The density of an aqueous solution containing 25.0 percent of ethanol (C2H5OH) by mass is 0.950 g/mL. (a) Calculate the molality of this solution. m (b) Calculate its molarity. M (c) What volume of the solution would contain 0.275 mole of ethanol

Answer:

[tex]\%m/m=76.1\%[/tex]

Explanation:

¡Hola!

En este caso, considerando la información dada, entendemos que se haría primero necesario calcular el volumen del cono en metros cúbicos, teniendo en cuenta que 0.005 km son 5 m y 300 cm son 3 m:

[tex]V=\frac{1}{3} \pi r^2h\\\\V=\frac{1}{3} \pi *(5m)^2(3m)=78.5m^3\\\\[/tex]

Ahora, convertimos esta cantidad a gramos por medio de la densidad para conocer la masa de la solución:

[tex]m_{sol}=78.5m^3*\frac{1.2g}{1m^3} =94.2g[/tex]

Finalmente, aplicamos la definición de %m/m para obtener:

[tex]\%m/m=\frac{300g}{300g+94.2g}*100\%\\\\ \%m/m=76.1\%[/tex]

¡Saludos!

Answer:

The right answer is "3 g".

Explanation:

Given:

Initial mass substance,

[tex]M_0=24 \ g[/tex]

By using the relation between half lives and amount of substances will be:

⇒ [tex]M=\frac{M_0}{2^n}[/tex]

        [tex]=\frac{24}{2^3}[/tex]

        [tex]=3 \ g[/tex]

Thus, the above is the correct answer.

Molar mass of Acetone

Now

Moles of Acetone:-

Amount of heat:-

Answer:

Se

Explanation:

First of all, we must note that any element that we must choose is an element that is in group sixteen.

Elements of groups 16 have six electrons in their outermost shell which can be used for bond formation thereby yielding a total of twelve electrons on the valence shell.

However, this is only possible for the heavier members of the group 16 (from sulphur downwards) which are able to expand their octet.

Oxygen can not expand its octet hence it is not the answer.

Answer:

See explanation

Explanation:

Isomers are molecules which have the same molecular formula but different structural formulas. Sometimes, isomers may even contain different functional groups.

The formula C2H6O may refer to an ether or an alcohol. The compound could be CH3CH2OH(ethanol) or CH3OCH3(methoxymethane).

Hence, the functional isomers of the formula C2H6O are ethanol and methoxymethane.

Answer:

Explanation:

The reaction between dimethyl malonate which is an active methylene group with an (∝, β-unsaturated carbonyl compound) i.e methyl vinyl ketone is known as a Micheal Addition reaction. The reaction mechanism starts with the base attack on the β-carbon to remove the acidic ∝-hydrogens and form a carbanion. The carbanion formed(enolate ion) attacks the methyl vinyl ketone(i.e. a nucleophilic attack at the β-carbon) to give a Micheal addition product, this is followed by the protonation to give the neutral product.

Explanation:

Chemical reactions can exhibit different rate constants at differing:

i)initial concentrations

ii)volumes of container

iii) temperatures

iv)none of the above.

The rate constant of a reaction is constant at a particular temperature.

It is not depending on the initial concentration of the reactants. It varies with temperature.

Thus, among the given options the correct answer is Temperatures.

Chemical reactions can exhibit different rate constant at different temperatures. Hence, the correct option is temperature.

The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances.

Chemical reactions proceed at vastly different speeds depending on the nature of the reacting substances, the type of chemical transformation, the temperature, etc.

For a given reaction, the speed of the reaction will vary with the temperature, the pressure, and the amounts of reactants present.

The rate constant goes on increasing as the temperature goes up, but the rate of increase falls off quite rapidly at higher temperatures.

On the other hand, the volume of the container, initial concentration does not affect the rate constant.

Therefore, Chemical reactions can exhibit different rate constant at different temperatures. Hence, the correct option is temperature.

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Answer:

true

Explanation:

Answer:

The correct answer is - 9935 years approximately.

Explanation:

Let z be the age in years to be found:

(15300 disintegrations) x (1.0 g / 0.250 g) / (1.84×10^4 disintegrations)

= 3.3260

half life of carbon = (1/2)^(z/5730 yr)

Solve for z

3.3260 = (1/2)^(z/5730)

Take the log of both sides:

log 3.3260  = (z/5730) log (1/2)

log 3.3260 / log (1/2) = z/5730

z = 5730 log 3.3260 / log (1/2)

= 1.73378816*5730

= 9935 years approximately.

