Draw a Lewis structure for each covalent molecule. a. CH3F b. N2H4

Answer:

[tex]M=5.0M\\\\m=6.2m[/tex]

Explanation:

Hello,

In this case, 38 % is commonly a by mass concentration, meaning that we have 38 grams of solute (sulfuric acid) per 100 grams of solution (water+sulfuric acid):

[tex]38\%=\frac{m_{H_2SO_4}}{m_{H_2SO_4}+m_{H_2O}}[/tex]

Hence, we compute the moles of sulfuric acid in 38 grams by using its molar mass (98 g/mol):

[tex]n_{H_2SO_4}=38g*\frac{1mol}{98g}=0.39mol H_2SO_4[/tex]

Next, the volume of the solution in litres by using the density of the solution:

[tex]V_{solution}=100g*\frac{1mL}{1.29g}*\frac{1L}{1000mL} =0.0775L[/tex]

This is done since the molarity is defined as the ratio of the moles of the solute to the volume of the solution in litres, thus we have:

[tex]M=\frac{n}{V}=\frac{0.38mol}{0.0775L}=5.0M[/tex]

On the other hand, the molality is defined as the ratio of the moles of the solute to the mass of the solvent in kilograms, thus, we compute the mass of water (solvent) as shown below:

[tex]m_{H_2O}=100g-38g=62g*\frac{1kg}{1000g}=0.062kg[/tex]

So compute the molality:

[tex]m=\frac{n_{solute}}{m_{solvent}}=\frac{0.39mol}{0.062kg}=6.2m[/tex]

Regards.

1. The molarity of the solution is 5 M

2. The molality of the solution is 6.26 M

Let the mass of the solution be 100 g.

Therefore, the mass of 38% of H₂SO₄ in the solution is 38 g.

Next, we shall determine the mole of 38 g of H₂SO₄.

Mole = mass / molar mass

Mole of H₂SO₄ = 38 / 98

Mole of H₂SO₄ = 0.388 mole

Next, we shall determine the volume of the solution

Volume of solution =?

Volume = mass / density

Volume of solution = 100 / 1.29

Volume of solution = 77.52 mL

1. Determination of the molarity of the solution

Molarity =?

Molarity = mole / Volume

Molarity = 0.388 / 0.07752

Molarity = 5 M

2. Determination of the molality of the solution

Molality = mole / mass (Kg) of water

Molality = 0.388 / 0.062

Molality = 6.26 M

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Complete Question

The complete question is shown on the first uploaded image

Answer:

[tex]-SH \ \ A[/tex]

[tex]-NH_2 \ \ B[/tex]

[tex]-COOH \ \ C[/tex]

[tex]-CH_3 \ \ D[/tex]

Explanation:

Now to correctly assign priorities in the following set of substituents we need to consider the molecular weight of the atom that is been directly attached to  the carbon bond

Now an atom with a higher molecular weight will imply that the substituent will have the the highest priority

So  S (32.065 u)with a the highest molecular weight implies that -SH has the highest priority

Next is   N with molecular weight(14.0067 u) implies that -[tex]NH_2[/tex] will  have the next priority

Next is   O with molecular weight(15.999 u) implies that [tex]-COOH[/tex] will  have the next priority

The Last is  H with molecular weight(1.00784 u) implies that  [tex]-CH_3[/tex] will  have the next priority

Answer:

The energy in cabbage leaves comes from the light from the sun.

Explanation:

The plant keeps the light (in the leaves) as energy to help make it grow.

Answer:

it shows the breakdown of the atom

Explanation:

it will show it molecularly

Answer:

CuI₂ + Br₂

Explanation:

Answer:

a

Explanation:

Answer:

1.6

Explanation:

velocity=distance/time

=8/5

Answer:

Sodium Hydroxide turns blue litmus red .

Answer:

Explanation:

As an example, the following cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m) generates a cell voltage of +1.10 V under standard conditions. Calculate and enter delta G degree (with 3 sig figs) for this reaction in kJ/mol.

