What is the sum of the coefficient when the following equation is balanced:

__ H2SO4 + __ NaOH → __ H2O + __ Na2SO4

a. 4
b. 5
c. 6
d. no right answer

Answers

Answer 1

Answer:

The sum of the coefficient is: 1 + 2 +2 + 1 = 6 ( option c)

Explanation:

First we will balance on both sides Na

On the right side we have 2x Na but on the left side we have only 1x Na. So we have to multiply NaOH on the left side by 2.

This will give us:

H2SO4 + 2 NaOH → H2O + Na2SO4

Now we have on both sides 2x Na

We see that on the left side we have 4x H ( 2x H of H2SO4 and 2x H of NaOH), but on the right side we only have 2x H. So, we have to multiply H2O on the right side by 2.

This will give us:

H2SO4 + 2 NaOH → 2 H2O + Na2SO4

Now we have on both sides 2x Na and 4x H.

Also the number of O is on both sides equal, due to this. ( Both sides have 6x O).

Finally, we have this reaction:   H2SO4 + 2 NaOH → 2 H2O + Na2SO4

The sum of the coefficient is: 1 + 2 +2 + 1 = 6 ( option c)

Answer 2

Final answer:

The sum of the coefficients when the equation H2SO4 + NaOH → H2O + Na2SO4 is balanced is 6. The balanced equation is 1 H2SO4 + 2 NaOH → 2 H2O + 1 Na2SO4.

Explanation:

The sum of the coefficients when the chemical equation H2SO4 + NaOH → H2O + Na2SO4 is balanced is the total of the numbers that are used to balance the equation.

To balance the equation, we need to ensure that there is the same number of each type of atom on both the reactant and product sides of the equation. In this case, we balance the equation as follows: 1 H2SO4 (aq) + 2 NaOH (aq) → 2 H2O (l) + 1 Na2SO4 (aq). So, the coefficients in the balanced equation are 1, 2, 2, and 1 respectively.

Adding these coefficients, we get:

1 (for H2SO4) + 2 (for NaOH) + 2 (for H2O) + 1 (for Na2SO4) = 6.

The answer is option c: 6.


Related Questions

The distance from Earth to the Moon is approximately 240.000 mi Review Constants Periodic Table Part D Earth s ound the Sun at an average speed of 29.783 km/s Convert this speed to miles per hour Express your answer using live significant figures VOAEG V v * un x-100

Answers

Answer:

66622.653 mi/hr

Explanation:

The average speed at which the Earth rotates around the Sun = 29.783 km/s

Also, The conversion of km to miles is shown below:

1 km = 0.621371 miles

The conversion of s to hr is shown below:

1 s = 1 / 3600 hr

So,

[tex]29.783\ km/s=\frac {29.783\times 0.621371\ miles}{\frac {1}{3600}\ hr}[/tex]

Thus,

The average speed at which the Earth rotates around the Sun in miles per hour =  66622.653 mi/hr

Recognize Physical and Chemical Properties of Alcohols Question Which of the following statements about the -OH group of alcohols is NOT true? Select the correct answer below: O The hydrogen has a slight positive charge. O The hydrogen is slightly acidic. O The hydrogen cannot participate in hydrogen bonding with other molecules. O none of the above FEEDBACK MORE INSTRUCTION SUBMIT Content attribution

Answers

Answer:

The hydrogen cannot participate in hydrogen bonding with other molecules.

Explanation:

Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.

Thus, hydrogen bonding is present in alcohols as the oxygen atom is linked to hydrogen

Partially positive end of the hydrogen atom is attracted to partially negative end of the oxygen atom. It is strong force of attraction between the molecules. Thus, hydrogen acquires slight positive charge and become electron deficient or acidic.

Hence, option C is incorrect.

An FM radio transmitter broadcasts at 98.4 MHz with a power of
45 kW. How many photons does it generate per second?

Answers

To determine the number of photons emitted per second by a 45 kW FM radio transmitter at 98.4 MHz, the energy of one photon is calculated using Planck's constant and the frequency, and then the power output is divided by this energy. The result is approximately 6.90 × 1028 photons per second.

To calculate the number of photons emitted per second by an FM radio transmitter, we need to know the energy of one photon and the total energy emitted per second (power output of the transmitter).

The energy of a photon, E, can be calculated using the Planck's equation: E = h * f, where h is Planck's constant (6.626 × 10-34 J·s), and f is the frequency of the photon in hertz. In this case, the frequency (f) is 98.4 MHz, or 98.4 × 106 Hz.

The total power output in joules per second is given by the transmitter's power, which is 45 kW, or 45,000 J/s. By dividing the total energy emitted per second by the energy of one photon, we obtain the number of photons emitted per second.

