Given that S is the central atom, draw a Lewis structure of OSF4 in which the formal charges of all atoms are zero. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons.

Answer:

Monosodium Glutamate (MSG) is a chemical which is used in agricultural of food chemistry.

Explanation:

Monosodium Glutamate (MSG) is a chemical that is used in types of different  food as food additives. The functional group that is present in Glutamate are carboxylic acid and amine. This chemical is used in different types of foods which is responsible for enhancing the taste of the food. Monosodium Glutamate is safe if it is used in moderate dose but adversely affected when it is used in large amount.

Answer:

[tex]_1^{1} {P}[/tex] is symbol for proton emission in the nucleus.

[tex]_2^{4} {He}[/tex] symbolises alpha emission, equivalent to helium atom emission of a radioactive particle

[tex]+_1^{0} {e}[/tex] is the radiation symbol for positiron particle. which occurs when beta + radioactive decay occurs

Answer: The molecular formula for the given organic compound is [tex]C_{18}H_{20}O_2[/tex]

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=39.61g[/tex]

Mass of [tex]H_2O=9.01g[/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 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, [tex]\frac{12}{44}\times 39.61=10.80g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 9.01 g of water, [tex]\frac{2}{18}\times 9.01=1.00g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

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

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles[/tex]

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

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = [tex]\frac{0.9}{0.10}=9[/tex]

For Hydrogen = [tex]\frac{1}{0.10}=10[/tex]

For Oxygen = [tex]\frac{0.10}{0.10}=1[/tex]

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is [tex]C_9H_{10}O[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

[tex]n=\frac{268.34g/mol}{134g/mol}=2[/tex]

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

[tex]C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2[/tex]

Thus, the molecular formula for the given organic compound is [tex]C_{18}H_{20}O_2[/tex].

Answer:

No additional particle was produced during the decay.

Explanation:

The equation of decay is given as;

¹⁰₆C  + ⁰₋₁ e → ¹⁰₅B + x

To identify x, we have to calculate its atomic and mass number.

In the reactants side;

Atomic Number = 6 + (-1) = 5

Mass number = 10 + 0 = 10

In the products side;

Atomic Number = 5 + x

Mass Number = 10 + x

Generally, reactant = product

Atomic Number;

5 = 5 + x

x = 5 - 5 = 0

Mass Number;

10 = 10 + x

x = 10 - 10 = 0

This means no additional particle was produced during the decay.

Answer:10 grams of CO2

Explanation:

Yeild= exp. yeild÷ thoretical yeild × 100

Yeild= 73.3÷73.4 × 100

Yeild= 0.1 ×100

Yeild= 10

Answer:

In 0 group of the periodic table

Explanation:

So they will not react with other atoms because they have a full outer shell of electrons and an overall charge of 0.

Hope it helps.

Answer:

The answer is 5.

Answer:

balanced because the total number of oxygen atoms is 3.

Answer:

7.25 x 10^47

Explanation:

1.204 x 10^24 moles*6.022 x 10^23 avogadro's number= 7.25 x 10^47

Answer:

1.  2.28 g

2. 0.82 g

3. 1.35 g

4. 36 %

5. 59 %

6. 2.17 g

7. 95 %

Explanation:

Hello,

1. In this case, the mass of the unknown mixture is obtained by subtracting the mass of the dish plus sample and the mass of the dish:

m = 26.57 g- 24.29 g = 2.28 g

2. In this case, the mass of the NaCl recovered is obtained by subtracting the mass of the dish plus NaCl and the mass of the dish:

m = 68.66 g- 67.84 g = 0.82 g

3. In this case, the mass of the sand recovered is obtained by subtracting the mass of the filter paper plus sand and the mass of the filter paper:

m = 37.69 g- 36.34 g = 1.35 g

4. The percentage of NaCl is computed by considering its mass and the mass of the unknown mixture:

% NaCl = 0.82 g / 2.28 g * 100 % = 36 %

5. The percentage of sand is computed by considering its mass and the mass of the unknown mixture:

% sand = 1.35 g / 2.28 g * 100 % = 59 %

6. Here,we have to add the mass of NaCl and sand:

m = 0.82 g + 1.35 g = 2.17 g

7. Finally, the percent recovery is obtained by diving the total recovered mass by the total obtained mass of the mixture:

% recovery = 2.17 g / 2.28 g * 100 % = 95 %

Best regards.

Answer:

1. H4C-NH-CH3

2. H3C-NH-C-CH3

H2C-CH3

1 +

H3C-CH2-N-CH3

CH3

N

3. H

4.