Answer:

[tex]Age=1040.55 million\ years[/tex]

Explanation:

From the question we are told that:

Initial Rock atoms [tex]a_1=2200atoms of 235^U[/tex]

Final Rock atom [tex]a_2=800 of  235^U[/tex]

Age of half life 713 million years

Generally the equation for Age is mathematically given by

[tex]\frac{1}{2^n}=\frac{800}{2200}[/tex]

[tex]\frac{1}{2^n}=\frac{1}{2.750}[/tex]

[tex]n=1.46[/tex]

Therefore

[tex]Age=713*1.46[/tex]

[tex]Age=1040.55 million\ years[/tex]

Answer:

The rate of the forward reaction equals the rate of the reverse reaction.

Explanation:

To know which option is correct, it important that we know what dynamic equilibrium is all about.

We'll begin by defining chemical equilibrium.

A chemical system is said to be in chemical equilibrium when there is no observable change in the properperties of the chemical system.

Dynamic equilibrium on the other hand can be defined as a state in which the forward and backward reaction is occurring at the same time. Thus, in dynamic equilibrium, the rate of the forward reaction is equal to the rate of the backward (i.e reverse) reaction.

Answer:

The correct approach is "12.25°C".

Explanation:

Given:

Mass of lead,

mc = 245 g

Initial temperature,

tc = 300°C

Mass of Aluminum,

ma = 150 g

Initial temperature,

ta = 12.0°C

Mass of water,

mw = 820 g

Initial temperature,

tw = 12.0°C

Now,

The heat received in equivalent to heat given by copper.

The quantity of heat = [tex]m\times s\times t \ J[/tex]

then,

⇒ [tex]245\times .013\times (300-T) = 150\times .9\times (T-12.0) + 820\times 4.2\times (T-12.0)[/tex]

⇒             [tex]3.185(300-T) = 135(T-12.0) + 3444(T-12.0)[/tex]

⇒             [tex]955.5-3.185T=135T-1620+3444T-41328[/tex]

⇒                         [tex]43903.5 = 3582.185 T[/tex]

⇒                                  [tex]T = 12.25^{\circ} C[/tex]

Answer:

The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.

Explanation:

Thermodynamic work is called the transfer of energy between the system and the environment by methods that do not depend on the difference in temperatures between the two. When a system is compressed or expanded, a thermodynamic work is produced which is called pressure-volume work (p - v).

The pressure-volume work done by a system that compresses or expands at constant pressure is given by the expression:

W system= -p*∆V

Where:

In this case:

Replacing:

W system= -1.013*10⁶ Pa* (-0.018 m³)

Solving:

W system= 18234 J

The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.

Answer:

Cu+(aq)--->Cu2+(aq) + e- : oxidation

reason: there is loss of electrons.

I2(s) + 2e--->2I-(aq) : reduction

reason: There is reduction of electrons.

Answer:

hydrogen ions (H⁺) and chloride ions (Cl⁻)

Explanation:

Hydrochloric acid (HCl) is a strong acid. That means that the compound dissociates completely into ions when is dissolved in water, as follows:

HCl → H⁺ + Cl⁻

The equilibrium is completely shifted to the right side (products). Thus, it is considered that the concentration of the non-dissociated compound (HCl) is negligible, and the major specials present in the solution are the hydrogen ions (H⁺) and chloride ions (Cl⁻).

Answer:

15.0 g

Explanation:

15.0% =0.150

100.0 g × 0.150= 15.0g

Sodium nitrate is "an inorganic compound with the formula of NaNO₃.

Inorganic compound is "a chemical compound that lacks carbon–hydrogen bonds".

15% = 0.15

100.0 g × 0.15= 15g

Hence, 15g of Sodium nitrate are needed to prepare 100 gms of a 15% by mass sodium nitrate.

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Answer:

Al^3+

Explanation:

Solubility rules tell us what substances are soluble in water. Since NaCl was added and no precipitate was observed, the mercury II ion is absent.

Addition of Na2SO4 does not form a precipitate meaning that Ba^2+ is absent.

If a precipitate is formed when NaOH is added, the the ionic reaction is as follows;

Al^3+(aq) + 3 OH^-(aq) ------> Al(OH)3(s)

Answer:

5×6.02×1023

Explanation:

there are 5×6.02×1023 molecules of p4010 in 5mole. there are four P atoms in a single molecule of p4010

Answer:

1.27 × 10⁵ L

Explanation:

Step 1: Given data

Step 2: Convert the temperatures to the Kelvin scale

We will use the following expression.