Number of Photons per Second = Power Output (J/s) / Energy per Photon (J)

Now, let's perform the calculations:

Energy per Photon, E = (6.626 × 10-34 J·s) × (98.4 × 106 Hz) = 6.522 × 10-26 J

Photons per Second = 45,000 J/s / 6.522 × 10-26 J = 6.90 × 1028 photons/s

Calculate the area of a square with side length 1 = 2.2 cm. Use standard decimal notation for answer input.

Answers

Answer: The area of square is [tex]4.84cm^2[/tex]

Explanation:

To calculate the area of square, we use the formula:

[tex]A=l^2[/tex]

where,

A = area of square = ?

l = edge length of square  = 2.2 cm

Putting values in above equation, we get:

[tex]\text{Area of square}=(2.2)^2=4.84cm^2[/tex]

Hence, the area of square is [tex]4.84cm^2[/tex]


Non-electrolyte = A, 10.6 grams, dissolved in solvent B, 740 grams, the boiling point of the solution is higher than the boiling point of the solvent
Pure dissolved in 0.284 ° C. Determine the Kb value of solvent B. Set the molecular weight of A = 106.

Answers

Answer:

2.1 °C/m

Explanation:

Hello, for this exercise, consider the formula:

[tex]T_{solution}-T{solvent}=K_bm_solute[/tex]

Considering that the difference in the temperature is 0.284°C, and the given molality by:

[tex]m_{solute}=\frac{10.6g\frac{1mol}{106g}}{740g*\frac{1kg}{1000g} } \\m_{solute}=0.135m[/tex]

Now, solving for [tex]K_b[/tex], we get:

[tex]K_b=\frac{0.284C}{0.135m}\\K_b=2.1 C/m[/tex]

Best regards.

1 kg of Oxygen gas is present in a 500 L tank. Find the specific and molar volumes of the gas.

Answers

Answer:

specific volume O2 = 0.5 Kg/m³

molar volume O2 = 8 E-3 m³/mol

Explanation:

specific volume (Sv):

∴ Sv = 1 / ρ

∴ ρ = mass / volume

∴ V = 500 L * ( m³/1000 L) = 0.5 m³

∴ ρ = 1 Kg / 0.5m³ = 2 Kg/m³

⇒ Sv = 1 / 2 Kg/m³ = 0.5 m³/Kg

molar volume ( Vm ):

∴ Vm = volume/mol

∴ Mw O2 = 1000g O2 * ( mol/16g O2) = 62.5 mol O2

⇒ Vm = 0.5 m³ / 62.5 mol

⇒ Vm = 8 E-3 m³/mol

In a particular experiment at 300 ∘C, [NO2] drops from 0.0100 to 0.00800 M in 100 s. The rate of appearance of O2 for this period is ________ M/s. In a particular experiment at 300 , drops from 0.0100 to 0.00800 in 100 . The rate of appearance of for this period is ________ . 4.0×10−3 2.0×10−3 2.0×10−5 4.0×10−5 1.0×10−5

Answers

Answer:

[tex]-r_{O_2}=1\times{10}^{-5}\frac{M}{s}[/tex]

Explanation:

The decomposition of [tex]NO_2[/tex] follows the equation  

[tex]2NO_2\rightarrow2NO+O_2[/tex]

By definition, the rate of a chemical reaction can be expressed by  

[tex]-r_{NO_2}=\frac{d\left[NO_2\right]}{dt}=\frac{0.010-0.008\ M}{100\ s}=2\times{10}^{-5}\frac{M}{s}[/tex]

The rate of appearance of [tex]O_2[/tex] is related to the rate of disappearance of [tex]NO_2[/tex] by the stoichiometry. This means that, for each mole of [tex]O_2[/tex] that appears 2 moles of [tex]NO_2[/tex]  are consumed. So  

[tex]-r_{O_2}=-r_{NO_2}\times\frac{1\ mole\ \ O_2}{2\ mole\ \ {\rm NO}_2}=1\times{10}^{-5}\frac{M}{s}[/tex]

Final answer:

The rate of appearance of O2 for this period is -1.0x10^-5 M/s.

Explanation:

The rate of appearance of O2 for a given period can be determined using the change in concentration of NO2 over that period. In this case, the concentration of NO2 drops from 0.0100 to 0.00800 M in 100 s. First, calculate the change in concentration of NO2:


Δ[NO2] = [NO2]final - [NO2]initial = 0.00800 M - 0.0100 M = -0.00200 M


Since the reaction is 2NO2 → 2NO + O2, the stoichiometric ratio between NO2 and O2 is 2:1. Therefore, the change in concentration of O2 is half of the change in concentration of NO2:


Δ[O2] = -0.00200 M ÷ 2 = -0.00100 M


Finally, divide the change in concentration of O2 by the time period to find the rate of appearance of O2:


Rate = Δ[O2] ÷ time = -0.00100 M ÷ 100 s = -1.0x10-5 M/s

A culture of E.coli bacteria doubles every 40 minutes. If there are 50 bacteria initially, how many will there be after 12 hours? Round your answer to the nearest whole number

Answers

Answer:

There will be 13107200 number of bacteria after 12 hours

Explanation:

1 hour = 60 minutes

So, 12 hours = ([tex]12\times 60[/tex]) minutes = 720 minutes

Initially there are 50 bacteria.