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m)

ΔG = ΔG° + RTInQ

Q = 1

ΔG = ΔG°

ΔG = =nFE°

n=no of electrons transfered.

E° = 1.1v

ΔG° = -2 * 96500 * 1.10

= -212300J

ΔG° =-212.3kJ/mol

Answer:

Alpha decay: 4 2He

Explanation:

Considering the equation given in the question, we can see that the mass number of daughter nuclei (228 88Ra) reduce by 4 and the atomic number reduce by 2 when compared with the parent nucleus (232 90Th). This simply indicates that the parent nucleus is undergoing alpha decay.

Therefore, the missing nucleus is

4 2He i.e alpha decay.

Please see attachment photo for further details.

the answer is b

The posterior pituitary secretes two important endocrine hormones—oxytocin and antidiuretic hormone

Answer:

Explanation:

1) Clayey-loamy soil tends to be excellent at retaining water which is critical for proper growth of wheat thus this kind of soil is preferred.

2) Alex is showing honesty as he informing the villagers about the actual truth and reality (he is not basing his answers on judgements rather he is basing it on scientific knowledge and being honest about it)

Explanation:

Before proceeding to answering the questions, let us go over some definitions.

pH scale: The pH scale measures how acidic or basic a substance is. The pH scale ranges from 0 to 14. A pH of 7 is neutral. A pH less than 7 is acidic.

pH stands for Potential of Hydrogen. It refers to the hydrogen ion concentration in a solution.

The keywords being; Hydrogen Ion concentration.

A Bronsted-Lowry acid is a chemical species that donates one or more hydrogen ions in a reaction.

A Lewis acid is  any substance that can accept a pair of nonbonding electrons. It is an electron pair acceptor.

An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons.

Based on the definitions of given above, it is obseved that both Bronsted lowry and arrhenius acids deals with hydrogen ions. Hence both of this acids can be measured using the pH scale. The  lewis acid on the other hand do not necessarily contain hydrogen ions, hence the pH scale cannot be utilized for it.

Examples includes;

Arrhenius acid; Nitric Acid – HNO3 etc

Lewis acid; boron trifluoride (BF3) and aluminum fluoride (AlF3) etc

Bronsted lowry acid; HCl etc

Answer:

what are the options?

Explanation:

do you have multiple choice for this question?

Answer:

A book inside a vehicle, a running cat, and a falling meteorite.

Explanation:

I just took it

Answer:

1) [tex]-Cl[/tex]

2) [tex]-OH[/tex]

3) [tex]-CH_2SH[/tex]

4) [tex]-CH_2OH[/tex]

Explanation:

We have the substituents:

a) [tex]-Cl[/tex]

b) [tex]-OH[/tex]

c) [tex]-CH_2OH[/tex]

d) [tex]-CH_2SH[/tex]

If we remember that Cahn-Ingold-Prelog rules the highest priority is given by the atomic number. Therefore the highest priority is "Cl" (an atomic number equal to 17), the next one is "OH" due to the oxygen (an atomic number equal to 8). For c) and d) we have a carbon bonded to the chiral carbon, therefore we have to check the next atom. The difference between c) and d) are the "O" and "S" atoms, the atom with the highest atomic number is "S" (an atomic number equal to 16)  therefore the highest priority is for d) and then c). So finally, the priority is:

1) [tex]-Cl[/tex]

2) [tex]-OH[/tex]

3) [tex]-CH_2SH[/tex]

4) [tex]-CH_2OH[/tex]

I hope it hepls!

