, atomic weight, and the nature of the crystal structure. The calculation involves finding the volume of an atom from the atomic weight, density and Avogadro's number and converting this volume to a radius by considering the atom as a sphere.

The radius of a platinum atom can be calculated using the density, atomic weight, and the information about the crystal structure (in this case, face-centered cubic or FCC). Platinum (Pt) has a density (d) of 21.4 g/cm3 and an atomic weight (AW) of 195.08 g/mol.

First, let's calculate the number of atoms (n) in the unit cell of an FCC structure. An FCC unit cell has 4 atoms. Avogadro's number (Na) is 6.022 x 1023 mol-1.

Keep in mind that the values have to be converted to cm in the end.

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Answer: The quantity of every prefix is written below as a power of ten.

Explanation:

In the metric system of measurement, the name of multiples and subdivision of any unit is done by combining the name of the unit with the prefixes.

For Example: deka, hecto and kilo means 10, 100 and 1000 respectively. Deci, centi and milli means one-tenth, one-hundredth, and one-thousandth respectively.

The quantity of these prefixes are written as the power of 10.

For the given prefixes:

Nano:  The quantity will be [tex]10^{-9}[/tex]

Kilo:  The quantity will be [tex]10^3[/tex]

Centi:  The quantity will be [tex]10^{-2}[/tex]

Micro:  The quantity will be [tex]10^{-6}[/tex]

Milli:  The quantity will be [tex]10^{-3}[/tex]

Mega:  The quantity will be [tex]10^6[/tex]

Hence, the quantity of every prefix is written above as a power of ten.

Answer:

Kinetic Energy of the particle is half the mass of the particle multiplied by the square of the velocity of the particle

Explanation:

Kinetic Energy is given by 1/2mv^2

Where, m is the mass of the object and v is the velocity of the object

Answer: kinetic energy is equal to 1/2mv^2

The question is incomplete, here is the complete question:

A chemist makes 600. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a stock solution of 0.00154 mol/L magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.

Answer: The concentration of chemist's working solution is [tex]5.90\times 10^{-4}M[/tex]

Explanation:

To calculate the molarity of the diluted solution (chemist's working solution), we use the equation:

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the stock magnesium fluoride solution

[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of chemist's magnesium fluoride solution

We are given:

[tex]M_1=0.00154M\\V_1=230mL\\M_2=?M\\V_2=600mL[/tex]

Putting values in above equation, we get:

[tex]0.00154\times 230=M_2\times 600\\\\M_2=\frac{0.00154\times 230}{600}=5.90\times 10^{-4}M[/tex]

Hence, the concentration of chemist's working solution is [tex]5.90\times 10^{-4}M[/tex]

Answer:

5.242

Explanation:

Molar mass of Ca = 40.078 g/mol

Molar mass of S = 32.065 g/mol

Equation of reaction

Ca (s) + S(s) - CaS(s)

To find the limiting reagent

Mole of Ca = 8.54/40.078 (g/mol) = 0.213 mol

Mole of S = 2.33g / 32.065 (g/mol) = 0.072664 mol

From their mole ratio, sulphur is the limiting reagent

From the reaction

1 mole of calcium react with 1 mole of Sulphur to yield 1 mole of calcium sulphide

0.072664 mol of calcium will react with 0.072664 of sulphur to yield 0.072664 mole of CaS

Theoretical yield of CaS = 0.072664 × molar mass of CaS = 0.072664 × 72.143 g/mol = 5.242 g

corrected question: A chemist adds 135mL of a 0.21M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Round your answer to significant digits.

Answer:

5.37g

Explanation:

0.21M means ; 0.21mol/dm³

1dm³=1L , so we can say 0.21mol/L

if 0.21mol of Zinc nitrate is contained in 1L of water

x   will be contained in 135mL of water

x= 0.21*135*10³/1

=0.02835moles

number of moles=  mass/ molar mass

mass= number of moles *molar mas

molar mass of Zn(NO₃)₂=189.36 g/mol

mass= 0.02835 *189.36

mass=5.37g

Answer:

                     Triplet at 1.15 ppm

                     Singlet at 2.02 ppm

                     Quartet at 4.10 ppm

Explanation:

                     In ethyl acetate there are three different sets of protons having different surrounding as compared to each other. The predicted ¹H-NMR of ethyl acetate is attached below and following are the signals shown by this compound,

(i)  A singlet around 2.02 ppm is a characteristic peak shown by protons at carbon a. This is because there is no proton available on the adjacent carbon so there is no coupling.