As number of bacteria doubles in every 40 minutes therefore rate of increase in number of bacteria will be similar to a first order reaction.

Hence, number of bacteria after 720 minutes = [tex]50\times (2)^{\frac{720}{40}}[/tex] = 13107200

A sample of the compound MSO4 weighing 0.1131 g reacts with BaCl2 and yields 0.2193 g BaSO4. What is the elemental mass of M and its identity? Hint: All the SO42- from the MSO4 appears in the BaSO4

Answers

Answer:

The atomic mass of the given metal M is 24.3 g/mol.

Therefore, the given metal M is magnesium (Mg)

Explanation:

Reaction involved: MSO₄ + BaCl₂ → BaSO₄ + MCl₂

Atomic mass (g/mol): oxygen (O)=16, sulphur (S)=32,

Molar mass of SO₄²⁻ = 96 g/mol and molar mass of BaSO₄ = 233.38 g/mol

Let the atomic mass of M be m g/mol.

Therefore, molar mass of MSO₄= (m + 96) g/mol

If 0.1131 g of MSO₄ gives 0.2193 g of BaSO₄ on reaction

Then, (m + 96) g/mol of MSO₄ gives 233.38 g/mol of BaSO₄

Therefore, (m + 96) g/mol of MSO₄ = [(233.38 g/mol) × (0.1131 g)] ÷ (0.2193 g)

⇒(m + 96) g/mol = [26.395] ÷ (0.2193)

⇒(m + 96) g/mol = 120.36 g/mol

⇒m = 120.36 g/mol - 96 g/mol = 24.36 g/mol ≈ 24.3 g/mol

Since, the atomic mass of the given metal M is 24.3 g/mol.

Therefore, the given metal M is magnesium (Mg)

in carbon tetrachloride proceeds as follows: 2N2O5→4NO2+O2. The rate law is first order in N2O5. At 64 ∘C the rate constant is 4.82 ×10−3s−1. Part A Part complete Write the rate law for the reaction. Write the rate law for the reaction. rate=2.41×10−3s−1[N2O5] rate=4.82×10−3s−1[N2O5] rate=9.64×10−3s−1[N2O5] rate=4.82×10−3s−1[N2O5]2 Previous Answers Correct Part B What is the rate of reaction when [N2O5]= 2.80×10−2 M ?

Answers

Answer:

Part A

[tex]rate= 4.82*10^{-3}s^{-1} * [N2O5][/tex]

Part B

[tex]rate= 1.35*10^{-4}Ms^{-1}[/tex]

Explanation:

Part A

The rate law is the equation that relates the rate of the reaction, the kinetic constant and the concentration of the reactant or reactants.

For the given chemical reaction we can write a general expression for the rate law as follows:

[tex]rate= k * [N2O5]^{x}[/tex]

where k is the rate constant and x is the order of the reaction with respect of N2O5 concentration. Particularly, a first order reaction kinetics indicate that the rate of the reaction is directly proportional to the concentration of only one reactant. Then x must be 1.

Replacing the value of the rate constant given in the text we can arrive to the following expression for the rate law:

[tex]rate= 4.82*10^{-3}s^{-1} * [N2O5][/tex]

Part B

Replacing the value of the concentration of N2O5 given, we can get the rate of reaction:

[tex]rate= 4.82*10^{-3}s^{-1} *2.80*10^{-2}M[/tex]

[tex]rate= 1.35*10^{-4}Ms^{-1}[/tex]

What is the fluid pressure (in units of ܲܽPascals) a distance 3.2 meters below the surface of Lake Superior?

Answers

Answer:

fluid pressure = 132590 pascal

Explanation:

Fluid pressure can be  calculated by using following relation:

[tex]P_o - P_g = g\time p\times (Z_b - Z_a )[/tex]

Where

Pb = fluid pressure at a distance of 3.2 m below from surface

Pa = fluid pressure at surface = atmospheric pressure = 101325 Pascal

g = gravitational constant = 9.8 m/sec2

[tex]\rho = density of fluid = 997 kg/m³[/tex]

Z = location depth from surface of lake

Therefore , [tex]Z_a = 0 meter[/tex]

[tex]Z_b = 3.2 meter[/tex]

[tex]P_b - P_a =g\times\rhop\times (Z_a - Z_b)[/tex]

               [tex]= 9.8 \times 997\times (3.2-0)[/tex]

               [tex]= 31265 kg/m-sec^2 [/tex]

1 Pascal = 1 Kg/(m.Sec)

[tex]P_b = P_a +31265[/tex]

       = 101325 + 132590 Pascal

       =  132590 pascal

A patient is undergoing a blood test to determine her cholesterol levels. The sample tube holds 3.0 cc (cubic centimeters). What is this volume in liters?