Answer:

[tex]C_{18}H_{24}O_3[/tex]

Explanation:

Hello,

In this case, combustion analyses help us to determine the empirical formula of a compound via the quantification of the released carbon dioxide and water since the law of conservation of mass is leveraged to attain it. In such a way, as 36.86 g of carbon dioxide were obtained, this directly represents the mass of carbon present in the sample, thus, we first compute the moles of carbon:

[tex]n_{C}=36.86gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2} =0.838molC[/tex]

Then, into the water one could find the moles of hydrogen:

[tex]n_{H_2O}=10.06gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O} =1.12molH[/tex]

Now, we compute the moles of oxygen by firstly computing the mass of oxygen:

[tex]m_{O}=13.42g-36.86gCO_2*\frac{12gC}{44gCO_2} -10.06 gH_2O*\frac{2gH}{18gH_2O} =2.25gO[/tex]

[tex]n_O=2.25gO*\frac{1molO}{16gO} =0.141molO[/tex]

Then, we have the mole ratio:

[tex]C_{0.838}H_{1.12}O_{0.141}\rightarrow C_6H_8O[/tex]

Whose molar mass is 12x6+1x8+16=96 g/mol, but the whole compound molar mass is 288.38, the factor is 288.38/96 =3, Therefore the formula:

[tex]C_{18}H_{24}O_3[/tex]

Regards.

Answer:

Molecular mass = [tex]C_{6} H_{8} O_{16}[/tex]

Explanation:

[tex](\frac{36.86gCO_{2} }{44g/molCO_{2} })[/tex] [tex](\frac{10.06gH_{2}O}{18g/molH_{2}O} )[/tex]

(0.8377mol of CO2) (0.556 mol of H2O)

mole ratio of CO2 to H2O = 1.5 : 1 ≅ 3 : 2

[tex](CO_{2} )_{3} , (H_{2} O)_{2}[/tex]

⇒ [tex](C_{3} H_{4} O_{8} )_{x}[/tex] = 288.38

(12 × 3 + 1 × 4 + 16 × 8)x = 288.38

168x = 288.38

x = 1.71 ≅ 2

∴ Molecular mass = [tex]C_{6} H_{8} O_{16}[/tex]

by fractional distillation of liquefied air or the use of zeolites to remove carbon dioxide and nitrogen from air or the electrolysis of water

Answer:

282.7KPa

Explanation:

Step 1:

Data obtained from the question.

Number of mole of (n) = 1.5 mole

Volume (V) = 13L

Temperature (T) = 22°C = 22 + 273°C = 295K

Pressure (P) =..?

Gas constant (R) = 0.082atm.L/Kmol

Step 2:

Determination of the pressure exerted by the gas.

This can be obtained by using the ideal gas equation as follow:

PV = nRT

P = nRT /V

P = 1.5 x 0.082 x 295 / 13

P = 2.79atm.

Step 3:

Conversion of 2.79atm to KPa.

This is illustrated below:

1 atm = 101.325KPa

Therefore, 2.79atm = 2.79 x 101.325 = 282.7KPa

Therefore, the pressure exerted by the gas in KPa is 282.7KPa

Pressure is defined as the force exerted by the molecules colliding with the surface of the object and other surfaces. The ideal gas follows the general gas law equation, PV = nRT.

The pressure exerted by the gas is 282.7 KPa.

Given:

Pressure (P) = ?

Gas constant (R) = 0.082atm.L/Kmol  

Temperature (T) = 22°C = 22 + 273°C = 295K

Volume (V) = 13L

Number of mole of (n) = 1.5 mole

Now, using the ideal gas equation:

PV = nRT

Substituting the values, we get:

P x 13 = 1.5 x 0.082 x 295

P x 13 = 36.285

P = [tex]\dfrac{36.285}{13}[/tex]

Pressure = 2.79 atm.

Also, 1 atm = 101.32 KPa

2.79 atm = 2.79 x 101.32 = 282.7 KPa.

Thus, the pressure exerted by the gas molecules is 282.7 KPa.

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Answer:

1.76cm

Explanation:

We'll begin by calculating the mass of titanium that contain 3.10x10²³ atoms. This can be obtained as follow:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10²³ atoms. This implies that 1 mole of titanium also contains 6.02x10²³ atoms.

