Carbon dioxide and an unknown gas start to effuse from a container and the carbon dioxide takes 4.69 times as long to escape as the unknown gas. What is the identity of the unknown gas? 7) A) Br2 B) H2 C) HCl D) CO E) NO2

The simple answer is because of the law of conservation of mass. This law states that matter can not be created or destroyed only transfered or rearranged. You can't have 5 oxygens reacting and only 3 oxygens as the product. What happened to the other two oxygens? They couldn't have magically disappeared. How much you put into an equation is what you will get out of the equation. This is one if the guarantees of life.

Hope this helped!

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Answer:

ExplaChemical equations must be balanced to ensure that the number of atoms for each element is equal. Any imbalance would be a violation of the law of conservation of massnation:

Answer:

17 protons and 20 neutrons

Explanation:

Chlorine (Cl) is an element of the periodic table whose atomic number is 17 and enters the group of halogens.

Cl37 is a stable isotope (atom from an element with the same atomic number but different atomic mass) of chlorine.

Is an isotope of chlorine found in nature, representing ~24% of the total.

Answer:

Stable

Explanation:

An unstable star wouldn't stay the same, a dying star would decrease in size, and a protostar would increase in size.

A star is stable when its size remains constant over time.

An astronomical object known as a star has been made up of a bright plasma spheroid that would be held together using gravity.

Time is just the ongoing progression of existence and things that happen in what seems to be an irrevocable order from the past, present, and forward into the future.

A star is stable when its size remains constant over time.

To know more about time and star.

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Answer:

The maximum wavelength of light will be 773.46 nm.

Explanation:

Energy required to break a fluorine-fluorine single bond = 155 kJ/mol

For 1 mole of fluorine gas  = 155 kJ = 155000 J

For [tex]6.022\times 10^{23}[/tex] molecules = 155000 J

Then energy to break 1 molecule = [tex]\frac{155000 J}{6.022\times 10^{23}}=2.57\times 10^{-19} J[/tex]

Energy of the photon E =[tex]2.57\times 10^{-19} J[/tex]

Wavelength of the light =[tex]\lambda [/tex]

[tex]E=\frac{hc}{\lambda }[/tex]

[tex]2.57\times 10^{-19} J=\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{\lambda }[/tex]

[tex]\lambda =7.7346\times 10^{-7} m=773.46 nm[/tex]

The maximum wavelength of light will be 773.46 nm.

Answer : The value of [tex]\Delta E[/tex] of the reaction is, -553.8 KJ

Explanation :

Formula used :

[tex]\Delta E=\Delta H-\Delta n_g\times RT[/tex]

where,

[tex]\Delta E[/tex] = internal energy of the reaction = ?

[tex]\Delta H[/tex] = enthalpy of the reaction = -184.6 KJ/mole = -184600 J/mole

The balanced chemical reaction is,

[tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]

when the moles of [tex]H_2\text{ and }Cl_2[/tex] are 3 moles then the reaction will be,

[tex]3H_2(g)+3Cl_2(g)\rightarrow 6HCl(g)[/tex]

From the given balanced chemical reaction we conclude that,

[tex]\Delta n_g[/tex] = change in the moles of the reaction = Moles of product - Moles of reactant = 6 - 6 = 0 mole

R = gas constant = 8.314 J/mole.K

T = temperature = [tex]25^oC=273+25=298K[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta E=(-184600J/mole\times 3mole)-(0mole\times 8.314J/mole.K\times 298K)[/tex]

[tex]\Delta E=-553800J[/tex]

[tex]\Delta E=-553.8KJ[/tex]

Therefore, the value of [tex]\Delta E[/tex] of the reaction is, -553.8 KJ

The reaction H2(g) + Cl2(g) → 2HCl(g) is exothermic with a standard enthalpy change of -184.6 kJ/mol. For 3 moles each of H2 and Cl2, the energy released (ΔU) is -276.9 kJ. At 1 atm and 25°C, ΔH and ΔU are nearly the same due to insignificant work against atmospheric pressure.

The given reaction (H2(g) + Cl2(g) → 2HCl(g)) is an exothermic reaction with a standard enthalpy change (ΔH) of -184.6 kJ/mol. This means that this amount of energy is released for every 1 mole of H2 and 1 mole of Cl2 that react to form 2 moles of HCl. If we're considering 3 moles of H2 and 3 moles Cl2, we are effectively dealing with 1.5 times the standard reaction. Hence, the total energy released (ΔU) is -184.6 kJ/mol * 1.5 = -276.9 kJ. An important note is that ΔH and ΔU are approximately the same for reactions at constant pressure and low temperatures (like 1 atm and 25°C), due to minimal work done against atmospheric pressure.

