Compare a 3s orbital to a 3p orbital. Which of the following are true? I. The orbitals have the same shape. II. The orbitals have different shapes. III. The orbitals have the same energy. IV. The orbitals have different energies.

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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Answer:

NaBr

Explanation:

Bromine's symbol is Br, not B, for B is for Boron

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:

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:

45.6 mm

Explanation:

1 centimeter = 10 millimeter

4.56 cm = x mm

x = 4.56 × 10

= 45.6 mm

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]

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 option C, that is, 1.36 L.

Explanation:

The reaction mentioned in the question is:  

Zn (s) + 2HCl (aq) ⇒ H2 (g) + ZnCl2 (aq)

It is clear that one mole of zinc is generating one mole of hydrogen gas as seen in the reaction. The mass of zinc mentioned in the question is 3.566 grams, the no of moles can be determined by using the formula,  

n = mass / molecular mass

The molecular mass of zinc is 65.39 g/mol, now putting the values in the formula we get,  

n = 3.566 g/ 65.39 g/mol

= 0.0545 mol

Based on the question, the partial pressure of water vapor is 18.65 mmHg and the atmospheric pressure is 752 mmHg. Therefore, the pressure of hydrogen gas will be,  

Pressure of hydrogen gas (H2) = 752 mmHg - 18.65 mmHg

= 733.35 mmHg

The liters of hydrogen gas produced can be calculated by using the equation, PV =nRT

R is the gas constant, having the value 62.36 L mmHg/K/mol and T is the temperature (273 + 21 = 294 K).  

Now putting the values in the equation we get,  

733.35 mmHg * V = 0.0545 mol * 62.36 mmHg/K/mol * 294 K

= 999.19 L / 733.35

= 1.36 L

Hence, the volume of hydrogen gas is 1.36 L.  

Answer:

3.17 V

Explanation:

The cell is operating under standard conditions. These standard conditions include; that the reaction takes place at 298 Kelvin (room temperature), the pressure of the system is 1 atmosphere (standard pressure), and the solutions have a Molarity of 1.0 M for both the anode and cathode solutions. All these conditions are satisfied in the cell under review in the question.

Hence;

E°anode (magnesium)= -2.37 V

E°cathode (silver) = 0.80 V

E°cell= E°cathode -E°anode

E°cell= 0.80-(-2.37)

E°cell= 0.80 + 2.37

E°cell= 3.17 V

Hence the standard cell potential of this cell at 25°C is 3.17 V

Answer:

The standard potential at 25ºC is 3.17 V.

Explanation:

The anode in  a galvanic cell is the electrode at which oxidation occurs and the cathode is the electrode  at which reduction occurs.

The overall cell reaction  will be the sum of two half-cell reactions. The standard reduction potentials are:

Mg²⁺ (1.0 M) + 2e⁻  →  Mg (s)          Eº = -2.37

Ag⁺ (1.0 M) + e⁻ → Ag (s)                 Eº= +0.80

Since the reactants are in their standard states (1.0 M) and at 25ºC we can write the half-cell reactions as follows:

Anode (oxidation):                        Mg (s) → Mg²⁺ (1.0 M) + 2e⁻

Cathode (reduction):     2Ag⁺ (1.0 M) + 2e⁻ → 2Ag (s)

Overall:            Mg (s) +2 Ag⁺ (1.0 M) + 2e⁻ → 2Ag (s) + Mg²⁺ (1.0 M) + 2e⁻

In order to balance the overall equation we multiply the reduction of  Ag⁺ by 2. We can do so because, as an intensive property, E° is not affected by  this procedure.

The standard emf of the cell, E°cell , which is composed of a contribution  from the anode and a contribution from the cathode, is given by:

[tex] Eº cell = Eº cathode - Eº anode [/tex]

[tex] Eº cell = EºAg⁺/Ag - Eº Mg²⁺/Mg [/tex]

[tex] Eº cell =0.80 V - (-2.37 V) [/tex]

Eº cell = 3.17 V

Answer:

Isobutanol (IUPAC nomenclature: 2-methylpropan-1-ol)

Explanation:

Isobutanol (IUPAC nomenclature: 2-methylpropan-1-ol) is an organic compound with the formula (CH3)2CHCH2OH (sometimes represented as i-BuOH). Hope this helped!