Answers

Answer:

The sample tube holds 3.0 cc which is equal to 0.0030 Liters.

Explanation:

Volume of the blood sample in the sample tube = 3.0 cc

cc = cubic centimeter ; A unit to measure a volume

1 cubic centimeter is equal to 0.001 Liters of volume.

[tex]1 cm^3=0.001 L[/tex]

So in 3.0 cubic centimeter there will be:

[tex]3.0 cm^3=3.0\times 0.001 L=0.0030 L[/tex]

The sample tube holds 3.0 cc which is equal to 0.0030 Liters.

Hydrocyanic acid has a Ka of 4.0
x10-10. What is the percent ofionization of
a 1.0 molar solution?

Answers

Answer:

0.002 %

Explanation:

Given that:

[tex]K_{a}=4.0\times 10^{-10}[/tex]

Concentration = 1.0 M

Consider the ICE take for the dissociation of Hydrocyanic acid as:

                                      HCN    ⇄     H⁺ +        CN⁻

At t=0                            1.0                -              -

At t =equilibrium        (1.0-x)                x           x            

The expression for dissociation constant of Hydrocyanic acid is:

[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {CN}^- \right ]}{[HCN]}[/tex]

[tex]4.0\times 10^{-10}=\frac {x^2}{1.0-x}[/tex]

x is very small, so (1.0 - x) ≅ 1.0

Solving for x, we get:

x = 2×10⁻⁵  M

Percentage ionization = [tex]\frac {2\times 10^{-5}}{1.0}\times 100=0.002 \%[/tex]

You opened up a new 500. mL bottle of RobitussinTM. The dosage is 1.1 tablespoon (TBS).

How many doses are in the bottle?


1 tsp. = 4.93 mL
3 tsp. = 1 TBS.

b)What is the mass of 35.2 mL of a saline (salt water) solution that has a density of 1.12 g cm−3?

Answers

Answer:

30.73 Doses and 39.42 g

Explanation:

It is necessary to know the conversion factors in that manner change the initial 500 mL to the number of doses as follow:

[tex]500 ml * \frac{1tsp}{4.93mL} * \frac{1TBS}{3tsp} * \frac{1 Dose }{1.1 TBS} = 30.7 doses\\[/tex]

And the for b. it is such necessary to take in account that 1 mL = 1 [tex]cm^{3}[/tex]

35.2 mL * 1.12 [tex]\frac{g}{cm^{3} }[/tex] = 39.42 g

Calculate the volume of 42.0 kg of a substance whose density is 8.96 g/mL. Express your answer in milliliters using the correct number of significant figures. Do not enter your answer using scientific notation.

Answers

Answer:

4687.5 mL

Explanation:

Given that the density of the substance = 8.96 g/mL

Mass = 42.0 kg

The conversion of kg into g is shown below as:

1 kg = 1000 g

So, mass = 42.0 ×  1000 g = 42000 g

Volume = ?

So, volume:  

[tex]Volume=\frac {{Mass}}{Density}[/tex]  

[tex]Volume=\frac {42000\ g}{8.96\ g/mL}[/tex]  

The volume of the 42.0 kg of the substance = 4687.5 mL

Give the electron configurations for the following ions: P, P. Ar, and Te plain what interaction is in the hydrogen fluoride (HF) and hydrogen chloride (HCI. Which one do you espect to have a higher bolling emperature? Why?

Answers

Explanation:

Atomic number of P = 15

Common oxidation state of P = -3, +3 and +5

Electronic configuration of P^3+: [tex]1s^2 2s^22p^63s^2[/tex]

Electronic configuration of P^3-: [tex]1s^2 2s^22p^63s^23p^6[/tex]

Electronic configuration of P^5+: [tex]1s^2 2s^22p^6[/tex]

Atomic number of Ar = 18

Argon has stable octet, so it generally does not exist as ions.

Electronic configuration of Ar: [tex]1s^2 2s^22p^63s^23p^6[/tex]

Atomic number of Te = 52

Common oxidation states of Te = -2, +2, +4, +6

Electronic configuration of Te^2-: [tex][Kr]4d^10 5s^2 5p^6[/tex]

Electronic configuration of Te^2+: [tex][Kr]4d^10 5s^2 5p^2[/tex]

Electronic configuration of Te^4+: [tex][Kr]4d^10 5s^2[/tex]

Electronic configuration of Te^6+: [tex][Kr]4d^10[/tex]

Boiling point of HF is more as compared to HCl

Reason:

Molecular weight of HF is low, but is boiling point is high because of presence of hydrogen bonding. Whereas in case of HCl, its molecular weight is high but has only weak van der Waals intercations. As hydrogen bonding is stronger than van der Waals interactions, therefore, boiling point of HF is more.

Calculate the linear momentum of photons of wavelength
(a) 600 nm,
(b) 70 pm,
(c) 200 m.