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Answer:

Explanation:

1) Reactants:

The reactants are:

2) Stoichiometric coefficients:

        Cl₂ (g) + 3F₂ (g) →

3) Product:

       Cl₂ (g) + 3F₂ (g) → 2Cl F₃ (g)

        As you see, that last equation si balanced: 2 atoms of Cl and 6 atoms of F on each side, and you conclude that the formula of the product is ClF₃.

Answer:

ΔG° =-990.17 kJ /mol

Explanation:

The equation is:

ΔG° = Δ H° -TΔS°  ....................equation 1

Δ H°rxn = Δ H°f (products)- Δ H°f(reactants)

Δ H°rxn = [[2XΔ H°f(SO₂)]+[2Δ H°f(H₂O)]-[2Δ H°f(H₂S)]+[3Δ H°f(O₂)]

Δ H°rxn = [2X(-296.8)+2(-241.8)]-[2X(-20.6)] = -1036

Δ S°rxn = Δ S°f (products)- Δ S°f(reactants)

Δ S°rxn = [[2XΔ S°f(SO₂)]+[2Δ S°f(H₂O)]-[2Δ S°f(H₂S)]+[3Δ S°f(O₂)]

Δ S°rxn = [2(248.2)+2(188.8)]-[2(205.8)+3(205.2)] = -153.8 J/mol K

      = -0.1538 KJ /mol

Putting values in equation 1  ( T = 298 K)

ΔG° = Δ H° -TΔS° = (-1036)-[(298)(-0.1538)

ΔG° =-990.17 kJ /mol

To calculate the standard change in free energy, ΔG°rxn, use the equation: ΔG°rxn = ΔH°rxn - TΔS°rxn. Calculate ΔH°rxn using the enthalpy of formation values and ΔS°rxn using the entropy values. Substitute the calculated values into the equation to find ΔG°rxn.

To calculate the standard change in free energy, ΔG°rxn, we can use the equation: ΔG°rxn = ΔH°rxn - TΔS°rxn. From the given information, we can calculate the values for ΔH°rxn and ΔS°rxn. First, calculate ΔH°rxn by using the enthalpy of formation values: ΔH°rxn = (2 mol × ΔHf° SO2(g) + 2 mol × ΔHf° H2O(g)) - (2 mol × ΔHf° H2S(g) + 3 mol × ΔHf° O2(g)). Then, calculate ΔS°rxn by using the entropy values: ΔS°rxn = (2 mol × S° SO2(g) + 2 mol × S° H2O(g)) - (2 mol × S° H2S(g) + 3 mol × S° O2(g)). Finally, with the calculated values, substitute them into the equation to find the value for ΔG°rxn.

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Answer:

132 g/mol

Explanation:

(NH₄)₂SO₄ has 2 nitrogen atoms, 8 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms.

From the periodic table, the molar mass of each element is:

N: 14.0 g/mol

H: 1.01 g/mol

S: 32.1 g/mol

O: 16.0 g/mol

So the molar mass of the compound is:

2N × (14.0 g/mol N) + 8H × (1.01 g/mol H) + 1S × (32.1 g/mol S) + 4O × (16.0 g/mol O)

= 28.0 + 8.08 + 32.1 + 64.0 g/mol

= 132 g/mol

If you need more precision, use more significant figures for the element molar masses.

Answer:

Percent ionic character is defined as the ratio of a bond’s actual dipole moment to the dipole moment it would have if the electron were completely transferred from one atom to the other, multiplied by 100

Explanation:

The percent ionic character seeks to establish the amount of electrovalency in a particular compound. It simply compares the covalency i.e extent of shared electrons to the the electrons that are transferred.

It is given as the ratio of the acutal dipole moment to the dipole moment due to ionic character of the bond multiplied by 100:

Percent ionic character = μ_obs/μ_ionic x 100

Where μobs is the actual dipole moment and μ ionic is the dipole moment if the bonds were 100% ionic.

Option C. Na+ has the inert gas structure of argon is false. Sodium has 11 protons and 11 electrons. When it loses an electron and forms Na+, it has 11 protons and 10 electrons. The inert gas that has 10 electrons is Neon not Argon. Argon has 18 electrons.