K = °C + 273.15

K = 21 °C + 273.15 = 294 K

K = -48 °C + 273.15 = 225 K

Step 3: Calculate the final volume of the balloon

We will use the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

V₂ = P₁ × V₁ × T₂/ T₁ × P₂

V₂ = 745 Torr × 1.41 × 10⁴ L × 225 K/ 294 K × 63.1 Torr

V₂ = 1.27 × 10⁵ L

Answer:

0.000000000000000969 atm

Explanation:

Based on the reaction:

CO(g) + Cl2(g) ⇄ COCl2(g)

Where equilibrium constant, Kp, is defined as:

Kp = 1.49x10⁻⁸ = PCOCl2 / PCO * PCl2

As PCO = PCl2 = 2.55x10⁻⁴atm:

1.49x10⁻⁸ = PCOCl2 / PCO * PCl2

1.49x10⁻⁸ = PCOCl2 / 2.55x10⁻⁴atm * 2.55x10⁻⁴atm

9,69x10⁻¹⁶atm = PCOCl2

In decimal form:

Answer:

Using the Iodoform test, we can differentiate both compounds.

Explanation:

Benzaldehyde (C6H5CHO -an aldehyde) and Acetophenone (C6H5COCH3 - a methyl ketone) can be differentiated by reacting both compounds with iodine in a basic (NaOH) solutions.

The methyl ketone (acetophenone) gives a pale yellow precipitate of triiodomethane (iodoform) while the aldehyde (benzaldehyde) would not react.

This is known as the IODOFORM test and is indicative for methyl ketones

Answer:

See explanation

Explanation:

Sugar is a polar crystalline substance. The sugar crystal is capable of dissolving in water since it is polar.

When sugar dissolves in water, a sugar solution is formed. If I want to separate the sugar from the water in the solution, I have to boil the solution to a very high temperature.

When I do that, the water in the sugar solution is driven off and the pure sugar crystal is left behind.

Answer:

no

Explanation:

Fluoride is not nucleophilic (having the tendency to donate electrons) enough to allow for the use of HF to cleave ethers in protic media(protic solvents are polar liquid compounds that have dissociable hydrogen atoms). The rate of reaction is comparably low, so that heating of the reaction mixture is required.

Answer:

The correct solution is "3.28 m".

Explanation:

According to the question,

Mol fraction of solvent,

= 0.0558

Molar mass of water,

= 18 g/mol

Mol of H₂O in 1000 g water,

= 55.55 mol

Now,

Let the mol of solute will be "x mol".

Total mol in solution will be "55.55 + x".

As we know,

⇒ The mol fraction of solvent = [tex]\frac{x}{55.55+x}[/tex]

                                        [tex]0.0558=\frac{x}{55.55+x}[/tex]

                                                [tex]x=0.0558[55.55+x][/tex]

                                                [tex]x=3.09969+0.0558x[/tex]

                               [tex]x-0.0558x=3.09969[/tex]

                                               [tex]x=\frac{3.09969}{0.9442}[/tex]

                                                  [tex]=3.38 \ m[/tex]

Answer:

[tex]V_2= 736mL[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the combined gas law:

[tex]\frac{P_2V_2}{T_2}= \frac{P_1V_1}{T_1}[/tex]

Thus, we solve for the final volume by solving for V2 as follows:

[tex]V_2= \frac{P_1V_1T_2}{T_1P_2}[/tex]

Now, we plug in the variables to obtain the result in milliliters and making sure we have both temperatures in Kelvins:

[tex]V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}\\\\V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}=736mL[/tex]

Regards!

Answer:

Theoretical yield of 1-Phenyl-1-propanol in grams = 0.871 g

Explanation:

A Grignard reagent is any of the numerous organic derivatives of magnesium (Mg) which are commonly represented by the general formula RMgX (where R is a hydrocarbon radical e.g. methyl, ethyl, propyl, etc.; and X is a halogen atom, e.g. chlorine, bromine, or iodine)

A Grignard reaction converts an aldehyde to a secondary alcohol. In the grignard reaction involving benzaldehyde as in this experiment, the grignard reagent used is ethyl magnesium bromide, EtMgBr, and the resulting product is 1-Phenyl-1-propanol, a secondary alcohol. The reaction is shown in the figure attached below.