Answers

Answer:

a) p = 1.10 * 10⁻²⁷ kg·m/s

b) p = 9.46 * 10⁻²⁴ kg·m/s

c) p = 3.31 * 10⁻³⁶ kg·m/s

Explanation:

To solve this problem we use the de Broglie's equation, which describes the wavelenght of a photon with its momentum:

λ=h/p

Where λ is the wavelength, h is Planck's constant (6.626 * 10⁻³⁴ J·s), and p is the linear momentum of the photon.

Rearrange the equation in order to solve for p:

p=h/λ

And now we proceed to calculate, keeping in mind the SI units:

a) 600 nm= 600 * 10⁻⁹ m

p=(6.626 * 10⁻³⁴ J·s) / (600*10⁻⁹m) = 1.10 * 10⁻²⁷ kg·m/s

b) 70 pm= 70 * 10⁻¹² m

p=(6.626 * 10⁻³⁴ J·s) / (70*10⁻¹²m) = 9.46 * 10⁻²⁴ kg·m/s

c) 200 m

p=(6.626 * 10⁻³⁴ J·s) / (200m) = 3.31 * 10⁻³⁶ kg·m/s

Why does ammonium nitrate (NH4NO3) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol? Why does ammonium nitrate (NH4NO3) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol? The vapor pressure of the water decreases upon addition of the solute. The osmotic properties of the system lead to this behavior. The overall enthalpy of the system decreases upon addition of the solute. The overall entropy of the system increases upon dissolution of this strong electrolyte. The overall enthalpy of the system increases upon dissolution of this strong electrolyte.

Answers

Answer: Option (c) is the correct answer.

Explanation:

Entropy is defined as the degree of randomness that is present within the particles of a substance.

As [tex]NH_{4}NO_{3}[/tex] is ionic in nature. Hence, when it is added to water then it will readily dissociate into ammonium ions ([tex]NH^}{+}_{4}[/tex]) and nitrate ions ([tex]NO^{-}_{3}[/tex]).

Therefore, it means that ions of ammonium nitrate will be free to move from one place to another. Hence, there will occur an increase in entropy.

Thus, we can conclude that ammonium nitrate ([tex]NH_{4}NO_{3}[/tex]) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol because the overall entropy of the system increases upon dissolution of this strong electrolyte.

Charles' law relates the way two gas properties change when another property remains the same. What are the two changing properties in Charles' law?
Pressure and temperature
Pressure and volume
Pressure, temperature, and volume
Temperature and volume

ik its not B

Answers

Answer:

option D= temperature and volume

Explanation:

Definition:

Charle's law stated that ,

" At constant pressure, the volume of given amount of gas is directly proportional to its absolute temperature"

V∝ T

V= kT

OR

V/T = k     (k is proportionality constant)

if the temperature is changed from T1 to T2 then the volume of gas is also changed from V1 to V2. Then expression will be:

V1/T1= k  and   V2/T2=k

V1/T1=V2/T2

suppose a cylinder is filled with a gas having volume V1 at temperature T1. When the gas is heated its temperature raises from T1 to T2 and its volume also increased with increase of temperature from V1 TO V2.

       

Final answer:

In Charles' law, the two changing properties are temperature and volume, with pressure maintained constant. The law indicates a direct proportionality between temperature and volume for a given amount of gas.

Explanation:

The two changing properties in Charles' law are temperature and volume. Charles' law states that for a given amount of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature. This means if the temperature of the gas increases, its volume increases as well, provided the pressure and the amount of gas remain constant. In other words, doubling the absolute temperature at constant pressure will double the volume. It is important to note that in gas laws, temperatures must always be expressed in kelvins.

prob.: Consider the combustion of butane (C4H10):
2C4H10(g) + 13O2(g) ==> 8CO2(g) + 10 H2O(L)
IN A PARTICULAR REACTION, 5.0 MOLES OF C4H10 ARE REACTED
WITHAN EXCESS OF O2. CALCULATE THE NUMBER OF MOLES OF CO2
FORMED.

Answers

When 5.0 moles of C₄H₁₀ react with an excess of O2, 20.0 moles of CO₂ are formed.

According to the balanced chemical equation:

2C₄H₁₀(g) + 13O₂(g) ==> 8CO₂(g) + 10H₂O(L)

The stoichiometric coefficient of CO₂ is 8. This means that for every 2 moles of C₄H₁₀ reacted, 8 moles of CO₂ are formed. Therefore, when 5.0 moles of C₄H₁₀  react, the number of moles of CO₂ formed can be calculated using the stoichiometric ratio:

Number of moles of CO₂ = (5.0 moles C₄H₁₀) × (8 moles CO₂ / 2 moles C₄H₁₀ )

                       = 20.0 moles CO₂

So, when 5.0 moles of C₄H₁₀  react with an excess of O₂, 20.0 moles of CO₂ are formed.

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Final answer:

In the combustion of butane, 2 moles of C4H10 react to produce 8 moles of CO2, making a 1:4 ratio. Therefore, when 5.0 moles of butane reacts it produces 20 moles of CO2 due to this stoichiometric ratio.