Further Explanation:

A. Na+ electron configuration is like the stable inert gas configuration - TRUE

Na has an electron configuration of 2,8,1 (or [tex]1s^2 \ 2s^2 \ 2p^6 \ 3s^1[/tex]). When it loses its valence electron, its electron configuration becomes 2,8 (or [tex]1s^2 \ 2s^2 \ 2p^6 [/tex]). This is similar to the electron configuration of Neon which has a full valence shell. A full valence shell has 8 electrons.

B. To remove the second electron, the increased attraction of protons to electrons must be overcome - TRUE

Ionization energy is the amount of energy needed to remove a valence electron from a gaseous atom. This energy is required to overcome the attraction of the valence electron to the nucleus. The closer the orbital (i.e. the electron), the stronger the attraction of the electrons to the nucleus. Hence, removing a second electron, which is now closer to the nucleus than the first, will require more energy to overcome its stronger attraction to the protons in the nucleus.

C. Na+ has the inert gas structure of Argon - FALSE

The electron structure of Na+ is 2,8 (or [tex]1s^2 \ 2s^2 \ 2p^6 [/tex] ).

The electron structure of Ar is 2,8,8 ( or [tex]1s^2 \  2s^2 \ 2p^6 \ 3s^2 \ 3p^6[/tex]).

D. Both A and C are False - FALSE

Only C is false.

E. Items A, B, and C are False - FALSE

A and B are true. Only C is false.

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Keywords: ionization energy, electron configuration

Answer: A) 4C2+S8 -> 4CS2

Explanation:

The chemical equation

4C2+S8 -> 4CS2 is flawed because the equation is not balanced i.e the carbon atom at the reactant is not equal to that of the product and for a chemical reaction to be a balanced equation, the number of atoms of elements at the reactants must be equal to that of the product.

As you can see, there are 8 atoms of carbon(C) at the reactant while we have only 4 carbon atoms at the product therefore making the equation unbalanced.

Answer: 160 ml

Explanation:

The expression used will be :

[tex]C_1V_1+C_2V_2=C_3V_3[/tex]

where,

[tex]C_1[/tex] = concentration of Ist alcohol solution= 10%

[tex]C_2[/tex] = concentration of 2nd alcohol solution= 55%

[tex]V_1[/tex] = volume of Ist alcohol solution = 200 ml

[tex]V_2[/tex] = volume of  2nd alcohol solution= v ml

[tex]C_3[/tex] = concentration of resulting alcohol solution= 30%

[tex]V_2[/tex] = volume of resulting alcohol solution= (v+200) ml

Now put all the given values in the above law, we get the volume of added.

[tex](10\times 200)+(55\times v)=(30\times (v+200)ml)[/tex]

By solving the terms, we get :

[tex]v=160ml[/tex]

Therefore, the volume of 55% mixture  needed to be added to obtain the desired solution is 160 ml.

Answer:

A. Nuclear energy

Explanation:

Nuclear energy is the energy created when light atoms combine in a fusion reaction or when heavier atoms split in a fission reaction.

Answer : The heat released per gram of the compound reacted with oxygen is, 71.915 kJ

Solution :

The balanced chemical reaction is,

[tex]2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{H_2O}\times \Delta H_{H_2O})+(n_{B_2O_3}\times \Delta H_{B_2O_3})]-[(n_{B_5H_9}\times \Delta H_{B_5H_9})+(n_{O_2}\times \Delta H_{O_2})][/tex]

where,

n = number of moles

Now put all the given values in this expression, we get

[tex]\Delta H=[(9\times -285.83)+(5\times -1271.94)]-[(2\times 73.2)+(12\times 0)]\\\\\Delta H=-9078.57kJ[/tex]

Now we have to calculate the heat released per gram of the compound reacted with oxygen.

As we know that,

1 mole of [tex]B_5H_9[/tex] has 63.12 grams of mass

So, 2 mole of [tex]B_5H_9[/tex] has [tex]2\times 63.12=126.24[/tex] grams of mass

As, 126.24 g of [tex]B_5H_9[/tex] release heat = 9078.57 kJ

So, 1 g of [tex]B_5H_9[/tex] release heat = [tex]\frac{9078.57}{126.24}=71.915kJ[/tex]

Therefore, the heat released per gram of the compound reacted with oxygen is, 71.915 kJ

The heat released per gram of the compound reacted with oxygen is -71.92 KJ/mol per gram of B5H9 reacted.