Mass of benzaldehyde in 0.650 mL = density × volume

Density of Benzaldehyde = 1.044 g/mL

Mass of benzaldehyde = 1.044 g/mL × 0.650 mL = 0.6786 g

Molar mass of benzaldehyde = 106 g/mol

Molar mass of 1-Phenyl-1-propanol = 136 g/mol

Mass of = mass of benzaldehyde × mole ratio of 1-Phenyl-1-propanol and benzaldehyde

Mass of 1-Phenyl-1-propanol = 0.6786 g × (136 g/mol)/(106 g/mol) = 0.871 g

Therefore, the theoretical yield of 1-Phenyl-1-propanol in grams = 0.871 g

Answer:

725.15 L

Explanation:

The balanced chemical equation for the reaction between Na₂O₂ and CO₂ is the following:

Na₂O₂ + CO₂ → Na₂CO₃ + 1/2 O₂

From the stoichiometry of the reaction, 1 mol of Na₂O₂ reacts with 1 mol CO₂. So, the stoichiometric ratio is 1 mol Na₂O₂/1 CO₂.

Now, we convert the mass of reactants to moles by using the molecular weight (Mw) of each compound:

Mw (Na₂O₂) = (23 g/mol x 2) + (16 g/mol x 2) = 78 g/mol

moles Na₂O₂ = 96.7 g/(78 g/mol) = 1.24 mol Na₂O₂

Mw (CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol

moles CO₂ = 0.0755 g/(44 g/mol) = 1.71 x 10⁻³ mol CO₂

Now, we calculate the number of moles of CO₂ we need to completely react with the mass of Na₂O₂ we have:

1.24 mol Na₂O₂ x 1 mol CO₂/1 mol Na₂O₂ = 1.24 mol CO₂

In 1 L of respired air we have 1.71 x 10⁻³ mol CO₂ (0.0755 g), so we need the following number of liters to have 1.24 mol of CO₂:

1.24 mol CO₂ x 1 L/1.71 x 10⁻³ mol CO₂ = 725.15 L

Answer:

3) About 0.35 grams of hydrogen gas.

4) About 65.2 grams of aluminum oxide.

Explanation:

Question 3)

We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.

Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:

[tex]\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}[/tex]

To balance it, we can simply add another sodium atom on the left. Hence:

[tex]\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}[/tex]

To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.

The molar mass of sodium is 22.990 g/mol. Hence:

[tex]\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}[/tex]

From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:

[tex]\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}[/tex]

And the molar mass of hydrogen gas is 2.016 g/mol. Hence:

[tex]\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}[/tex]

Given the initial value and the above ratios, this yields:

[tex]\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}[/tex]

Cancel like units:

[tex]=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}[/tex]

Multiply. Hence:

[tex]=0.3463...\text{ g H$_2$}[/tex]

Since we should have two significant values:

[tex]=0.35\text{ g H$_2$}[/tex]

So, about 0.35 grams of hydrogen gas will be released.

Question 4)

Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:

[tex]\displaystyle \text{O$_2$} + \text{Al} \rightarrow \text{Al$_2$O$_3$}[/tex]

To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:

[tex]\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}[/tex]

To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.

The molar mass of aluminum is 26.982 g/mol. Thus:

[tex]\displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}[/tex]

According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:

[tex]\displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}[/tex]

And the molar mass of aluminum oxide is 101.961 g/mol. Hence: [tex]\displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}[/tex]

Using the given value and the above ratios, we acquire:

[tex]\displaystyle 34.5\text{ g Al}\cdot \displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}\cdot \displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}[/tex]

Cancel like units:

[tex]\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}[/tex]

Multiply:

[tex]\displaystyle = 65.1852... \text{ g Al$_2$O$_3$}[/tex]

Since the resulting value should have three significant figures:

[tex]\displaystyle = 65.2 \text{ g Al$_2$O$_3$}[/tex]

So, approximately 65.2 grams of aluminum oxide is produced.

Answer:

Solution given:

3.

[tex]2Na+H_2O→Na _2O+H_2[/tex]

2Na=2*23g.

2O=18g.

[tex]Na_2O[/tex]=62g

[tex]H_2[/tex]=2 g

we have

2*23g of Na produce 2g of [tex]H_2[/tex]

Now

7.9 g of Na produce 2*7.9/(2*23)

=0.34g of [tex]H_2[/tex]

4.

we have

[tex]3O_2+4Al→2Al_2O_3[/tex]

[tex]3O_2[/tex]=3*16g*2g

4Al=4*27g

[tex]2Al_2O[/tex]= 2*27*2g+2*16*3g

4*27g of Al produces 204g of [tex]Al_2O_3[/tex]

34.5g of Al produces 204g*34.5/(4*27)

When electron goes from a higher shell to lower shell then it loses energy .

So, when an electron moved from shell n = 2 to shell n = 1 then a photon of energy is released.