.

.

.

.

.

.

Answer:

66.0 atm

Explanation:

We can calculate the osmotic pressure (π) using the following expression.

[tex]\pi = i \times M \times R \times T[/tex]

where,

Step 1: Calculate i

Sodium sulfate completely dissociates according to the following equation.

Na₂SO₄ ⇒ 2 Na⁺ + SO₄²⁻

Since it produces 3 ions, i = 3.

Step 2: Calculate M

We can calculate the molarity of Na₂SO₄ using the following expression.

[tex]M = \frac{mass\ of\ solute }{molar\ mass\ of\ solute\ \times liters\ of\ solution} = \frac{65.0g}{142.04g/mol \times 0.500L} =0.915M[/tex]

Step 3: Calculate T

We will use the following expression.

K = °C + 273.15

K = 20°C + 273.15 = 293 K

Step 4: Calculate π

[tex]\pi = 3 \times 0.915M \times \frac{0.08206atm.L}{mol.K} \times 293K =66.0 atm[/tex]

Answer:

25,50,25

Explanation:

Answer:

Boron

Explanation:

Answer: a. boron

Explanation: bc the Boron Family is named after the first element in the family. atoms in this family have 3 valence electrons. this family includes a metalloid (boron), and the rest are metals.

Answer:

3.011x10^24 molecules

Explanation:

1 mole=6.022x10^23 molecules

5 moles*6.022x10^23 molecules/mole=3.011x10^24 molecules

Answer:

THE HEAT REQUIRED TO CHANGE 2 KG OF ETHANOL FROM 26 °C TO THE BOILING POINT AND TO VAPOR AT THAT TEMPERATURE IS 1965.175 KJ.

Explanation:

Boiling point of ethanol = 78.29 °C = 78.29 + 273 K = 351.29 K

Mass = 2 kg = 2000 g

Final temp. = 26.0 °C = 26 + 273 K= 299 K

Change in temperature = (78.29 - 26) °C = 52.29 °C

1. Heat required to raise the temperature from 26 °C to the boiling point?

Heat = mass * specific heat * change in temperature

Heat = 2000 * 2.44 * 52.29

Heat = 255 175.2 J

2. Heat required to change the liquid to vapor at that temperature?

Heat = mass * enthalphy of vaporization

Heat = 2000 * 855

Heat =1 710000 J

The total heat required to raise the temperature of 2 kg of ethanol from 26 °C to the boiling point and then to change the liquid to vapor at that temperature will be:

Heat = mcT + m Lv

Heat = 255 175.2 J + 1710000 J

Heat = 1965175.2 J

Heat = 1965.175 kJ of heat.

Answer:

they would repel

Explanation:

unlike charges attract while like ones repel.

Answer:

D) it is the heat required to change a gram of substance from a liquid to a gas

Explanation:

idk i think its correct but if its wrong just let me know

Answer:

[tex]m=14,105.71 g Fe[/tex]

Explanation:

Hello,

In this case, the first step is to compute the volume of the block considering the length, height and width:

[tex]V=L \times W \times H =12.5 in\times 3.50 in \times 2.50 in =109.375 in^3[/tex]

Then, we compute the volume in cubic centimetres:

[tex]V=109.375in^3\times \frac{16.3871 cm^3}{1in^3} =1792.34cm^3[/tex]

Finally, as the density is given by:

[tex]\rho =\frac{m}{V}[/tex]

We solve for the mass:

[tex]m=\rho \times V= 7.87\frac{g}{cm^3} \times 1792.34 cm^3\\\\m=14,105.71 g Fe[/tex]

Best regards.

Answer:

327.89g/mol

Explanation:

Step 1:

The following data were obtained from the question:

Van 't Hoff factor (i) = 1 (since the molecule is non-electrolyte)

Temperature (T) = 25°C = 25°C + 273 = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Mass of molecule = 241mg

Volume of water = 250mL

Molarity (M) =?

Osmotic pressure (Π) = 0.072 atm

Step 2:

Determination of the molarity of the molecule.

This can be obtained as follow:

Π = iMRT

0.072 = 1 x M x 0.0821 x 298

Divide both side by 0.0821 x 298

M = 0.072 / (0.0821 x 298)

M = 2.94×10¯³ mol/L

Step 3:

Determination of the number of mole of the molecule. This can be obtained as follow:

Molarity = 2.94×10¯³ mol/L

Volume of water = 250mL = 250/1000 = 0.25L

Mole of molecule =..?