The equation goes as follows;

2B5H9(l) + 12O2(g) → 5B2O3(s) + 9H2O(l)

We have the following information;

ΔH°f  B5H9(l) =  73.2 kJ/mol

ΔH°f  B2O3(s) = −1271.94 kJ/mol

ΔH°f  H2O(l) = −285.83 kJ/mol

Note that;

ΔHrxn = ∑ΔH°f (products) - ΔH°f (reactants)

ΔHrxn =  ∑(5 × ( −1271.94 kJ/mol)) + (9 × ( −285.83 kJ/mol)) - ∑(2 × (73.2 kJ/mol) + (12 × 0)

ΔHrxn = -9078.57  kJ/mol

Since 1 mole of B5H9 = 63.12 g/mol

Two moles of B5H9 reacted so 2 moles × 63.12 g/mol = 126.24 g

Heat released per gram of B5H9 reacted = -9078.57  kJ/mol/126.24 g

= -71.92 KJ/mol per gram of B5H9 reacted.

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Answer:

Explanation:

Given parameters:

Mass of ionic compound = 0.3257g

Mass of AgBr precipitate = 0.7165g

Unknown:

Percent mass of Br in the original compound.

Solution

The percent mass of Br in original compound = [tex]\frac{mass of  Br  in the sample}{mass of sample}[/tex]

Now we have to find the mass of Br⁻:

We must note that the same mass of Br⁻ would move through the ionic sample to form the precipitate.

Mass of Br in AgBr = [tex]\frac{Atomic mass of Br}{Molar mass of AgBr}  x mass of precipitate[/tex]

Mass of Br =  [tex]\frac{80}{80 + 108}[/tex]  x 0.7165

Mass of Br = 0.426 x 0.7165 = 0.305g

Percent mass of Br = [tex]\frac{0.305}{0.3257}[/tex] x 100 = 93.7%

Answer:

Explanation:

1.  The reaction has at least two reactants and one product

Synthesis

Synthesis is a sort of combination reaction in which a single product is formed from  two or more reactants:

               A + B → C

2.  The reaction has one reactant and at least two products

Decomposition

In decomposition reaction, a compound breaks down into individual elements or compounds.

           A → B + C

3.  Oxygen is one of the reactants and a large amount of energy is released

Combustion

In combustion reaction, oxygen combines with a fuel source to produce energy.

4.  One or more atoms replaces a part of a compound

Replacement

In this type of reaction, one substance replaces another we say it is a single replacement reaction. Some other types involves the actual exchange of partners.

5. The products are always salt and water

Neutralization

Neutralization reaction is a reaction in which acids reacts with bases to produce salts and water.

Final answer:

Chemical reactions are matched with their properties: Synthesis is a reaction with multiple reactants and one product, Decomposition has one reactant and multiple products, Combustion involves oxygen and energy release, Replacement consists of atoms substituting part of a compound, and Neutralization results in salt and water.

Explanation:

When matching the chemical reactions in Column B with their properties listed in Column A, the following connections can be made:

Therefore, the correct matches are:

First, in this case, we define the K constant as the solubility of the solute in water divided by the solubility of the solute in ether.

K = (X grams of solute / 75 mL of ethyl ether) / (5 g of solute / 100 mL of water)

K = (X / 75) / (5 / 100)

1 = (X / 75) / (5 / 100)

5 / 100 = X / 75

0,05 = X /75

X = 0,05 × 75 = 3.75 g of solute that will be extracted by the ether

Answer:

The correct answer is 2.44 grams.

Explanation:

The partition coefficient, K = concentration of solute in ether / concentration of solute in water

As partition coefficient is 1, therefore, the concentration of solute in both the solvents would be similar.  

Thus, when 25 milliliters of ether is added to 100 ml of water Wether/25 = Wwater/100

However, Wwater + Wether = 5.00 g

Wwater = 5.00 g - Wether

So, Wether/25 = 5.00 - Wether/100

Wether = 1.00 g

Thus, Wwater = 5.00 - 1.00 = 4.00 grams

So, for the first time, the solute extracted by ether is 1.0 gram

Now add 25 milliliters of ether to 4.00 grams of solute of 100 ml water,

Wwater + Wether = 4.00 g

Wwater = 4.00 g - Wwater

So, Wether/25 = 4.00 - Wether/100

Wether = 0.8

Thus, Wwater = 4.0 - 0.8 = 3.2 grams

So, for the second time the amount of solute extracted by ether is 0.8 gram.  

