A boy reached his home by riding motorcycle within 5 minutes by travelling 8 km distance ,them what is his average velocity?​

Answer: The molecular formula for the given organic compound X is [tex]C_6H_{8}O_7[/tex]

Explanation:

We are given:

Mass of [tex]CO_2=4.13g[/tex]

Mass of [tex]H_2O=1.13g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 4.13 g of carbon dioxide, =[tex]\frac{12}{44}\times 4.13=1.13g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 1.13 g of water, [tex]\frac{2}{18}\times 1.13=0.125g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (3.00) - (1.13+ 0.125) = 1.75 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.13g}{12g/mole}=0.094moles[/tex]

Moles of Hydrogen =[tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.125g}{1g/mole}=0.125moles[/tex]

Moles of Oxygen =[tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.75g}{16g/mole}=0.109moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles

For Carbon = [tex]\frac{0.094}{0.094}=1[/tex]

For Hydrogen = [tex]\frac{0.125}{0.094}=1.33[/tex]

For Oxygen = [tex]\frac{0.109}{0.094}=1.16[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1: 1.33: 1.16

Converting them into whole number ratios by multiplying by 6:

The ratio of C : H : O = 6: 8: 7

Hence, the empirical formula for the given compound is [tex]C_6H_8O_7[/tex]

Empirical mass = [tex]6\times 12+8\times 1+7\times 16=192g[/tex]

The equation used to calculate the valency is :

[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]

Putting values in above equation, we get:

[tex]n=\frac{192g/mol}{192g/mol}=1[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_6H_8O_7\times 1=C_6H_{8}O_7[/tex]

Thus molecular formula for the given organic compound X is [tex]C_6H_{8}O_7[/tex]

Answer:

Explanation:

Precision generally refers to the repeatability of measurements. That is, the closeness of repeated meaurements in values. Accuracy, on the other hand, refers to the closeness of a measured value to the true value.

For apprentice X, the three measurements taken (26.5, 26.6, 26.4) were close in values with an average of 26.5 and a mean deviation of 0.07. Hence, the apprentice is said to be precise in measurement. However, the average of the measurement is quite far from the true length of meaurement. Hence, the apprentice is inaccurate.

Apprentice Y had a mean measurement value of 27.6 and a mean deviation of 0.2. The measurements are far from each other and the apprentice is therefore said to be far from being precise. The mean measurement value is also far from true value. Hence the apprentice is also inaccurate in his/her measurement.

Apprentice Z has a mean measurement of 27.07 and a mean deviation of 0.11.  The repeated measurements were close enough to be precise while the average is also close to the true length, and hence, the measurement is said to be accurate.

Answer:

ai) Rate law,  [tex]Rate = k [CH_3 Cl] [Cl_2]^{0.5}[/tex]

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

[tex]CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)[/tex]

If [tex]A + B \rightarrow C + D[/tex], the rate of backward reaction is given by:  

[tex]Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}[/tex]

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

[tex]0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1[/tex]

Using the information provided in lines 3 and 4 of the table and insering the value of a:

[tex]0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\[/tex]

[tex]0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5[/tex]

The rate law is: [tex]Rate = k [CH_3 Cl] [Cl_2]^{0.5}[/tex]

The rate constant [tex]k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}[/tex] then becomes:

[tex]k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25[/tex]

b) Overall order of reaction =  a + b

Overall order of reaction = 1 + 0.5

Overall order of reaction = 1.5

The rate law is for this reaction is    [tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]

where the rate constant is k = [tex]1.25 M^{-0.5} s^{-1}[/tex] and the overall order of the reaction is 1.5.

Let's consider the following reaction.

CH₃Cl(g) + 3 Cl₂(g) → CCl₄(g) + 3 HCl(g)

The rate law for a chemical reaction is an equation that relates the reaction rate with the concentrations of the reactants.