(ii) A triplet around 1.15 ppm is a peak shown by protons at carbon c. This is because there are two proton present on the adjacent carbon so, according to (n+1) rule it will give a triplet peak.

(iii) A quartet around 4.10 ppm is a peak shown by protons at carbon b. This is because there are three proton present on the adjacent carbon so, according to (n+1) rule it will give a quartet peak. Also, as it is directly attached to oxygen atom hence, the chemical shift is downfielded or deshielded.

Answer: The oxidation and reduction half reactions are written below.

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

[tex]X\rightarrow X^{n+}+ne^-[/tex]

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

[tex]X^{n+}+ne^-\rightarrow X[/tex]

For the given chemical reaction:

[tex]FeSO_4(aq.)+Zn(s)\rightarrow Fe(s)+ZnSO_4(aq.)[/tex]

The half reactions for the given reaction follows:

Oxidation half reaction:  [tex]Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-[/tex]

Reduction half reaction: [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]

Hence, the oxidation and reduction half reactions are written above.

The balanced half-reaction for oxidation in the reaction is Zn (s) -> Zn2+ (aq) + 2e-, signifying Zinc is being oxidised. The reduction half-reaction is Fe2+ (aq) + 2e- -> Fe (s) demonstrating Iron being reduced.

In this chemical reaction:FeSO4 (aq) + Zn (s) --> Fe (s) + ZnSO4 (aq) two processes occur simultaneously; oxidation and reduction. Oxidation involves loss of electrons, and in this case, it is the Zinc metal that gets oxidized. It unites with Sulphate forming ZnSO4 and in the process gains a positive charge by losing 2 electrons.

The balanced half-reaction for oxidation is: Zn (s) --> Zn2+ (aq) + 2e-

Reduction on the other hand involves gain of electrons. Iron in this reaction gains electrons to form metallic iron (Fe). This can be represented by the following balanced half-reaction:

Fe2+ (aq) + 2e- --> Fe (s).

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Complete question

A brick has a mass of 4.0 kg and the Earth has a mass of 6.0 X 10²⁷ g. Use this information to answer the questions below. Be sure your answers have the correct number of significant digits. What is the mass of 1 mole of bricks? Round your answer to 2 significant digits. How many moles of bricks have a mass equal to the mass of the Earth? Round your answer to 2 significant digits.

Answer:

The mass of 1 mole of bricks = 24 X 10²⁶ g (2 significant digits), and

2.5 moles of bricks (2 significant digits) have a mass equal to the mass of the Earth.

Explanation:

Given mass of bricks = 4.0 kg

4.0kg of brick = 4000g

1 mole of any substance always contain  6.022 X 10²³ Avogadro's number

1 mole of bricks contains 6.022 X 10²³ bricks

Therefore, mass of 1 mole of brick = 6.022 X 10²³ X 4000g

= 4000 X 6.022 X 10²³ g

= 24088 X 10²³ g

The mass of 1 mole of bricks = 24 X 10²⁶ g (2 significant digits)

-----------------------------------------------------------------------------------------------------

⇒How many moles of bricks have a mass equal to the mass of the Earth?

Given mass of Earth = 6.0 X 10²⁷ g

24088 X 10²³ g --------> mass of 1 mole of bricks

6.0 X 10²⁷ g --------------> ?

= (6.0 X 10²⁷ g)/(24088 X 10²³ g)

= 2.4909 moles of bricks

≅ 2.5 moles of bricks ( 2 significant digits)

Therefore, 2.5 moles of bricks have a mass equal to the mass of the Earth

Answer

2.4x10^27 g

2.5 mol

Explanation:

Atomic number of carbon is 6 and its electronic distribution is 2, 4. As there are 4 valence electrons present in a carbon atom. Hence, it belongs to group 14 of the periodic table.