Now, add 25 ml of ether to 3.2 grams of solute of 100 ml water

Wwater + Wether = 3.20 g

Wwater = 3.20 - Wether

So, Wether/25 = 3.20 - Wether/100

Thus, Wwater = 3.20 - 0.64 = 2.56 grams

So, for the third time, the amount of solute extracted by ether is 0.64 grams.  

Therefore, total weight of solute extracted by ether is 1.00 + 0.80 + 0.64 = 2.44 grams.  

Answer:

a. 392

Explanation:

According to Boyles law, the pressure of a fixed mass of a gas at constant temperature is inversely proportional to the volume.

P1V1=P2V2

P1=488mm Hg

P2=757.8mm Hg

V1=609 mL

Therefore we use the values above in the formula and get:

488 mm Hg×609 mL=757.8 mm Hg×V2

V2=(488mmHg×609 mL)/757.8 mm Hg

=392.177 mL

Answer:

Percentage composition of phosphorus is 43.66%.

Percentage composition of oxygen is 56.66%.

[tex]P_2O_5[/tex]is the empirical formula for this compound.

Explanation:

Moles of phosphorus =[tex]\frac{35.07 g}{31 g/mol}=1.1312 mol[/tex]

Moles of phosphorus oxide =[tex]\frac{71.00 g}{31x+16 y g/mol}=z[/tex]

1.1312 moles of moles of phosphorus gives z moles of phosphorus oxide.

The from 1 mol phosphorus :

[tex]\frac{z}{1.1312 }[/tex] moles of phosphorus oxide...(1)

[tex]2xP+yO_2\rightarrow 2P_xO_y[/tex]

According to reaction, 2x moles of phosphorus gives 2 mol of phosphorus oxide.

The from 1 mol phosphorus :

[tex]\frac{2}{2x}[/tex] moles of phosphorus oxide...(2)

(1)=(2)

[tex]\frac{z}{1.1312 }=\frac{2}{2x}=\frac{71.00 g}{31x+16 y g/mol}[/tex]

[tex]x=\frac{2}{5}y[/tex]

The molecular formula of the phosphorus oxide :[tex]P_xO_y[/tex]:

[tex]P_{\frac{2y}{5}}O_{y}=P_2O_5[/tex]

The empirical formula is simplest ratio of elements in a compound.

[tex]P_2O_5[/tex]is the empirical formula for this compound.

Percentage composition of phosphorus =

[tex]P\%=\frac{2\times 31 g/mol}{2\times 31 g/mol+5\times 16g/mol}\times 100[/tex]

[tex]P\%=43.66\%[/tex]

Percentage composition of oxygen=

[tex]O\%=\frac{5\times 16 g/mol}{2\times 31 g/mol+5\times 16g/mol}\times 100[/tex]

[tex]O\%=56.33\%[/tex]

The percent composition of the phosphorus oxygen compound is 49.4% phosphorus and 50.6% oxygen. The empirical formula, based on the simplest ratio of moles of each component, is P2O4.

Firstly, to obtain the percent composition of the phosphorus oxide compound, we divide the mass of phosphorus by the total mass of the compound, then multiply the quotient by 100. Thus, using the given masses: (35.07 g P/71.00 g P-Oxide) * 100 = 49.4% Phosphorus.

By subtraction, we can find the percentage of Oxygen: 100% - 49.4 % = 50.6% Oxygen. This gives us the percent composition of the phosphorus oxygen compound.

Next, we need to identify the empirical formula from these percentages. We convert percentages to grams (which doesn't change our values because percentages are based on a total of 100), and then we convert from grams to moles using the atomic masses of phosphorus (P) and oxygen (O): 35.07 g P * (1 mol P/30.97 g) = 1.13 mol P and 35.93 g O * (1 mol O/16.00 g) = 2.24 mol O.

To find the empirical formula, we then divide each mole quantity by the smallest obtained mole value (1.13), to generate the simplest mole ratio: P1O1.98, which we round to P1O2 due to natural variation. Therefore, the empirical formula for this phosphorus oxygen compound is P1O2 or more commonly written as P2O4.

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Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = [tex]1.83g/cm^3=1.83g/ml[/tex]

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

[tex]\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml[/tex]

Now we have to calculate the molarity of solution.

[tex]Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L[/tex]

Now we have to calculate the molality of the solution.