The  rate law for this reaction is:

r = k [CH₃Cl]ᵃ [Cl₂]ᵇ

where,

If we write the ratio r₂/r₁, we get:

r₂/r₁ = k [CH₃Cl]₂ᵃ [Cl₂]₂ᵇ / k [CH₃Cl]₁ᵃ [Cl₂]₁ᵇ

r₂/r₁ = {[CH₃Cl]₂/ [CH₃Cl]₁}ᵃ

(0.029 M/s)/(0.014 M/s) = {0.100M/0.050 M}ᵃ

a ≈ 1

If we write the ratio r₃/r₂, we get:

r₃/r₂ = k [CH₃Cl]₃ᵃ [Cl₂]₃ᵇ / k [CH₃Cl]₂ᵃ [Cl₂]₂ᵇ

r₃/r₂ = {[Cl₂]₃/[Cl₂]₂}ᵇ

(0.041 M/s)/(0.029 M/s) = {0.100 M/0.050 M}ᵇ

b ≈ 0.5

So far, the rate law is:

[tex]r = k [CH_3Cl] [Cl_2]^{0.5}[/tex]

Let's use the values of the first experiment to find the value of the rate constant.

[tex]k = \frac{r}{[CH_3Cl][Cl_2]^{0.5} } = \frac{0.014 M/s}{(0.050 M)(0.050 M)^{0.5} } = 1.25 M^{-0.5} s^{-1}[/tex]

The final rate law is:

[tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]

It is the sum of the individual orders of reaction.

a + b = 1 + 0.5 = 1.5

The rate law is for this reaction is    [tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]

where the rate constant is k = [tex]1.25 M^{-0.5} s^{-1}[/tex] and the overall order of the reaction is 1.5.

Learn more about the rate law here: https://brainly.com/question/14945022

Answer:

Explanation:

2 HgCl₂ + C₂O₄²⁻   =  2 Cl⁻ + Hg₂Cl₂ + 2CO₂

1 )

Rate of reaction [tex]= k[HgCl_2]^m[C_2O_4^{-2}]^n[/tex]

             [HgCl₂]        [C₂O₄²⁻ ]           Rate  

1 .              .124             .115               1.61 x 10⁻⁵

2 .             .248             .115             3.23 x 10⁻⁵

3 .              .124             .229              6.4 x 10⁻⁵

4 .              .248             .229            1.28 x 10⁻⁴

comparing 1 and 3 , when concentration of HgCl₂ remains constant and concentration of C₂O₄²⁻  becomes twice , rate becomes 4 times so rate is proportional to square of concentration of C₂O₄²⁻  .

Hence n = 2

comparing 1 and 2 , when concentration of HgCl₂ becomes twice  and concentration of C₂O₄²⁻  remains constant  , rate becomes 2 times so rate is proportional to simply  concentration of C₂O₄²⁻  .

Hence m = 1

Putting the data of  1 in the rate equation found

 1.61 x 10⁻⁵ = k x .124 x  .115²

k = 11.3 x 10⁻⁴ M⁻² s⁻¹

hey there!

atomic mass cobalt = 58.9332 amu

therefore:

1 mol Co ------------ 58.9332 g

moles ------------ 50 g

moles = 50 x 1 / 58.9332

moles  = 50 / 58.9332

 = 0.8484 moles of Co

Hope this helps!

0.8484 moles are in 50 grams of cobalt.

A mole is defined as 6.02214076 × [tex]10^{23}[/tex] of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Atomic mass cobalt = 58.9332 amu

Therefore:

1 mol Co ------------ 58.9332 g

moles ------------ 50 g

moles = 50 x 1 ÷ 58.9332

moles  = 50 ÷ 58.9332

= 0.8484 moles of Co

Learn more about moles here:

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Answer:

ATGGCCTA

Explanation:

For this we have to keep in mind that we have a specific relationship between the nitrogen bases:

-) When we have a T (thymine) we will have a bond with A (adenine) and viceversa.