Explanation:

In the balanced chemical equation for the combustion of butane, we see that '2 moles of C4H10' react to produce '8 moles of CO2'. This is a ratio of 1:4. So, for each mole of butane that reacts, 4 moles of carbon dioxide are produced.

If in a particular reaction, we have '5.0 moles of C4H10', then according to the stoichiometric ratio (1:4) we can calculate the number of moles of CO2 formed. By multiplying the moles of C4H10 by 4, i.e., 5.0 x 4, we get '20 moles of CO2'. Therefore, when 5.0 moles of butane reacts with excess oxygen, 20 moles of CO2 are produced.

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Suppose that NaCl is added to hexane (C6H14) instead of water. Which of the following intermolecular forces will exist in the system?

Check all that apply.

Ion-dipole force between Na+ ions and a hexane molecule
Ion-ion force between Na+ and Cl− ions
Dipole-dipole force between two hexane molecules
Hydrogen bonding between Na+ ions and a hexane molecule
London dispersion force between two hexane molecules

Answers

Answer:

Ion-ion force between Na+ and Cl− ions

London dispersion force between two hexane molecules

Explanation:

"Ion-dipole force between Na+ ions and a hexane molecule " does not exist since hexane has only non-polar bonds and therefore no dipole.

"Ion-ion force between Na+ and Cl− ions " exists since both are ions.

"Dipole-dipole force between two hexane molecules " does not exist since hexane molecules do not have a dipole.

"Hydrogen bonding between Na+ ions and a hexane molecule " does not exist since the hydrogen in the hydrogen bond must be bonded directly to an electronegative atom, which hexane does not have since it is a hydrocarbon.

"London dispersion force between two hexane molecules" exist since hexane is a molecular compound.

Answer:

Ion-ion force between Na⁺ and Cl⁻ ions.

London dispersion force between two hexane molecules.

Explanation:

Suppose that NaCl is added to hexane (C₆H₁₄) instead of water. Which of the following intermolecular forces will exist in the system?

Check all that apply.

Ion-dipole force between Na⁺ ions and a hexane molecule. NO. Hexane is nonpolar so it can interact through ion-dipole forces.Ion-ion force between Na⁺ and Cl⁻ ions. YES. Na⁺ is a cation (positive ion) and interacts with the anion (negative ion) Cl⁻ through ion-ion forces.Dipole-dipole force between two hexane molecules. NO. Since hexane molecules are not dipoles, they do not show dipole-dipole interaction.Hydrogen bonding between Na⁺ ions and a hexane molecule. NO. Hydrogen bonding requires H bonded to a very electronegative atom, such as N, O or F, which does not happen in hexane.London dispersion force between two hexane molecules. YES. The kind of interaction between nonpolar molecules is London dispersion force.

A citrus dealer in Florida sells boxes of 100 oranges at a roadside stand. The boxes routinely are packed with one to three extra oranges to help ensure that customers are happy with their purchases. The average weight of an orange is 7.2 ounces, and the average weight of the boxes in which the oranges are packed is 3.2 pounds. Determine the total weight of five of these 100-orange boxes.

Answers

Answer:

[tex]\large \boxed{\text{250 lb}}[/tex]

Explanation:

1. Weight of oranges

An average box contains 102 oranges, so

Five boxes contain 510 oranges

[tex]\text{Weight of oranges} = \text{510 oranges} \times \dfrac{\text{7.2 oz}}{\text{1 orange}} = \text{3672 oz}\\\\\text{Weight} = \text{3672 oz} \times \dfrac{\text{1 lb}}{\text{16 oz}} = \text{230 lb}[/tex]

2. Weight of boxes

[tex]\text{Weight} = \text{5 boxes} \times \dfrac{\text{3.2 lb}}{\text{1 box}} = \text{16 lb}[/tex]

3. Total weight

[tex]\begin{array}{rcr}\text{Oranges} & = & \text{230 lb}\\\text{Boxes} & = & \text{16 lb}\\\text{TOTAL} & = & \textbf{250 lb}\\\end{array}\\\text{On average, the total weight is $\large \boxed{\textbf{250 lb}}$}[/tex]

Note: The answer can have only two significant figures because that is all you gave for the average weight of an orange.

What happens to the outlet temperatures of the hot and cold fluid when the both the hot fluid is at a maximum and the cold fluid is at a minimum flowrate?

a- Remains same as that of inlet temperature

b- Hot water temperature decreases and cold water temperature increase

c- Hot water remains almost same but cold water temperature increases drastically.

Answers

Answer:

c- Hot water remains almost same but cold water temperature increases drastically.

Explanation:

In a process of heat exchange where a hot and a cold fluid are present, the amount of heat that is transfered from one to another, the heat lost by the hot fluid, is gained by the cold fluid. That can be calculated by the equation : Q = m * Cp * dT.