So, being a non-metal carbon atom tends to form covalent bonds. Also, the electronegativity value of carbon atom is mid-way between those of metals and non-metals.

In order to complete its octet, carbon atom tends to gain or share its 4 valence electrons. Therefore, it forms four bonds in all its compounds.

The Si-Si bond will be less stronger than C-C bond due to the larger size of silicon atom. As a result, there will be less overlapping between the silicon atoms due to which the bond formed will be weak in nature.

Whereas C-C bond is more stronger as a carbon atom is smaller in size as compared to silicon atom. As a result, more will be the overlapping between the carbon atoms. Hence, more stronger will be C-C bond than Si-Si bond.

As there will be covalent bonds present in a carbon chain and these bonds are weak as compared to ionic bonds. Hence, relatively little heat is released when a C chain reacts and one bond replaces the other.

Thus, we can conclude that out of the given options statements of I, II, III and V are true.

Answer:

energy required is 0.247kJ

Explanation:

The formula to use is Energy = nRdT;

Where n is number of mole

R is the molar gas constant

dT is the change in temperature

n = reacting mass of mercury / molar mass of mercury = 27.4/200.59 = 0.137

dT = final temperature - initial temperature = 376.20 - 158.30 = 217.90K

R = 8.314Jper mol per Kelvin

Energy = 0.137 x 8.314 x 217.90 = 247.12J

Energy in kJ= 247.12/1000= 0.247kJ

Answer:

ρ = 0.439 g/mL

Explanation:

Given data

0.000285 L × (1000 mL/ 1 L) = 0.285 mL

The density (ρ) of a liquid is equal to its mass (m) divided by its volume (V).

ρ = m/V

ρ = 0.125 g/0.285 mL

ρ = 0.439 g/mL

The density of the liquid is 0.439 g/mL.

The density of a liquid, defined as mass per unit volume, with given mass of 0.125 grams and volume of 0.000285 liters, can be calculated using the formula: Density = mass/volume. After converting liters to milliliters, the density is approximately 0.439 g/mL.

The question is asking for the calculation of the density of a liquid given its mass and volume. Density is a property of matter defined as mass per unit volume. It can be calculated using the formula: Density = mass/volume. In this case, the mass of the liquid is 0.125 grams and the volume is 0.000285 liters. But please note that the density should be presented in g/mL. Therefore, we need to convert the volume from liters to milliliters. 1 liter equals 1000 milliliters, so 0.000285 liters equals 0.285 milliliters. Using the formula, Density = 0.125 g / 0.285 mL which gives us a density of approximately 0.439 g/mL.

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Answer:a. 1-propanol,

Explanation:

Boiling point depends on the relative molecular mass of the compound and the kind of intermolecular interactions present in the substance. 1-propanol has a relative molecular mass of 60 and posses strong intermolecular hydrogen bonding. The presence of hydrogen bonding is known to lead to an increase in boiling point. As a result of this, it is obvious that 1-propanol has the highest boiling point among other compounds in the list.

Final answer:

1-propanol has the highest boiling point among the mentioned compounds due to its ability to form hydrogen bonds, which significantly elevate its boiling point compared to the other options.

Explanation:

The question pertains to determining which of the listed compounds would have the highest boiling point. When comparing boiling points, factors such as molecular mass, polarity, and the presence of hydrogen bonding play crucial roles. Among the options, 1-propanol (C3H7OH) stands out because it is capable of hydrogen bonding, which significantly increases its boiling point compared to the other compounds listed, such as dimethyl ether (CH3OCH3) and ethyl methyl ether (CH3OCH2CH3), which primarily exhibit dipole-dipole interactions and London dispersion forces. Ethanol (CH3CH2OH) also exhibits hydrogen bonding but has a shorter carbon chain than 1-propanol, leading to a slightly lower boiling point.

Answer:

P(total) = 46.08 atm

Explanation:

Given data:

Mass of NH₂COONH₄ = 6.8 g

Kc = 0.008

Temperature = 202°C = 202+273 = 475 K

Total pressure at equilibrium = ?