-) When we have C (Cytosine) we will have a bond with G (Guanine) and viceversa.

Therefore if we have: TACCGGAT. We have to put the corresponding nitrogen base, so:

TACCGGAT

ATGGCCTA

I hope it helps!

Answer:

a) Based on the given information, the volume of the solvent given is 999 ml and the density of water given is 0.9982 gram per ml.  

The mass of solvent can be calculated by the formula mass = density * volume

mass = 0.9982 * 999 = 0.997 Kg

The molarity of the solution given is 2.300 * 10^-2 M

Molality of the glycerol solution can be calculated by using the formula,  

Molality = molarity/solvent (kg) = 2.300 * 10^-2 / 0.997 = 0.023 m

b) Molarity or the moles of the solute given is 2.300 * 10^-2 moles

The moles of solvent can be determined by using the formula, n = mass of solvent/mol.wt = 997/18 (mol.wt of solvent is 18 g/mol), now putting the values we get,  

n = 997/18 = 55.4

The mole fraction of the glycerol will be = 2.300 * 10^-2 M/(2.300 * 10^-2)+55.40

= 4.15 * 10^-4

c) The mass percent of glycerol can be determined by using the formula,  

mass of solute/mass of solution * 100% ----- (i)

The mass of solute can be determined by using the formula,

n = mass of solute/mol.wt of solute

The n or the no. of moles is 2.300 * 10^-2 and the molecular weight of glycerol is 92.09 g/mol. Now putting the values we get,  

mass = 2.300 * 10^-2 * 92.09 = 2.12 grams

Now putting the values in equation (i) we get,  

mass percent = 2.12 / 997 * 100% = 0.21%

d) Based on the above calculation, the mass of solute (glycerol) is 2.12 g or 2.12 * 1000 mg

The volume of water is 999 ml or 999 * 10^-3 L

The concentration of the glycerol solution will be,  

Concentration = 2.12 * 10^3 mg/999 * 10^-3 L  

= 2.12/999 * 10^6 mg/L

= 2122.1 ppm

Considering the solution of molality, mole fraction, mass percentage and ppm, you obtain that:

a) The molality of the glycerol solution is 0.02306 molal.

b) The mole fraction of glycerol in the solution is 4.15×10⁻⁴.

c) The percent by mass is 0.212%.

d) The concentration of the glycerol solution is 2118.07 ppm.

Molality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.

The Molality of a solution is determined by the expression:

[tex]molality=\frac{number of moles of solute}{kilogramsof solvent}[/tex]

In this case, you have a 2.300×10⁻² M solution of glycerol (C₃H₈O₃) in water. The sample was created by dissolving a sample of C₃H₈O₃ in water and then bringing the volume up to 1.000 L.

So, being the molarity the number of moles of solute that are dissolved in a certain volume, the number of moles of glycerol can be calculated as:

number of moles of glycerol= 2.300×10⁻² M× 1 L

number of moles of glycerol= 2.300×10⁻² moles

On the other side, the volume of water needed was 999 mL and the density of water at 20.0∘C is 0.9982 [tex]\frac{g}{mL}[/tex]. So, the mass of water needed can be calculated as:

999 mL×0.9982[tex]\frac{g}{mole}[/tex] = 997.2 grams of water= 0.9972 kg of water

Then, the molality of the solution is:

[tex]molality=\frac{2.300x10^{-2} moles}{0.9972 kg}[/tex]

molality= 0.02306 molal

Finally, the molality of the glycerol solution is 0.02306 molal.

The molar fraction is a way of measuring the concentration that expresses the proportion in which a sustance is found with respect to the total moles of the solution.