1 mole of titanium = 48g

Now, if 48g of titanium contains 6.02x10²³ atoms,

Then Xg of titanium will contain 3.10x10²³ atoms i.e

Xg of titanium = (48x3.10x10²³)/6.02x10²³

Xg of titanium = 24.72g

Next, we shall determine the volume of the titanium. This is illustrated below:

Density of titanium = 4.50g/cm³

Mass of titanium = 24.72g

Volume of titanium =..?

Density = Mass /volume

Volume = Mass /Density

Volume of titanium = 24.72g/4.50g/cm³

Volume of titanium = 5.49cm³

Finally we shall determine the edge length of the titanium cube as follow:

Volume = L³

5.49cm³ = L³

Take the cube root of both side

L = 3√(5.49cm³)

L = 1.76cm

Therefore, the edge length is 1.76cm

Answer:

 (V) 1s 2 2s 2 2p 6 3s 2 3p 5

Explanation:

Electron Affinity can be defined as the energy associated with a neutral atom, when an electron is added to form a negative ion.

1s 2 2s 2 2p 6 3s 2 3p 5 , is the electronic configuration of Chlorine.

The electron affinity is positive because it is an exorthermic reaction, meaning that, energy was released during the addition of an electron to the atom.

Cl (g)  +  e-   ------->  Cl- (g)    = -349KJ/mol

Answer:

0.0560 mol

Explanation:

You need to use the ideal gas law equation:

[tex]PV = nRT[/tex]

P = 750 torr

V = 1.35 L

n = moles

R = 62.364 L torr [tex]mol^{-1}[/tex] [tex]K^{-1}[/tex]

T = 17.0 ˚C + 273.15 = 290.15 K

Rearranging the equation for n:

[tex]n= \frac{PV}{RT}[/tex]

[tex]n= \frac{(750)(1.35)}{(62.364)(290.15)}[/tex]

n = 0.0560 mol

Answer:

n=PV/RT

Explanation:

on edge 2020

Answer:

1.40 × 10⁻²⁰ J

Explanation:

Step 1: Calculate the mass of 1 atom of argon

The molar mass of argon is 39.95 g/mol, that is, 6.02 × 10²³ atoms of Ar have a mass of 39.95 g. We can use this relation to find the mass of 1 atom of Ar.

[tex]\frac{39.95g}{1mol} \times \frac{1mol}{6.02 \times 10^{23}atom } =6.64 \times 10^{-23}g/atom[/tex]

Step 2: Convert the mass of 1 atom of argon to kilograms

We will use the relationship 1 kg = 1,000 g.

[tex]6.64 \times 10^{-23}g \times \frac{1kg}{1,000g} =6.64 \times 10^{-26}kg[/tex]

Step 3: Calculate the kinetic energy of 1 atom of Ar moving at 650 m/s

[tex]E=\frac{1}{2} \times m \times v^{2} = \frac{1}{2} \times 6.64 \times 10^{-26}kg \times (650m/s)^{2} = 1.40 \times 10^{-20}J[/tex]

The kinetic energy of an Ar atom at a speed of 650 m/s is 1.40 × 10⁻²⁰ J

Kinetic Energy:

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity:

K.E = [tex]\frac{1}{2}[/tex] x [tex]m[/tex] x [tex]v^{2}[/tex]

Step 1: Calculate the mass of 1 atom of argon

The molar mass of argon is 39.95 g/mol, that is, 6.02 × [tex]10^{-23}[/tex] atoms of Ar have a mass of 39.95 g. We can use this relation to find the mass of 1 atom of Ar.

[tex]\frac{39.95 g}{1 mol} * \frac{1 mol}{1,000 g} = 6.64 * 10^{-26} g/ atom[/tex]  

Step 2: Convert the mass of 1 atom of argon to kilograms

We will use the relationship 1 kg = 1,000 g.