Solution:

Equilibrium equation:

NH₂COONH₄   ⇄   2NH₃ + CO₂

Formula:

Kp = Kc(RT)³

Kp = 0.008 (0.0821 atm.L/mol.K )³(475K)³

Kp = 471.56

                                                                  NH₃       CO₂

            Initial concentration                       0          0

            Change in concentration             2x           x

            equilibrium concentration            2x           x

AS

471.56 = 2x (x)

471.56 = 2x²

x² = 471.56 /2

x² = 235.78

x = 15.36

Pressure of ammonia = 2x = 2(15.36) = 30.72 atm

Pressure of carbon dioxide = x = 15.36 atm

Total pressure:

P(total) = P(NH₃) + P(CO₂)

P(total) = 30.72 atm +  15.36 atm

P(total) = 46.08 atm

Answer:

1. 2.04 W/m²

2. 1.63°C

Explanation:

The radiative force that the Earth receives comes from the Sun. When the Sun rays come to the surface, some of them are absorbed and then it is reflected in the space. The greenhouse gases (like CO2) blocks some of these rays, and then the surface stays warm. The excessive amount of these gases makes the surface warmer, which unbalance the climate on Earth.

1. The variation of the radiative forcing can be calculated based on the concentration of the CO2 by the equation:

ΔF = 5.35*ln(C/C0)

Where C is the final concentration, and C0 is the initial concentration.

ΔF = 5.35*ln(410/280)

ΔF = 2.04 W/m²

2. The temperature change in the Earth's surface caused by the variation of the radiative forcing can be calculated by:

ΔT = 0.8*ΔF

ΔT = 0.8*2.04

ΔT = 1.63 K = 1.63°C

The increase in CO2 concentration from 280 ppm to 410 ppm today results in an extra radiative forcing of 2.2 W/m² on Earth's surface. The equivalent temperature change is approximately 1.7°C.

The increase in CO2 concentration from 280 ppm to 410 ppm today has resulted in an extra radiative forcing on Earth's surface. The radiative forcing is calculated using the formula ΔF (Wm^(-2)) = α ln(C/C0), where α = 5.35. By plugging in the values, the radiative forcing is approximately 2.2 W/m².

The equivalent temperature change can be calculated using the formula ΔT(K) = λ*ΔF, where λ = 0.8 per (Wm^(-2)). By multiplying the radiative forcing by the climate sensitivity, we get a temperature increase of about 1.7°C.

It is important to note that the temperature increase may catch up to a new equilibrium once oceans warm and ice melts, resulting in a higher temperature increase in the future.

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Answer:

a) No, it couldn't

b) 78.81 g

Explanation:

When a nonvolatile solute is added to a pure solvent, the melting point of the solvent must decrease, which is called cryoscopy. The temperature variation (melting point of pure solvent - the melting point of solution) can be calculated by:

ΔT = Kc.W.i

Where Kc is the cryoscopy constant (for water it's 1.86 °C/mol.kg), W is the molality, and i is the van't Hoff factor of the solute.

W =m1/m2*M1

Where m1 is the mass of the solute (in g), m2 is the mass of the solvent (in kg), and M1 is the molar mass of the solute (CaCl2 = 111.0 g/mol). The van't Hoff factor indicates how much of the solute ionizes in the solution.

a) The melting point of water is 0°C, so let's calculate the new melting point with the given information:

m1 = 70.1 g

m2 = 100 g = 0.1 kg

W = 70.1/(0.1*111) = 6.31 mol/kg

ΔT = 1.86*6.31*2.5

ΔT = 29.34°C

0 - T = 29.34

T = -29.34°C

Thus, at -33°C, the ice will not melt yet.

b) Let's then found out the value of m1 in this case:

ΔT = Kc.W.i

0 - (-33) = 1.86*W*2.5

4.65W = 33

W = 7.1 mol/kg

W = m1/M*m2

7.1 = m1/(111*0.1)

m1 = 78.81 g

Calcium chloride cannot melt the ice at -33°C. Cryoscopy is the method of determining a decrease in melting point due to dissolved substances.

The decrease in the melting point of a pure solvent when a non-volatile solute is added to the solution.