Where:

m= mass flowrate

Cp=specific-heat-capacity

dT= difference between the inlet and outlet temperatures

We can state the equality of this equation for both fluids:

(mass-flowrate * specific-heat-capacity * (temperature-in – temperature-out) ) for hot medium = (mass-flowrate * specific-heat-capacity * (temperature-out – temperature-in) ) for cold medium

Now if we increase the value of the mass flow rate of the hot fluid to its maximum and decrease the mass flowrate of the cold fluid to its minimum. That will mean that the difference between the inlet and outlet temperatures for the cold fluid must increase so as to compensate the increase on the other side of the equation and guarantee the equality. And taking in account that the inlet temperature of the cold fluid is known and what can change is the outlet one, the correct answer is c that says that the outlet temperature of the cold fluid increases drasically.

it is not from thermodynamic energy?
a) utilization
b) kinetic
c) potential
d) stored energy

Answers

Answer:

(a) Utilization

Explanation:

Thermodynamic energy are:

(1) Kinetic energy:

Kinetic energy is that type of energy of a body which occurs due to the motion of body. Kinetic energy is always positive. In a bound system, the system remains bound as long as the kinetic energy is less than the potential energy due to the interaction of the body.

                            [tex]K.E = \frac{1}{2}mv^{2}[/tex]

(2) Potential energy:

Potential energy is defined as the energy in which the energy is possessed by a body or a system for doing some work, by virtue of its position above the ground level.

Therefore,

Potential energy = PE = mgh

This potential energy is a result of gravity pulling downwards. The gravitational constant, g, is the acceleration of an object due to gravity. This acceleration is about 9.8 m s⁻².

(3) Stored energy:

Stored energy is a Potential energy.

Hence, the only energy which is not a thermodynamic energy is option (a) Utilization energy.

QUESTION 1 Choose the statement that best reflects the three basic steps of any recrystallization Dissolve the impure solid in a minimal amount of boiling solvent: cool the solution to form crystals, vacuum filter the solution to collect the pure crystals. Dissolve the impure solid in an excess of solvent at room temperature; cool the solution to form crystals: vacuum filter to collect the pure crystals O Dissolve the impure solid in an excess of boiling solvent: cool the solution to form crystals vacuum filter to collect the pure crystals. Dissolve the impure solid in a minimal amount of solvent at room temperature; cool the solution to form crystals, vacuum filter to collect the pure crystals.

Answers

Answer:

Dissolve the impure solid in a minimal amount of boiling solvent, cool the solution to form crystals, vacuum filter the solution to collect the pure crystals.

Explanation:

Recrystallization is a process when a solid with impurities is purified. To do this a solvent of the compound we want must be used. We need to use only the quantity necessary to dissolve the compound of interest, otherwise, the solvent will dissolve the impurities or it will interfere in the crystallization.

For most of the solids, the solubility increases with the increase of the temperature, so to speed up the process, heat must be added at the system, or the solvent must be boiling. Then, the solution will be cooled to form the crystals of the compound purified, and then it must be filtered in a vacuum because the crystals can slow down the filtration.

The vapor pressure of water is 28.3 mm Hg at 28 °C. What mass of water vapor, in mg, would be present in a vapor volume of 600. mL at 28°C? Selected Answer: B. 16.3 Correct Answer B. 16.3

Answers

Explanation:

The given data is as follows.

       P = 28.3 mm = [tex]\frac{28.3}{760}[/tex],            V = 600 mL = [tex]600 ml \frac{0.001 L}{1 ml}[/tex] = 0.6 L

      R = 0.082 L atm/mol K,                 T = (28 + 273) K = 301 K

Therefore, according to ideal gas law PV = nRT. Hence, putting the values into this equation calculate the number of moles as follows.

                                    PV = nRT

                 [tex]\frac{28.3}{760} \times 0.6 L[/tex] = [tex]n \times 0.082 L atm/mol K \times 301 K[/tex]            

                       n = [tex]9.03 \times 10^{-4}[/tex] mol  

As it is known that number of moles equal to mass divided by molar mass. Hence, mass of water vapor present will be calculated as follows. (molar mass of water is 18 g/mol)

               No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

        [tex]9.03 \times 10^{-4}[/tex] mol = [tex]\frac{mass}{18 g/mol}[/tex]

                            mass = [tex]162.5 \times 10^{-4}g[/tex]

                                      = [tex]163 \times 10^{-4}g[/tex] (approx)

Since, 1 g = 1000 mg. Therefore, [tex]163 \times 10^{-4}g[/tex] will be equal to  [tex]163 \times 10^{-4}g \times \frac{10^{3}mg}{1 g}[/tex]

                          = 16.3 mg

Thus, we can conclude that mass of water vapor present is 16.3 mg.