6.64 x [tex]10^{-23}[/tex] g x [tex]\frac{1 kg}{1,000g}[/tex] = 6.64 x [tex]10^{-26}[/tex] kg

Step 3: Calculate the kinetic energy of 1 atom of Ar moving at 650 m/s

E= [tex]\frac{1}{2}[/tex] x [tex]m[/tex] x [tex]v^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 6.64 x [tex]10^{-26}[/tex] kg x (650 m/s)² = 1.40 x [tex]10^{-20}[/tex] J

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The given question is incomplete, the complete question is:

When 238.g of benzamide (C7H7NO) are dissolved in 600.g of a certain mystery liquid X, the freezing point of the solution is 5.0°C lower than the freezing point of pure X. On the other hand, when 238.g of iron(III) chloride are dissolved in the same mass of X, the freezing point of the solution is 11.5°C lower than the freezing point of pure X.

Calculate the van't Hoff factor for iron(III) chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.

Answer:

The correct answer is 3.0.

Explanation:

Based on the given question, the weight of benzamide given is 238 grams, the molecular mass of benzamide is 121.14 gram per mole. The benzamide is dissolved in 600 grams of liquid X, therefore, the solution's total weight is,  

= 238 + 600 = 838 grams

In case of benzamide, ΔTf = kf × molality (It is given that the solution's freezing point is 5 degree C lesser in comparison to the pure X's freezing point). Now putting the values we get,  

5 = kf × 238 / (121.14×838) × 1 (Van't Hoff factor) -------- (i)

In case of benzamide, the van't Hoff factor will be 1, as it is neither associate nor dissociate.  

On the other hand, the mass of ferric chloride given is 238 grams getting dissolved in the similar mass of X, therefore again the total mass of the solution will be 838 grams. The freezing point of the solution is 11.5 degree C lesser than the pure X's freezing point. The molecular mass of ferric chloride is 162.2 gram per mol.  

For FeCl3,  

ΔTf = kf × molality

11.5 = kf × 238/ (162.2 × 838) × i (Van't Hoff factor)------ (ii)

Now dividing equation (ii) by (i) we get,  

11.5 / 5 = (kf/kf) × (121.14 / 162.2) × i  

i = 3.0

Answer:

The added mole will be "11.388 moles".

Explanation:

The given values are:

Initial volume,

V₁ = 4.11 L

Final volume,

V₂ = 11.3 L

Number of moles,

n₁ = 6.51

On applying Avogadro's law,

⇒  [tex]\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}[/tex]

On putting the estimated values, we get

⇒  [tex]\frac{4.11}{6.51} =\frac{11.3}{n_{2}}[/tex]

On applying cross-multiplication, we get

⇒  [tex]4.11 \ n_{2}=11.3\times 6.51[/tex]

⇒  [tex]4.11 \ n_{2}=73.563[/tex]

⇒  [tex]n_{2}=\frac{73.563}{4.11}[/tex]

⇒  [tex]n_{2}=17.898 \ moles[/tex]

So that the number of moles at added gas will be:

[tex]=17.898-6.51[/tex]

[tex]=11.388 \ moles[/tex]

Answer:

2.35 atm  

Explanation:

1 atm = 14.70 lb/in²

[tex]p = \text{34.5 lb/in}^{2} \times \dfrac{\text{1 atm}}{\text{14.70 lb/in}^{2}} = \textbf{2.35 atm}[/tex]

Answer:

The correct answer is is option B

b. 93.3 g

Explanation:

SEE COMPLETE QUESTION BELOW

Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)

How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?

a. 7.30 g

b. 93.3 g

c. 146 g

d. 150 g

e. 196 g

CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION

Answer: (Chlorine-35)

Element X is Ca

Explanation:

Number of proton plus the number of electron gives the mass number.

Answer:

THE STANDARD HEAT OF SOLUTION OF SODIUM HYDROXIDE IN WATER IS -7.68 KJ PER MOLE.