[tex]\Delta T = K_c. W.i[/tex]

Where

[tex]K_c[/tex]  - cryoscopy constant= 1.86 °C/mol.kg for water

[tex]W[/tex]  - molality = 6.31 mol/kg

[tex]i[/tex]  - Van't Hoff factor of the solute = 2.5

Put the values in the formula,

[tex]\Delta T = 1.86\times 6.31\times 2.5\\\\\Delta T = \rm \ 29.34^oC[/tex]

Therefore, Calcium chloride cannot melt the ice at -33°C.

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Answer: density =3.377g/cm³

Explanation:

Density =( molecular weight × effective number of atoms per unit cell) / (volume of unit cell × avogadro constant)

D= (M ×n) /(V×A)

M= 137g/mol

n= 2 (For BCC)

V=a³ , where a= 4r/√3

a= (4×222)/√3

a=512.69pm

a= 512.69×10^-10cm

V= ( 512.69×10^-10)^3

V= 1.3476×10^-22cm³

D= (137×2)/(1.3476×10^-22 × 6.02^23)

D= 3.377g/cm³

Therefore the density of barium is 3.377g/cm³

Given:

Radius of barium,

We know,

→ [tex]a = \frac{4r}{\sqrt{3} }[/tex]

     [tex]= \frac{4\times 222}{\sqrt{3} }[/tex]

     [tex]= 512.69 \ pm[/tex]

     [tex]= 512.69\times 10^{-10} \ cm[/tex]

Now,

→ [tex]V = a^3[/tex]

      [tex]= (512.69\times 10^{-10})^3[/tex]

      [tex]= 1.3476\times 10^{-22} \ cm^3[/tex]

hence,

The density will be:

→ [tex]D = \frac{M\times n}{V\times A}[/tex]

      [tex]= \frac{137\times 2}{1.3476\times 10^{-22}\times 6.02\times 10^{23}}[/tex]

      [tex]= 3.377 \ g/cm^3[/tex]

Thus the response above is right.

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Answer:

CaAsHO₄

Explanation:

The data has a mistake in one of the values there. I believe the mistake is on the hydrogen. So, I'm going to assume the value of Hydrogen is 0.6%, so the total percent composition would be 100.1% (Something better). All you have to do is replace the correct value of H (or the value with the mistaken option) and do the same procedure.

Now, to calculate the empirical formula, we can do this in three steps.

Step 1. Calculate the amount in moles of each element.

In these case, we just divide the percent composition with the molar mass of each one of them:

Ca: 22.3 / 40.078 = 0.5564

As: 41.6 / 74.9216 = 0.5552

O: 35.6 / 15.9994 = 2.2251

H: 0.6 / 1.00794 = 0.5953

Now that we have done this, let's calculate the ratio of mole of each of them. This is doing dividing the smallest number of mole between each of the moles there. In this case, the moles of As are the smallest so:

Ca: 0.5564/0.5552 = 1.0022

As: 0.5552/0.5552 = 1

O: 2.2251/0.5552 = 4.0077

H: 0.5953/0.5552 = 1.0722

Now, we round those numbers, and that will give us the number of atoms of each element in the empirical formula

Step 3. Write the empirical formula with the rounded numbers obtained

In this case we will have:

Ca: 1

As: 1

O: 4

H: 1

The empirical formula would have to be:

CaAsHO₄

Answer: Thus chopping a log into sawdust is not an example of a chemical reaction.

Explanation:

Physical reaction is a reaction in which there is no rearrangement of atoms and thus no new substance is formed. There is only change in physical state of the substance.

Example: chopping a log into sawdust as the shape and size of wood changes.

Chemical reaction is a reaction in which there is rearrangement of atoms and thus new substance is formed. There may or may not be a change in physical state.

Example: burning a plastic water bottle : as the plastic would combine with oxygen to give oxidised products.

the production of hydrogen gas from water: the chemical composition changes

the tarnishing of a copper penny: oxidation of copper takes place to give copper oxide.

charging a cellular phone: the electrical energy is used for a chemical reaction

Chopping a log into sawdust is an example of a physical change because it changes the size not the composition of the substance.