[tex]Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg[/tex]

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

The molality of concentrated sulfuric acid is [tex]\( 0.500 \, \text{mol/kg} \)[/tex], and the molarity is [tex]\( 0.0183 \, \text{mol/L} \)[/tex].

1. Mass of sulfuric acid [tex](\( \text{H}_2\text{SO}_4 \))[/tex] in 100 g of solution:

[tex]\[ \text{Mass of } \text{H}_2\text{SO}_4 = 98.0 \% \times 100 \, \text{g} = 98.0 \, \text{g} \][/tex]

2. Moles of sulfuric acid [tex](\( \text{H}_2\text{SO}_4 \))[/tex]:

[tex]\[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{98.0 \, \text{g}}{98.086 \, \text{g/mol}} \approx 1.000 \, \text{mol} \][/tex]

3. Molality [tex](\( m \))[/tex]:

[tex]\[ \text{Molality} = \frac{1.000 \, \text{mol}}{2.0 \, \text{kg}} = 0.500 \, \text{mol/kg} \][/tex]

4. Molarity [tex](\( M \))[/tex]:

[tex]\[ \text{Volume of solution} = \frac{100 \, \text{g}}{1.83 \times 10^{-3} \, \text{kg/cm}^3} \approx 54.6 \, \text{L} \]\[ \text{Molarity} = \frac{1.000 \, \text{mol}}{54.6 \, \text{L}} \approx 0.0183 \, \text{mol/L} \][/tex]

So, the molality of concentrated sulfuric acid is [tex]\( 0.500 \, \text{mol/kg} \)[/tex], and the molarity is [tex]\( 0.0183 \, \text{mol/L} \)[/tex].

Answer:

Person one has conduction, person 2 has radiation, and person 3 has convention.

Explanation:

Person 1 is touching the mug to get warm, which is a transfer of heat.

Person 2 is exposed to the fire, which is radiation.

Person 3 is exposed to the warm air of convection.

Explanation:

A process that involves transfer of heat from a hot substance to a cold substance by coming in contact with each other is known as conduction.

For example, a person has his hands cupped around a ceramic mug of hot chocolate then heat is transferring from ceramic mug to the hands.

Whereas a process in which heat energy is transferred in a wave-like motion through the space is known as radiation.

For example, a person is toasting a marshmallow above the fire is getting heat energy in the form of radiation.

Also, third person who is roasting a hot dog above the glowing coals is getting heat energy in the form of radiation.

Hence, we can conclude that primary source of heat transfer for person 1 is conduction, and for both person 2 and 3 is radiation.

True, the kilogram is used as a unit for mass in the the international system of units (SI) for scientific measurements.

Answer: True

Explanation:

Mass is defined as the amount of matter contained in the body.

Its units are kg, gram, milligram which are inter convertible.

S.I or M.K.S system has seven fundamental units which are used to find derived units

1) Mass - Kilogram

2) Length - meter

3) Time - Seconds

4) Electric Current - Ampere

5) Amount of substance - Moles

6) Intensity of light - Candela

7) Temperature - Kelvin

Thus the kilogram is the SI unit for mass.

Final answer:

The heat of reaction for the combustion of a mole of titanium in this calorimeter is 15220 kJ/mol.

Explanation:

In order to calculate the heat of reaction for the combustion of a mole of Ti in this calorimeter, we need to use the equation q = mcΔT. First, we need to calculate the amount of heat absorbed by the calorimeter and its contents.

The heat capacity of the calorimeter is given as 9.84 kJ/K, and the temperature change is 91.60°C - 25.00°C = 66.60°C. Therefore, the heat absorbed by the calorimeter and its contents is:

q = (9.84 kJ/K)(66.60°C) = 654.24 kJ.

Next, we need to determine the amount of moles in 2.060 g of titanium. The molar mass of titanium is 47.867 g/mol, so:

moles of titanium = 2.060 g / 47.867 g/mol = 0.043 moles.

Finally, we can calculate the heat of reaction for the combustion of a mole of Ti:

heat of reaction = 654.24 kJ / 0.043 moles = 15220 kJ/mol.

Answer:

The activation energy of the reaction is 1.152 kJ/mol.

Explanation:

Activation energy is the minimum amount which is absorbed by the reactant molecules to undergo chemical reaction.