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:

[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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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

the answer is b

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

Answer:

[tex]m_{CaBr_2}=115gCaBr_2[/tex]

Explanation:

Hello,

In this case, we remember molarity is defined as the ratio of the moles of the solute to the volume of the solution in litres, therefore, for the given concentration and volume, we should solve the moles of calcium bromide to subsequently compute the mass, as shown below:

[tex]M=\frac{n}{V}\\ \\n=M*V=1.15\frac{mol}{L} *500mL*\frac{1L}{1000mL}=0.575mol[/tex]

Next, since molar mass of calcium bromide is 200 g/mol, we can compute the mass:

[tex]m_{CaBr_2}=0.575mol*\frac{1mol}{200g} \\\\m_{CaBr_2}=115gCaBr_2[/tex]

Regards.

Answer:

CuI₂ + Br₂

Explanation:

Answer:

a

Explanation:

Answer:

3.77 mg of K-40 decayed into Ar-40.

Data:

1) K-40, Ca-40, Ar-40: all three have the same atomic mass

2) 90% of the potassium-40 will decay into calcium-40

3) 10% of the potassium-40 will decay into argon-40.

4) K-40 inside the rock = 0.81 mg

5) Ar-40 trapped = 0.377 mg

Soltuion:

1) 0.377 mg of Ar-40 is the 10% of the mass of the K-40 that decayed

=> x * 10% = 0.377 mg => x * 0.1 = 0.377mg

=> x = 0.377 mg / 0.1 = 3.77 mg

That means that 3.77 mg of K-40 decayed into Ar-40. And this is the answer to the question.

Additionaly, you can analyze the content of all K-40 and Ca-40, to understand better the case.

2) The mass of the K-40 that decayed into Ca-40 is 9 times (ratio 9:1) the amount that decayed into Ar-40 =>

mass of K-40 that decayed into Ca-40 = 9 * 0.377 = 3.393 mg

3) Total amount of K-40 that decayed = amount that decayed into Ar-40 + amount that decayed into Ca-40 = 0.377mg + 3.393mg = 3.77 mg

4) Original amount of K-40 = amount of K-40 that decayed + amount of K-40 present in the rock = 3.77mg + 0.81 mg = 4.58 mg

5) amount of K-40 that decayed into Ar-40 as percent

% = [3.77 mg / 4.58mg] * 100 = 82.31 %.

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.

To know more about ideal gas law, refer to the following link:

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Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

Where K' = K1 * K2

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just mechanism A and B are consistent with observed rate law

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just mechanism A is consistent with the observation of J. H. Sullivan

Answer:

d

Explanation:

Answer:

9.2x10²g

Explanation:

Data obtained from the question include the following:

Density = 0.92g/ml

Volume = 1L = 1 x 1000 = 1000mL

Mass =..?

Density is simply defined as the mass of the substance per unit volume of the substance. Mathematically it can be represented as:

Density = Mass /volume.

Mass = Density x volume

Mass = 0.92 x 1000

Mass = 9.2x10²g.

Therefore, 1L of olive will weigh 9.2x10²g.

Answer:

1.6

Explanation:

velocity=distance/time

=8/5

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:

a. Impurities

c. Recrystallizing solvent

Explanation:

In this type of reaction the products are never considered totally pure, that is why as a final product it must always be taken into account that it is proportional, it will be a recrystallization solvent and other impurities with which the products were mixed.

Answer:

Its A! Gg and Gg

Explanation:

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:

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:

[tex]m_{NaCl}=33g[/tex]

Explanation:

Hello,

In this case, in order to compute the grams of sodium chloride starting by the molecules, the first step is to compute the moles contained in the given amount of molecules by using the Avogadro's number:

[tex]n_{NaCl}=3.4x10^{23}molecules*\frac{1mol}{6.022x10^{23}molecules} =0.56mol[/tex]

Then, by using the molar mass of sodium chloride (58.45 g/mol) we can directly compute the grams:

[tex]m_{NaCl}=0.56mol*\frac{58.45g}{1mol} \\\\m_{NaCl}=33g[/tex]

Regards.

Answer: The amount of energy required is 101 Joules

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 = ?

m= mass of gold = 30.0 g

c = specific heat capacity of gold = [tex]0.13J/g^0C[/tex]

Initial temperature of the water = [tex]T_i[/tex] = 15°C

Final temperature of the water = [tex]T_f[/tex]  = 41°C

Change in temperature ,[tex]\Delta T=T_f-T_i=(41-15)^0C=26^0C[/tex]

Putting in the values, we get:

[tex]Q=30.0g\times 0.13J/g^0C\times 26^0C[/tex]

[tex]Q=101J[/tex]

The amount of energy required is 101 Joules

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.