In this case, the number of moles of glycerol= 2.300×10⁻² moles

Having 997.2 grams of water, the moles of solvent can be determined knowing that the molar mass of water is 18[tex]\frac{g}{mole}[/tex]:

[tex]number of moles of water=997.2 gramsx\frac{1 mole}{18 grams}[/tex]

number of moles of water= 55.4 moles of water

Being the number of total moles the sum of the moles of grycerol and the numbers of moles of water, the mole fraction can be calculated as:

[tex]mole fraction=\frac{2.300x10^{-2} moles}{2.300x10^{-2}moles+55.4 moles}[/tex]

mole fraction= 4.15×10⁻⁴

In summary, the mole fraction of glycerol in the solution is 4.15×10⁻⁴.

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

The mass percentage of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The mass percentage is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

[tex]percent by mass=\frac{mass of solute}{mass of solution}x100[/tex]

In this case, remmeber that you have a number of moles of glycerol of 2.300×10⁻² moles .

Being the molar mass glycerol 92.09 [tex]\frac{g}{mole}[/tex], the mass of glycerol can be calculated as:

2.300×10⁻² moles×92.09 [tex]\frac{g}{mole}[/tex]= 2.11807 grams of glycerol

Remembering that you have 997.2 grams of water, the percent by mass is calculated as:

[tex]percent by mass=\frac{mass of glycerol}{mass of glycerol + mass of water}x100[/tex]

Solving:

[tex]percent by mass=\frac{2.11807 grams}{2.11807 grams + 997.2 grams}x100[/tex]

[tex]percent by mass=\frac{2.11807 grams}{999.31807 grams}x100[/tex]

percent by mass= 0.212%

Finally, the percent by mass is 0.212%.

Parts per million (ppm) is a unit of measurement of concentration that measures the number of units of substance in each million units of the whole. In this case, the concentration measurement refers to mg of glycerol per L of solution.

Being the mass of glycerol 2.11807 grams equal to 2118.07 mg (1 g=1000mg), the concentration is:

[tex]concentration=\frac{2118.07 mg}{1 L solution}[/tex]

Solving:

concentration= 2118.07 ppm

In summary, the concentration of the glycerol solution is 2118.07 ppm.

Learn more about:

Answer:

B. The principal quantum number (n) is 6, the angular momentum quantum number (ell) is 2, and there are 5 electrons in the subshell.

Explanation:

Given;

electronic configuration of 6d⁵

where;

6 in the configuration shows that, the principal quantum number (n) is 6

d is the subshell

d orbital has angular momentum quantum number (l) of 2

            s orbital, l = 0

            p orbital, l = 1

            d orbital, l = 2

            f orbital,  l = 3

5 in the configuration means that there are 5 electrons in the subshell

Therefore, 6d⁵ symbol means that "The principal quantum number (n) is 6, the angular momentum quantum number (ell) is 2, and there are 5 electrons in the subshell".

Answer: CO2 and H2O

Explanation: I already took the test it's right

Answer:

that is why co2 is in the power of 2ik

Answer:

building and managing their own clients, purchasing supplies and keeping records

hope this helps you

Answer:

ΔH° = -1255.8 kJ

Explanation:

Step 1: Data given

ΔHf∘ C2H2 = 227.4 kJ/mol

ΔHf∘ O2 = 0 kJ/mol

ΔHf∘ CO2 = -393.5 kJ/mol

ΔHf∘ H2O = -241.8 kJ/mol

Step 2: The balanced equation

2C2H2(g)+5O2(g) ⟶ 4CO2(g)+2H2O(g)

C2H2(g)+5/2O2(g) ⟶ 2CO2(g)+H2O(g)

Step 3: Calculate ΔH° of the reaction

ΔH° = (2*ΔHf∘ CO2 + ΔHf∘ H2O) - (ΔHf∘ C2H2)

ΔH° = (2* -393.5 kJ/mol + (-241.8) kJ/mol) - 227.4 kJ/mol

ΔH° = -787 - 241.8 - 227. kJ/mol

ΔH° = -1255.8 kJ

Answer: EXTINCTION

Explanation:

Answer:

extinction

Explanation:

Explanation:

H

|

H-C=C-C-OH

| | |

H H H

in this organic compound the functional groups are OH and alkene

prop-2-enol

since it has a double bond, alkene is the functional group. and also it has OH group so hydroxyl group also the other functional group

Explanation:

Moles=mass/molar mass

moles × molar mass = mass

0.206 x 119= mass

Mass= 24. 51grams

Answer:

24.51 g to the nearest hundredth

Explanation:

1 mole of SOCl2 =  32 + 16 + 2 *35.5 = 119 g

0.206 mol = 0.206 * 119

=  24.51 g.