Initial temperature of reaction = [tex]T_1=100 K[/tex]

Final temperature of reaction = [tex]T_2=200 K[/tex]

Initial rate of the reaction at 100 k = [tex]K_1=k[/tex]

Final rate of the reaction at 200 k = [tex]K_2=2k[/tex]

Activation energy is calculated from the formula:

[tex]\log\frac{K_2}{K_1}=\frac{E_a}{2.303\times R}(\frac{T_2-T_1}{T_1T_2})[/tex]

R = Universal gas constant = 8.314 J/ K mol

[tex]\log\frac{2k}{k}=\frac{E_a}{2.303\times 8.314 J/K mol}(\frac{200 K-100K}{200 K\times 100K})[/tex]

[tex]E_a=1,152.772 J/mol=1.152 kJ/mol[/tex]

The activation energy for this chemical reaction can be calculated using the Arrhenius equation and the provided information about the doubling of the rate constant as the temperature increases from 100 K to 200 K. An alternative two-point form of the Arrhenius equation can also be used for this calculation.

The activation energy of a reaction can be calculated by utilizing the Arrhenius equation, which relates the effect of temperature on the rate constant, k of a reaction: k = Ae−Ea/RT. In this equation, R is the ideal gas constant (8.314 J/mol/K), T is the temperature in Kelvin, Ea is the activation energy in joules per mole, A is a constant known as the frequency factor, and e is a mathematical constant.

Given the data that when the temperature increases from 100 K to 200 K the rate constant (k) doubles, we can use an alternative approach to calculate the activation energy. This involves rearranging the Arrhenius equation in a two-point form and inserting the provided temperatures and rate constants. Substitution and subsequent calculation will yield the activation energy value in kJ/mol.

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Answer:

The answer is False.

Explanation:

When mixed with an acidic liquid etc. a violent reaction happens creating Hydrogen Gas.

The statement that sodium borohydride is stable in acidic aqueous solutions is false.

Sodium borohydride is a very useful reducing agent in chemistry. It is able to carry out many important organic transformations where reduction reactions are involved.

However, sodium borohydride is unstable in acid solutions. It decomposes in acidic and even neutral solutions to release hydrogen gas. Therefore, the statement that sodium borohydride is stable in acidic aqueous solutions is false.

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Answer:

Final temperature = 48.6867 °C

Explanation:

The expression for the heat of combustion in bomb calorimeter is:

ΔH = -C ΔT

where,

ΔH is the enthalpy of the reaction

C is the heat capacity of the bomb calorimeter

ΔT is the temperature change

Given in the question:

ΔH°rxn = -1.28 x 10⁴ kJ

From the balanced reaction, 4 moles of aniline reacting with oxygen. Thus, enthalpy change of the reaction in kJ/mol is:

ΔH = -12800 kJ/4 = -3200kJ/mole

Given:

Mass of aniline combusted = 6.55 g

Molar mass of aniline = 93.13 g/mol

Thus moles of aniline = 6.55 / 93.13 moles = 0.0703 moles

The total heat released from 0.0703 moles of aniline is

ΔH = -3200kJ/mole x 0.0703 moles = -224.96 kJ

Given: Heat capacity of calorimeter is 14.25 kJ/°C

T₁ (initial) = 32.9°C

T₂ (final) = ?

From the above formula:

-224.96 kJ = -14.25kJ/°C (T₂ - 32.9)

Solving for T₂ , we get:

T₂ = 48.6867 °C

Answer:

CH₄ < CH₃Cl < CH₃OH

Explanation:

In all three compounds, the central C atom has four bonds, so the molecules all have a tetrahedral molecular geometry.

The only difference is in one bond:  CH₃-Cl, CH₃-H, and CH₃-OH.

CH₃-Cl: The C-Cl bond is polar, so the strongest intermolecular forces are dipole-dipole.

CH₄: the molecule is symmetrical, so the strongest intermolecular forces are London dispersion forces.

CH₃-OH: The OH group can form hydrogen bonds.

The order of strength of intermolecular forces is

London dispersion forces < dipole-dipole < hydrogen bonds

The order of boiling points is

CH₄ < CH₃Cl < CH₃OH

The boiling points of compounds CH3Cl, CH4, and CH3OH increase in the order CH4, CH3Cl, and CH3OH due to the increasing strength of intermolecular forces namely London dispersion forces, dipole-dipole interactions, and hydrogen bonding, respectively.