Explanation:

Variables:

Mass of NaOH = 3.25 g

Mass of water = 50 g

Initial temperature of water = 22°C = 22 + 273 K = 295 K

Final temperature of the reaction mixture = 24.8 °C = 24.8 + 273 K = 297.8 K

Assuming that:

1. specific heat of water = 4.184 J/g °C

2. total mass of the reaction mixture = 50 g + 3.25 g = 53.25 g

3. the rise in temperature = (297.8 K - 295 K ) = 2.8 K

4. Molar mass of sodium hydroxide = ( 23 + 16 + 1) = 40 g/mol

5. number of mole of sodium hydroxide = mass / molar mass

n = 3.25 g / 40 g/mol

n = 0.08125 moles

The rise in temperature for the reaction mixture produces how much of heat:

Heat = mass * specific heat * change in temperature

Heat = 53.25 * 4.184  * 2.8

Heat = 623.8344 J of heat.

Equation of reaction:

NaOH + H2O -------> NaOH + H2O + Heat

This is not a reaction but a dissolution as sodium hydroxide is very soluble in water and this reaction is exothermic where heat is given off.

So since 3.25 g having 0.08125 moles produces 623.8344 J of heat, 1 mole of the sodium hydroxide used will produce:

0.08125 mole of sodium hydroxide = 623.8344 J of heat

1 mole of sodium hydroxide = ( 623.8344 / 0.08125 J of heat

= 7677.96 J of heat per mole of sodium hydroxide.

= 7.68 kJ of heat

So therefore, the standard heat of solution of sodium hydroxide in water is -7.68 kJ of heat since its an exothermic reaction.

Answer: 1.67 kg

Explanation:

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

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat absorbed=[tex]4.31\times 10^1kJ[/tex] = [tex]43100J[/tex]   (1kJ=1000J)

m= mass of substance = ?

c = specific heat capacity = [tex]0.385J/g^0C[/tex]

Change in temperature ,[tex]\Delta T=T_f-T_i=6.71\times 10^1^0C=67.1^0C[/tex]

Putting in the values, we get:

[tex]43100J=m\times 0.385J/g^0C\times 67.1^0C[/tex]

[tex]m=1670g=1.67kg[/tex]   (1kg=1000g)

Thus the mass (in kg) of the copper sample is 1.67

Answer:

There are 0.09996826 moles per liter of the solution.

Explanation:

Molar mass of HNO3: 63.02

Convert grams to moles

0.63 grams/ 63.02= 0.009996826

Convert mL to L and place under moles (mol/L)

100mL=0.1 L

0.009996826/0.1= 0.09996826 mol/L

The concentration of the HNO₃ solution is 10⁻¹ moles per liter.

Molar Concentration is known as Molarity. It is the number of moles of solute present in per liter of the solution.

Molar concentration in terms of Molecular weight

C = [tex]\frac{m}{V} \times \frac{1}{M.W}[/tex]

where,

C is the molar concentration in mol/L

m is the mass or weight of solute in grams

V is volume of solution in liters

M.W is the molecular weight in g/mol

Molar concentration of HNO₃ = [tex]\frac{\text{Mass of}\ HNO_3}{\text{Liter of solution}} \times \frac{1}{\text{Molecular weight of}\ HNO_3}[/tex]

Molecular weight of HNO₃ = Atomic weight of H + Atomic weight of N + 3 (Atomic weight of O)

                                  = 1 + 14 + 3 (16)

                                  = 15 + 48

                                  = 63 g/mol

Convert mL into L

1 mL = 0.001 L

100 mL = 100 × 0.001

            = 0.1 L

Now, put the value in above formula we get

Molar concentration of HNO₃ = [tex]\frac{\text{Mass of}\ HNO_3 / \text{Liter of solution}}{\text{Molar mass of}\ HNO_3}[/tex]

                                                   =  [tex]\frac{0.63\ g / 0.1\ L}{63\ g/mol}[/tex]

                                                   = [tex]\frac{6.3}{63}[/tex]

                                                  = 0.1

                                                   = 10⁻¹ moles per liter

Thus, we can say that 10⁻¹ moles per liter is the concentration of the HNO₃ solution.

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