Final answer:

To find the mass of water vapor in a given vapor volume at a certain temperature, you can use the ideal gas law. Convert the temperature from Celsius to Kelvin, calculate the number of moles using the ideal gas law equation, and then convert moles to grams using the molar mass of water. Finally, convert grams to milligrams. The mass of water vapor in a vapor volume of 600 mL at 28 °C is 5097 mg.

Explanation:

To find the mass of water vapor, we can use the ideal gas law equation: PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R represents the ideal gas constant, and T represents temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin by adding 273.15. Then, we can rearrange the equation to solve for n, the number of moles. Once we have the moles, we can use the molar mass of water to convert it to mass in grams. Finally, we can convert grams to milligrams by multiplying by 1000.

Given:

Vapor pressure of water = 28.3 mm Hg
Temperature = 28 °C
Vapor volume = 600 mL

Converting the temperature to Kelvin:

T = 28 °C + 273.15 = 301.15 K

Using the ideal gas law equation to find the number of moles (n):

n = PV / RT

Substituting the values:

n = (28.3 mm Hg * 600 mL) / (62.36 L mm/mol K * 301.15 K)

n = 0.2829 mol

Using the molar mass of water (18.015 g/mol) to find the mass:

mass = n * molar mass

mass = 0.2829 mol * 18.015 g/mol

mass = 5.097 g

Converting grams to milligrams:

mass = 5.097 g * 1000 mg/g

mass = 5097 mg

Therefore, the mass of water vapor in a vapor volume of 600 mL at 28 °C is 5097 mg.

One liter of ocean water contains 35.06 g of salt. What volume of ocean water would contain 1.00 kg of salt? Express your answer in L using the correct number of significant figures. Do not enter your answer using scientific notation.

Answers

Answer:

28.52 L

Explanation:

First, let's calculate the density of the ocean, which is the mass divided by the volume:

d = m/V

d = 35.06/1

d = 35.06 g/L

So, for a mass of 1.00 kg = 1000.00 g

d = m/V

35.06 = 1000.00/V

V = 1000.00/35.06

V = 28.52 L

How all the data are expressed with two significant figures, the volume must also be expressed with two.

How many grams of 48.0 wt% NaOH (FM 40.00) should be diluted to 1.00 L to make 0.11 M NaOH? (Enter your answer using two digits.)

Answers

Answer:

The answer is 916.67 g

Explanation:

48.0 wt% NaOH means that there are 48 g of NaOH in 100 g of solution. With this information and the molecular weight of NaOH (40 g/mol), we can calculate the number of mol there are in 100 g of this solution:

[tex]\frac{48 g NaOH}{100 g solution}[/tex] x [tex]\frac{1 mol NaOH}{40 g}[/tex] = 0.012 mol NaOH/100 g solution

Finally, we need 0.11 mol in 1 liter of solution to obtain a 0.11 M NaOH solution.

0.012 mol NaOH ------------ 100 g solution

0.11 mol NaOH------------------------- X

X= 0.11 mol NaOH x 100 g/ 0.012 mol NaOH= 916.67 g

We have to weigh 916.67 g of 48.0%wt NaOH and dilute it in a final volume of 1 L of water to obtain a 0.11 M NaOH solution.

how does erucamide make substances scratch resistant?

Answers

Answer:

Explained

Explanation:

Erucamide is plastic additive used as anti slip agent and used as scratch resistant. It is used as scratch resistance because

- It reduces coefficient of friction between the polymer layer.

- Slip additives in Erucamide form coalesce along the surface.

-These additives move from the bulk to the part surface to form closed pack layers.

Two hundred kg of liquid contains 30% butane, 40% pentane, and the rest hexane (mass %) Determine: The mole fraction composition of the liquid The mass fraction composition on hexane free basis 1. 2.

Answers

Answer:

The mole fraction composition of the liquid is :

Mole fraction of butane, pentane and hexane are 0.3638,0.3908 and 0.2454 respectively.

Explanation:

Mass of the liquid mixture = 200 g

Percentage of butane = 30%

Mass of butane = [tex]\frac{30}{100}\times 200 g=60 g[/tex]

Moles of butane = [tex]n_1=\frac{60 g}{58 g/mol}=1.0345 mol[/tex]

Percentage of pentane= 40%

Mass of pentane= [tex]\frac{40}{100}\times 200 g=80 g[/tex]

Moles of pentane= [tex]n_2=\frac{80 g}{58 g/mol}=1.1111 mol[/tex]

Percentage of hexane = 100% - 30% - 40% = 30%

Mass of hexane = [tex]\frac{30}{100}\times 200 g=60 g[/tex]

Moles of hexane = [tex]n_2=\frac{60 g}{86 g/mol}=0.6977 mol[/tex]

Mole fraction of butane, pentane and hexane : [tex]\chi_1, \chi_2 \& \chi_3[/tex]

[tex]\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{1.0345 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3638[/tex]

[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}=\frac{1.1111 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3908[/tex]

[tex]\chi_3=\frac{n_1}{n_1+n_2+n_3}=\frac{0.6977 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.2454[/tex]

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