The boiling points of these compounds, CH3Cl, CH4, and CH3OH, vary due to their different types of intermolecular forces. Methane (CH4) is a nonpolar molecule that only has London dispersion forces, therefore, it is expected to have the lowest boiling point. Chloromethane (CH3Cl) is slightly polar due to the large electronegativity difference between Carbon and Chlorine, and it will have stronger London dispersion forces than CH4 as well as minor dipole-dipole interactions, giving it a higher boiling point than CH4. Methanol (CH3OH) has the strongest intermolecular forces due to the presence of hydrogen bonding. Therefore, CH3OH is expected to have the highest boiling point. The ordering from lowest to highest boiling point is therefore CH4 < CH3Cl < CH3OH.

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Answer:

adding 750 grams of water at 60° Celsius .

Explanation:

Q = m.c.ΔT,

Where, Q is the amount of heat lost by water (Q = ??? J).

m is the mass of water (m = 750.0 g).

c is the specific heat capacity of the water (c = 4.18 J/g.°C).

ΔT is the temperature difference (ΔT = final T - initial T = - 10.0°C, the temperature of water is lowered by 10.0°C).

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(- 10.0°C) = - 31350.0 J = -31.350 kJ.

Now, we can calculate the Q that is gained by the different added amounts of water:

ΔT = 70.0°C - 50.0°C = 20.0°C,

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(20.0°C) = 62700.0 J = 62.70 kJ.

ΔT = 70.0°C - 60.0°C = 10.0°C,

∴ Q = m.c.ΔT = (325.0 g)(4.18 J/g.°C)(10.0°C) = 13585.0 J = 13.585 kJ.

ΔT = 70.0°C - 60.0°C = 10.0°C,

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(10.0°C) = 31350.0 J = 31.350 kJ.

ΔT = 70.0°C - 55.0°C = 15.0°C,

∴ Q = m.c.ΔT = (1000.0 g)(4.18 J/g.°C)(15.0°C) = 62700.0 J = 62.70 kJ.

adding 750 grams of water at 60° Celsius

Answer:

0.185moles of Al₂O₃

Explanation:

Mass of Al = 10g

Mass of O₂ = 19g

Equation of the reaction: 4Al + 3O₂ → 2Al₂O₃

This is the balanced reaction equation.

Solution

From the given parameters, the reactant that would determine the extent of the reaction is Aluminium. It is called the limiting reagent. Oxygen is in excess and it is in an unlimited supply.

Working from the known mass to the unknown, we simply solve for the number of moles of Al using the mass given.

Then from the equation, we can relate the number of moles of Al to that of Al₂O₃ produced:

 Number of moles of Al = [tex]\frac{mass}{molar mass}[/tex]

                                        =   [tex]\frac{10}{27}[/tex]

                                        = 0.37mol

From the equation:

         4 moles of Al produced 2 moles of Al₂O₃

    0.37 mole will yield:  [tex]\frac{2 x 0.37}{4}[/tex] = 0.185moles of Al₂O₃

0.74 moles of aluminium oxide can be produced from the reaction of 10.0 g of aluminium and 19.0 of oxygen.

The equation of the reaction between aluminium and oxygen gas to produce aluminium oxide is given below as follows:

[tex]4Al + 3O₂ → 2Al₂O₃[/tex]

From the equation of reaction 4 moles of aluminium reacts with 3 moles of oxygen to produce 2 moles of aluminium oxide.

The moles of reactants is calculated using:

molar mass of aluminium = 27.0 g

molar mass of oxygen = 32.0 g

moles of aluminium = 10/27 = 0.37 moles

moles of oxygen = 19/32 = 0.59

Aluminium is the limiting reactant

Thus, moles of aluminium oxide produced = 0.37 × 2 = 0.74 moles

Therefore, 0.74 moles of aluminium oxide can be produced from the reaction of 10.0 g of aluminium and 19.0 of oxygen.

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Answer:

Gay-Lussac's law, Amontons' law or the pressure law was found by Joseph Louis Gay-Lussac in 1809.

Explanation:

It states that, for a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.

Answer:

Gay-Lussac's law

Explanation:

Gay-Lussac's second law, which is called isochoric transformation or isovolumetric transformation, is related to the behavior of gases when subjected to a constant volume. The elaboration of this law was attended by the French scientist Jacques Alexandre Cesar Charles.

According to Lussac, when a gas is placed in a container at a constant volume, it is found that if a pressure is exerted on that gas, a proportional increase in the absolute temperature of that gas will occur.

Generally, under Gay-Lussac law, a gas's pressure and temperature will always be directly proportional as long as the volume is constant. Thus, increasing the pressure increases the temperature; decreasing the pressure decreases the temperature.