Molarity = mole /Volume

2.94×10¯³ = mole / 0.25

Cross multiply

Mole of molecule = 2.94×10¯³ x 0.25

Mole of molecule = 7.35×10¯⁴ mole.

Step 4:

Determination of the molar mass of the molecule.

Mole of molecule = 7.35×10¯⁴ mole.

Mass of molecule = 241mg = 241×10¯³g

Molar mass of molecule =..?

Mole = Mass /Molar Mass

7.35×10¯⁴ = 241×10¯³/ Molar Mass

Cross multiply

7.35×10¯⁴ x molar mass = 241×10¯³

Divide both side by 7.35×10¯⁴

Molar Mass = 241×10¯³/7.35×10¯⁴

Molar Mass = 327.89g/mol

Therefore, the molar mass of the molecule is 327.89g/mol

Answer:

[tex]0.745~M[/tex]

Explanation:

In this case, we have a dilution problem. So, we have to use the dilution equation:

[tex]C_1*V_1=C_2*V_2[/tex]

Now, we have to identify the variables:

[tex]C_1~=~14.9~M[/tex]

[tex]V_1~=~25~mL[/tex]

[tex]C_2~=~?[/tex]

[tex]V_2~=~0.5~L[/tex]

Now, we have different units for the volume, so we have to do the conversion:

[tex]0.5~L\frac{1000~mL}{1~L}=~500~mL[/tex]

Now we can plug the values into the equation:

[tex]C_2=\frac{14.9~M*25~mL}{500~mL}=0.745~M[/tex]

I hope it helps!

Answer:

[Ne] 3s² 3p⁴

Electrons per shell: 2,8,6

Answer : 2,8,6 (Sulphur Atom)

                2,8,8 (Sulphur Ion)

Answer:

B) It can have different compositions, but it has a set of characteristics that does not change.

Explanation:

On e d g e n u i t y

I believe the answer is d lmk if  im  wrong or right

Answer:

The 1st and 4th options are correct

I.the oxidized form has a higher affinity for electrons

IV. the greater the tendency for the oxidized form to accept electrons

Explanation:

Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.

Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.

(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  ammonia

Explanation:

The mixture contains two base compound which are

           ammonia,

and     diethylamine

Now the addition of HCl which is  a strong acid in step 1  will cause the protonation of  the  two base compound , which makes the soluble hence resulting in them being extracted to the aqueous layer as represented in below

       [tex]NH_3 + HCl\to NH_4 ^{+} + Cl^-[/tex]

and

     [tex](CH 3CH 2) 2NH + HCl \to (CH 3CH 2) 2NH_2^{+} + Cl[/tex]

Answer:

Explanation:

Boiling.

Use water filter

Use Ultraviolet Light.

Use chlorine drops

I would recomade boiling as the main

because its the easiest and cheapest Or water filter if you have one

Answer:

The percentage of the R-enantiomer is 26.18% while the percentage of the S-enantiomer is 73.82%

Explanation:

If the specific rotation of R enantiomer = -28.4, then the specific rotation of S = +28.4

Now, let us have x = % R, thus

% S = 100-x =y

Hence;

{- 28.4x + 28.4( 100 -x)}/100= 14.1

Thus;

-28.4x + 2840 -28.4x = 1410

-56.8x + 2840 = 1410

-56.8x = 1410-2840

-56.8x = -1430

x = 1430/56.8

x = 26.18%

y = 100-26.18% = 73.82%

Answer:

(B.) and (C.)

b. So that clean NMR spectra can be obtained that do not contain solvent peaks.

c. So that the yield can be determined.

Explanation:

The solvent used in Nuclear Magnetic Resonance (NMR) spectrometer is Trimethyl silane (TMS), a neutral solvent which doesn't give off any signals. Other solvents could have interactions with the radiation, and disrupt the spectra.

Furthermore, for accurate determination of the actual yield and overall percentage yield, solid must be separated from the solvent, dried and weighed.

I hope this was explanatory enough.

Answer:

A. ethane > butane > decane > dodecane > heptadecane

Explanation:

In fractionating column, crude oil is separated by means of fractional distillation due to the wide range of boiling point of the crude products such as ethane, propane, butane pentane etc.

The product with the least weight rises to top height while the product with highest weight will move down.

For the given hydrocarbon products, the ranking according to their molecular weight, starting with the lighter product to heavier product is

ethane (C2), butane (C4), decane(C10), dodecane (C12), heptadecane(C17).

Thus, the correct ranking, starting with the product that will rise highest is ethane > butane > decane > dodecane > heptadecane