Answer:

d

Explanation:

it's called the hydrocarbon

Answer:

A hydrocarbon

Explanation:

Answer:

1. Increase in the temperature of the water

2. Increasing the surface area of the lithium

Explanation:

1. Increase in the temperature of the water

The activation energy for the lithium water reaction is +161 kJ/mol while the activation energy for the sodium is +109 kJ/mol, hence for increased reaction rate, the water temperature will be raised to enable more lithium atoms enter into reaction with the water molecules as their energy is increased lowering the activation energy required for the reaction.

2. Increasing the surface area of the lithium

As the lithium floats on the water, due to its low temperature and the heat evolved from the reaction of lithium with the cold water is below the melting point of lithium, the reaction rate can be increased by increasing the surface area of lithium sample by grinding so as to increase the number of lithium water reaction sites.

Answer: Thus the final temperature for both the mercury and the water is [tex]28.8^0C[/tex]

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]heat_{released}=heat_{absorbed}[/tex]

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]-[m_1\times c_1\times (T_{final}-T_1)]=[m_2\times c_2\times (T_{final}-T_2)][/tex]

Q = heat absorbed or released

[tex]m_1[/tex] = mass of mercury= 452 g

[tex]m_2[/tex]= mass of water = 145 g

[tex]T_{final}[/tex]  = final temperature = ?

[tex]T_1[/tex]   = temperature of mercury = [tex]85.0^0C[/tex]

[tex]T_2[/tex] = temperature of water = [tex]23.0^0C[/tex]

[tex]c_1[/tex]   = specific heat of mercury  = [tex]0.138J/g^0C[/tex]

[tex]c_2[/tex]   = specific heat of water= [tex]4.184J/g^0C[/tex]

Now put all the given values in equation (1), we get

[tex]-[452\times 0.138\times (T_f-85.0)^0C]=145\times 4.184\times (T_f-23.0)^0C[/tex]

[tex]T_f=28.8^0C[/tex]

Thus the final temperature for both the mercury and the water is [tex]28.8^0C[/tex]

Answer:

The metal surface becomes more positive as electrons are lost from it.

Explanation:

Let us note that photoelectric effect refers to a phenomenon in which electrons are ejected from a clean metal surface irradiated with light of appropriate frequency. This photon must possess a frequency above the threshold frequency of the metal and its energy must exceed the work function of the metal. When these conditions are met, electrons are emitted from a clean metal surface, having a constant kinetic energy as long as the frequency of the incident photon remains constant.

However, as photoelectric effect progresses and electrons are lost from the metal surface, the metal surface becomes more positive. The more positive the surface, the greater the attraction of the positive surface for the emitted electrons. This reduces the kinetic energy of the emitted photons even though the frequency of incident photons is held constant.

Answer:

P = 20.1697 atm

Explanation:

In this case we need to use the ideal gas equation which is:

PV = nRT (1)

Where:

P: Pressure (atm)

V: Volume (L)

n: moles

R: universal gas constant (=0.082 L atm / K mol)

T: Temperature

From here, we can solve for pressure:

P = nRT/V  (2)

According to the given data, we have the temperature (T = 20 °C, transformed in Kelvin is 293 K), the moles (n = 125 moles), and we just need the volume. But the volume can be calculated using the data of the cylinder dimensions.

The volume for any cylinder would be:

V = πr²h  (3)

Replacing the data here, we can solve for the volume:

V = π * (17)² * 164

V = 148,898.93 cm³

This volume converted in Liters would be:

V = 148,898.93 mL * 1 L / 1000 mL

V = 148.899 L

Now we can solve for pressure:

P = 125 * 0.082 * 293 / 148.899

The question is incomplete, the complete question is;

Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 28.2 g of carbon dioxide is produced from the reaction of 15.1 g of methane and 81.2 g of oxygen gas, Calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.

Answer:

71.1%

Explanation:

The balanced reaction equation must first be written;

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

Let us obtain the number of moles of carbon dioxide corresponding to 28.2 g

Number of moles = mass/ molar mass= 28.2/44.01 g/mol = 0.64 moles of CO2

Next, we obtain the limiting reactant. This is the reactant that yields the least amount of product.

For methane;

Number of moles in 15.1 g= mass/molar mass= 15.1/16gmol-1 = 0.9 moles

From the reaction equation;

1 mole of methane yields 1 mole of carbon dioxide

Hence 0.9 moles of methane yields 0.9 moles of carbon dioxide.

For oxygen

Number of moles of oxygen corresponding to 81.2 g of oxygen= mass/ molar mass= 81.2g/32gmol-1 = 2.5 moles of oxygen

From the reaction equation;

2 moles of oxygen gas yields 1 mole of carbon dioxide

2.5 moles of oxygen gas yields 2.5 × 1 /2 = 1.25 moles of carbon dioxide.

Methane is the limiting reactant.

Theoretical yield of carbon dioxide= 0.9 moles

Actual yield of carbon dioxide= 0.64 moles

% yield = actual yield/ theoretical yield × 100

% yield= 0.64/0.9 ×100

% yield = 71.1%

Answer:

3.80*10⁻³ moles of HCl can be produced from 0.226 g of SOCl₂

Explanation:

The balanced reaction is:

SOCl₂ + H₂O ----> SO₂ + 2 HCl

By stoichiometry of the reaction they react and produce:

Being:

the molar mass of the compounds participating in the reaction is:

Then, by stoichiometry of the reaction, the following amounts of mass react and are produced:

Then the following rule of three can be applied: if by stoichiometry of the reaction 118.9 grams of SOCl₂ produce 2 moles of HCl, 0.226 grams of SOCl₂ how many moles of HCl do they produce?

[tex]moles of HCl=\frac{0.226 grams of SOCl_{2}*2 mole of HCl }{118.9 grams of SOCl_{2} }[/tex]

moles of HCl= 3.80*10⁻³

3.80*10⁻³ moles of HCl can be produced from 0.226 g of SOCl₂

Answer:

The interaction between sodium ion and the partial negative charge on the oxygen is stronger.

Explanation:

The predominant interaction that exists between sodium ion and the partial negative charge on the oxygen is ion-dipole interaction.

The predominant interaction that exists between two ethanol molecules is hydrogen bonding interaction.

The order of strength of the intermolecular interactions in decreasing order:

Ionic bond > ion-dipole interaction > hydrogen bonding > dipole-dipole interaction > ion-induced dipole interaction > induced dipole- dipole interaction > london force

So, ion-dipole interaction is stronger than hydrogen bonding.

Hence, the interaction between sodium ion and the partial negative charge on the oxygen is stronger.

Answer:

Calcium hydride

Explanation:

Answer:

Yes it is true

Explanation:

This is because the Clausius theorem states that The cyclic integral always has two defined results. The results include it being less than or equal to zero under certain conditions.

When the system consists of only reversible processes, the cyclic integral is equal to zero. If it consists of and irreversible processes, the integral is usually less than zero.

Answer:

The  pKa  of  Glu is   [tex]pK_a = 4.25[/tex]

Explanation:

From the question we are told that

  The [tex]K_a[/tex] value of  Glu is [tex]K_a = 5.62*10^{-5}[/tex]

The [tex]pKa[/tex] is mathematically represented as

         [tex]pK_a = -log (K_a )[/tex]

         substituting values

          [tex]pK_a = -log (5.62 *10^{-5} )[/tex]

          [tex]pK_a = 4.25[/tex]

Answer:

solubility is 1.984x10⁻⁹M

Explanation:

When CaCO₃ is in water, the equilibrium that occurs is:

CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)

Kps = [Ca²⁺] [CO₃²⁻] = 4.96x10⁻⁹

If you have a 0.250M solution of Na₂CO₃, [CO₃²⁻] = 0.250M:

[Ca²⁺] [0.250M] = 4.96x10⁻⁹

Assuming you are adding an amount of CaCO₃:

[X] [0.250 + X] = 4.96x10⁻⁹

Where X is the amoun of CaCO₃ you can add, that means, solubility

X² + 0.250X - 4.96x10⁻⁹ = 0

Solving for X:

X = -0.25M → False answer, there is no negative concentrations.

X = 1.984x10⁻⁹M.

That means, solubility is 1.984x10⁻⁹M

Answer:

[tex]1.984x10^{-8}M[/tex]

Explanation:

Hello,

In this case, the equilibrium reaction for the solubility of calcium carbonate is:

[tex]CaCO_3(s) \rightleftharpoons Ca^{2+}(aq)+CO_3^{-2}(aq)[/tex]

In such a way, since 0.250 M sodium carbonate solution is the solvent, we assume an initial concentration of carbonate anion to be also 0.250 M since sodium carbonate is completely dissolved, for that reason the equilibrium equation turns out:

[tex]Ksp=[Ca^{2+}][CO_3^{2-}]\\\\4.96x10^{-9}=x*(0.250+x)[/tex]

Hence, solving for [tex]x[/tex] we have:

[tex]x=1.984x10^{-8}M[/tex]

Which corresponds to the molar solubility if calcium carbonate as well.

Regards.

Answer:

THE MOLECULAR FORMULA FOR THE COMPOUND IS C2 Cl2

Explanation:

mass = 0.04380 g

Volume = 10 mL = 10 / 1000 L = 0.010 L

Pressure = 1.10 atm

R = 0.082 L atm mol^-1 K^-1

Temperature = 290.5 K

Carbon = 25.305 %

Chlorine = 74.695 %

Atomic mass of carbon = 12

Atomic mass of chlorine = 35.5

First calculate the empirical formula by following these steps:

1. Write the percentage composition of the elements involved and divide by its atomic mass

Carbon = 25.305 / 12 = 2.10875

Chlorine = 74.695 /35.5 = 2.1040

2. Divide each by the smaller value

Carbon = 2.10875 / 2.1040 = 1.002

Chlorine = 2.1040 / 2.1040 = 1

3. Round up to the whole number

The empirical formula is C Cl

Next is to calculate the molar mass of the compound using ideal gas equation

PV = mRT/ MM

Mm = mRT / PV

Mm = 0.04380 * 0.082 * 290.5 / 1.10 * 0.010

Mm = 1.0433 / 0.011

Mm = 94.845 g/mol

Mm = 94.85 g/mol

Now that we know the molar mass, we can go on to calculate the molecular formula:

(C Cl) n = Molar mass

( 12 + 35.5) n = 94.85

(47.5)n = 94.85

n = 94.85 / 47.5

n = 1.9968

n ~ 2

The molecular formula can then be written as C2Cl2.

Image is not given in the question, so the image for the question is given below.

Answer:

Fischer projection is a two-dimensional representation of a three-dimensional organic molecule by projection.

Wedge and dash bonds are used to represent the three-dimensional structure of a molecule, in which wedges indicates bonds towards the viewer, solid lines indicates bonds in the plane of the image and dashed lines indicates bonds away from the viewer.

Wedge and dash bonds structure of D-glyceraldehyde is attached below.