Chemical reactions involve such kind change that leads to the formation of a new product (compound).

For example- the production of hydrogen gas from water, the burning of plastic produce gases.

These reactions refer to the rearrangements of the atoms in a substance but do not to the formation of a new product such as temperature, phase, size, etc.

For example- chopping a log into sawdust

Therefore, chopping a log into sawdust is an example of a physical change because it changes the size not the composition of the substance.

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To calculate the amount of heat needed to boil ethanol, use the formula Q = m * c * ΔT. Convert the temperature to Kelvin, calculate the change in temperature, and then apply the formula to find the amount of heat. The amount of heat needed to boil 183.g of ethanol is 20 kJ.

To calculate the amount of heat needed to boil ethanol, we can use the equation:

Q = m * c * ΔT

Where Q is the amount of heat, m is the mass of ethanol, c is the specific heat capacity of ethanol, and ΔT is the change in temperature.

First, we need to calculate the change in temperature:

Now we can calculate the amount of heat:

Therefore, the amount of heat needed to boil 183.g of ethanol is 20 kJ.

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#SPJ3

Answer:

Pressure of alcohol after reaching equilibrium the second time = 2.62atm

Explanation:

Detailed step by step with explanation is as shown in the attachment.

Explanation:

While working in a laboratory it is necessary to be free from any kind of distraction. This is because a distraction while working in a laboratory can lead to serious consequences.

For this it is important to put all your electronic devices like mobile phones, tablets etc in your backpack. Also, you should not carry any kind of eatable, water bottle or anything into the lab.

This is because chemicals do react with food items. Hence, then food does not remain fit for consumption.

Thus, we can conclude that pen or pencil and lab manual personal items are permitted to be at your workstation in the organic laboratory.

Answer:

a > c > b

Explanation:

As higher is the strength and stiffness of the bond between two atoms, more stable it is, and more difficult is to these bonds vibrate. So, the stretching vibration decreases when the strength and stiffness increases.

As more bonds are done between the atoms, more strength, and stiffness they have. So, the order of increase is:

simple bond > double bond > triple bond

And the increased frequency of vibration is:

triple bond > double bond > simple bond

An alkane is a hydrocarbon that has only simple bonds between carbons, an alkene is a hydrocarbon with one double bond between carbon, and an alkyne is a hydrocarbon with one triple bond. So, the increase in vibration of them is:

alkyne (a) > alkene (c) > alkane (b)

Answer: (A) < (C) < (B)

Ranking in order of increasing frequency.

Explanation:

Answer:

4.7 × 10¹⁴ g/cm³

Explanation:

The mass of the sun (and of the neutron star) is 1.98 × 10³⁰ kg = 1.98 × 10³³ g.

The diameter of the neutron star is 20 km and its radius is 1/2 × 20 km = 10 km = 10⁴ m = 10⁶ cm

If we consider the neutron star to be approximately spherical, we can calculate its volume using the following expression.

V = 4/3 × π × r³

V = 4/3 × π × (10⁶ cm)³

V = 4.19 × 10¹⁸ cm³

The density (ρ) is equal to the quotient between the mass and the volume.

ρ = 1.98 × 10³³ g / 4.19 × 10¹⁸ cm³ = 4.73 × 10¹⁴ g/cm³ ≈ 4.7 × 10¹⁴ g/cm³

Answer: 0.39mol of the steam are in the cooker.

Explanation:Please see attachment for explanation

Given:

                                   = 115+273

                                   = 388 K

We know the relation,

→ [tex]PV = nRT[/tex]

or,

→     [tex]n = \frac{PV}{RT}[/tex]

By substituting the values, we get

          [tex]= \frac{4\times 3.1}{0.082\times 388}[/tex]

          [tex]= 0.39 \ mol[/tex]

          [tex]= 0.39 \ mol[/tex]

Thus the above answer is correct.

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Answer:

Explanation:

In gel filtration chromatography a porous matrix is used as the stationary phase. The smaller molecules are able to diffuse into this matrix while the larger ones are not. For this reason the larger molecules will elute first, followed by the smaller ones in decreasing order of size.

In other words the largest molecule will elute first and the smallest molecule will elute last: