What is the Ke of the 20 gram shoe as it falls to the ground at 6 m/s?


PLEASE HELP ITS 7TH GRADE SCIENCE!!!

QUESTION :WHICH OF THE FOLLOWING IS AN EXAMPLE OF A CONTROLLED EXPERIMENT TO TEST THIS?

ANSWER:

D. The temperatures of five breakers of 250 mL of water are varied, and 10 g of sugar is added to each breaker.

Answer:

A 1 liter volumetric flask should be used.

Explanation:

First we convert 166.00 g of KI into moles, using its molar mass:

Molar mass of KI = Molar mass of K + Molar mass of I = 166 g/mol

Then we calculate the required volume, using the definition of molarity:

Liters = moles / molarity

Answer:

0.01228s⁻¹ = rate constant

Half-life = 56.4s

Explanation:

The first order reaction follows the equation:

ln[A] = -kt + ln[A]₀

Where [A] is amount of reactant after time t = 45.0%, k is rate constante and [A]₀ initial amount of reactant = 100%

ln[45%] = -k*65s + ln[100%]

-0.7985 = -k*65s

0.01228s⁻¹ = rate constant

Half-life is:

Half-life = ln2 / k

Answer:

c. Kc and Kp are equal when the sum of the stoichiometric coefficients for the products equals the sum of the stoichiometric coefficients for the reactants.

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to write the relationship between Kc and Kp in terms of the temperature and the change in the stoichiometric coefficients:

[tex]Kp=Kc(RT)^{\Delta \nu}[/tex]

Thus, we can see that the Kp=Kc just when Δυ=0, which is the sum of the coefficients of products minus that of reactants; and therefore, the correct answer will be c. Kc and Kp are equal when the sum of the stoichiometric coefficients for the products equals the sum of the stoichiometric coefficients for the reactants.

Regards!

Answer: The mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g

Explanation:

Molarity is calculated by using the equation:

[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)  

Molarity of [tex]H_2SO_4[/tex] = 3.0 M

Volume of solution = 25.0 L

Putting values in equation 1, we get:

[tex]\text{Moles of }H_2SO_4=(3.0mol/L\times 25.0L)=75mol[/tex]

The ideal gas equation is given as:

[tex]PV=nRT[/tex] .......(2)

where,

P = pressure of the gas = 0.68 atm

V = volume of gas = [tex]3.1\times 10^3L[/tex]

n = number of moles of gas = ? moles

R = Gas constant = 0.0821 L.atm/mol.K

T = temperature of the gas = 298 K

Putting values in equation 2, we get:

[tex]0.68atm\times 3.1\times 10^3L=n\times 0.0821L.atm/mol.K\times 298K\\\\n=\frac{0.68\times 3.1\times 10^3}{0.0821\times 298}=86.16mol[/tex]

For the given chemical equation:

[tex]NH_3(g)+H_2SO_4(aq)\rightarrow (NH_4)_2SO_4(aq)[/tex]

By stoichiometry of the reaction:

If 1 mole of [tex]H_2SO_4[/tex] reacts with 1 mole of [tex]NH_3[/tex]

So, 75 moles of [tex]H_2SO_4[/tex] will react with = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex]NH_3[/tex]

As the given amount of [tex]NH_3[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent

Thus, [tex]H_2SO_4[/tex] is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 1 mole of [tex]H_2SO_4[/tex] produces 1 mole of [tex](NH_4)_2SO_4[/tex]

So, 75 moles of [tex]H_2SO_4[/tex] will produce = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex](NH_4)_2SO_4[/tex]

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We know, molar mass of [tex](NH_4)_2SO_4[/tex] = 132.14 g/mol

Putting values in above equation, we get:

[tex]\text{Mass of }(NH_4)_2SO_4=(75mol\times 132.14g/mol)=9910.5g[/tex]

Hence, the mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g

Answer:

Mixture of a Strong Acid and a Strong Base

On mixing a strong acid and strong base neutralization (pH = 7) takes place. The resulting solution may be an acid or base depending on the Concentration. Say, N1, V1 is the strength and volume of the strong acid and N2, V2 is the strength and volume of the strong base

Explanation:

Most introductory chemistry books will teach that the reaction between an acid and a base is called neutralization, and the products formed are water and a salt

Answer:

Magnesium reacts with dilute hydrochloric acid in a conical flask which is ... One student can add the magnesium ribbon to the acid and stopper the flask, ... 50 cm3 of 1M hydrochloric acid is a six-fold excess of acid.

Answer:

10764 J

Explanation:

Remember the equation for specific heat::

q = mcΔT

q = 100 x 0.897 x 120

q = 10764

Answer:

a) 0.11

b)76.9

c) 8.8

d) 1.7*10^-4

Explanation:

Step 1: Data given

K = 1.3 * 10^-2 for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)

Step 2: Formula of K

aA(g) + bB(g) ⇌ cC(g) + dD(g)

K = [C]^c *[D]^d  / [A]^a * [B]^b

K = 1.3 * 10^-2 = [NH3]² / [H2]³*[N2]

Step 3:

a) 1/2N2 + 3/2H2(g) ⇌ NH3(g)

N2(g) + 3H2(g) ⇌ 2NH3

1/2N2 + 3/2H2(g) ⇌ NH3(g)    =>K' =  [tex]\sqrt{K}[/tex]

K' = [tex]\sqrt{1.3*10^-2}[/tex] = 0.11

b. 2NH3(g) ⇌ N2(g) + 3H2(g)

N2(g) + 3H2(g) ⇌ 2NH3

2NH3(g) ⇌ N2(g) + 3H2(g)    =>K' = 1/K

K' = 1/(1.3*10^-2) = 76.9

c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)

N2(g) + 3H2(g) ⇌ 2NH3

NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)    

=>K' = [tex]\frac{1}{\sqrt{K} }[/tex]

K' = [tex]\frac{1}{\sqrt{1.3*10^-2} }[/tex]

K' = 8.8

d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)

N2(g) + 3H2(g) ⇌ 2NH3

2N2(g) + 6H2(g) ⇌ 4NH3(g)

K' = K²

K' = (1.3*10^-2)²

K' = 1.7 *10 ^-4

Values of equilibrium constant at given temperature for the following reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.

Equilibrium constant is define as the ration of the concentrations of product to the concentrations of reactant with respect to the exponent of their coefficients.

Given chemical reaction is:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Equilibrium constant for this reaction is:

K = [NH₃]² / [N₂][H₂]³

K = 1.3 × 10⁻² (given)

1/2N₂(g) + 3/2H₂(g) ⇌ NH₃(g)

K₁ = √K

Because concentration of all given species is 1/2 of the given reaction, so value of K₁ will be written as:
K₁ = √(1.3 × 10⁻²) = 0.11

2NH₃(g) ⇌ N₂(g) + 3H₂(g)

K₂ = 1/K

Because concentration of reactant and products are reciprocal from the concentration of original given reaction, so value of K₂ will be written as:
K₂ = 1/1.3 × 10⁻² = 76.9

NH₃(g) ⇌ 1/2N₂(g) + 3/2H₂(g)

K₃ = 1/√K

Because concentrations of given species is reciprocal as well as half of the given original reaction, so value of K₃ will be written as:
K₃ = 1/√(1.3 × 10⁻²) = 8.8

2N₂(g) + 6H₂(g) ⇌ 4NH₃(g)

K₄ = K²

Because concentrations of given species is double of the given original reaction, so value of K₄ will be written as:
K₄ = (1.3 × 10⁻²)² = 1.7 × 10⁻⁴

Hence, the value of K for given reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.

To know more about equilibrium constant, visit the below link:
https://brainly.com/question/12858312

Answer:

For (a): The moles of Ar is 16.94 moles

For (b): The moles of [tex]Cl_2[/tex] is 16.94 moles

For (c): The total number of moles in a tank is 23.47 moles

For (d): The mole fraction of Ar is 0.722

For (e): The mole fraction of [tex]Cl_2[/tex] is 0.278

For (f): The partial pressure of Ar is 2.888 atm

For (g): The partial pressure of [tex]Cl_2[/tex] is 1.112 atm

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of Ar = 675.5 g

Molar mass of Ar = 39.95 g/mol

Plugging values in equation 1:

[tex]\text{Moles of Ar}=\frac{675.5g}{39.95g/mol}=16.91 mol[/tex]

Given mass of [tex]Cl_2[/tex] = 465.0 g

Molar mass of [tex]Cl_2[/tex] = 70.9 g/mol

Plugging values in equation 1:

[tex]\text{Moles of }Cl_2=\frac{465.0g}{70.9g/mol}=6.56 mol[/tex]

Total moles of gas in the tank = [16.91 + 6.56] mol = 23.47 mol

Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]        .....(2)

where n is the number of moles

Moles of Ar = 16.94 moles

Total moles of gas in the tank = 23.47 mol

Putting values in equation 2, we get:

[tex]\chi_{Ar}=\frac{16.94}{23.47}\\\\\chi_{Ar}=0.722[/tex]

Total mole fraction of the system is always 1

Mole fraction of [tex]Cl_2[/tex] = [1 - 0.722] = 0.278

Raoult's law is the law used to calculate the partial pressure of the individual gases present in the mixture.

The equation for Raoult's law follows:

[tex]p_A=\chi_A\times p_T[/tex]                  .....(3)

where [tex]p_A[/tex] is the partial pressure of component A in the mixture and [tex]p_T[/tex] is the total partial pressure of the mixture

We are given:

[tex]\chi_{Ar}=0.722\\p_T=4.00atm[/tex]

Putting values in equation 3, we get:

[tex]p_{Ar}=0.722\times 4.00atm\\\\p_{Ar}=2.888atm[/tex]

We are given:

[tex]\chi_{Cl_2}=0.278\\p_T=4.00atm[/tex]

Putting values in equation 3, we get:

[tex]p_{Cl_2}=0.278\times 4.00atm\\\\p_{Cl_2}=1.112atm[/tex]

Answer:

See explanation.

Explanation:

Hello there!

In this case, since this problem  is about gas laws, more specifically about the Gay-Lussac's one since the volume is said to be constant, we can use the following equation for its solution for the final pressure, P2:

[tex]\frac{P_2}{T_2} = \frac{P_1}{T_1}[/tex]

[tex]P_2= \frac{P_1T_2}{T_1}\\\\P_2 =\frac{12.0atm*450K}{300K}\\\\P_2= 18.0atm[/tex]

Thus, we fill in the table as follows:

                           Initial        Final

Pressure           12.0 atm     18.0 atm

Volume                4.0 L         4.0 L

Temperature      300K         450K

Regards!

Answer:it dissolves and evaporates

Explanation:

Answer:

the correct answer is c

Explanation:

becuase i had the same question

The answer ( 13 and 8 )

x²=5²+12²

x²=25+144

x²=169

x=13

x²=5²+6²

x²=25+36

x²=61

x=7.8

x=8

Answer:

Compound A, C8H10, absorbed 3 equivalents of H2 on catalytic hydrogenation over a Pd/C catalyst to give B (C8H16). On ozonolysis, compound A gave, among other things, a ketone which was identified as cyclopentanone. On treatment with NaNH2 in NH3, followed by addition of iodomethane, compound A gave a new hydrocarbon, C (C9H12). Draw the structure of Compound A and Compound B.

Explanation:

The degree of unsaturation in the given compound A  C8H10 is:

DU= (Cn+1)-Hn/2 -Xn/2 +Nn/2

DU=(8+1)-10/2

    =9-5

    =4

So, the given compound has either three double bonds and a ring or four double bonds or four rings.

Given,

compound A C8H10 absorbed three equivalents of H2 on catalytic hydrogenation that means compound A has three double bonds or it has one double bond and one triple bond.

The structure of compounds A, B and C along with the entire reaction is shown below:

Answer:

43.5

Explanation:

Hope that helps

Answer:

Hydrogen and fuel cell technologies power cars, buildings and more. But how ... Test your knowledge with this quiz! ... How do fuel cells generate electricity

Answer:

A

Explanation:

fuel cell is a device that converts the chemical energy from fuel into electricity via a chemical reaction with oxygen or another oxidizing agent. Batteries work in a closed system, while fuel cells require their reactants to be replenished.

Answer: Pressure of the gas is 0.129375 atm when the volume of the vessel increased by a factor of 16.00.

Explanation:

The formula for ideal gas equation is as follows.

[tex]PV = Nk_{b}T[/tex]

where,

[tex]k_{b}[/tex] = Boltzmann constant

N = number of moles

That can also be written as:

[tex]\frac{PV}{T} = constant[/tex]

As pressure and volume are inversely proportional to each other. So, if one of the state variable is increased then the other one will decrease or vice-versa.

So, if volume of the vessel increased by a factor of 16.00 then it means pressure is decreased by a factor of 16.00

Therefore, final volume is as follows.

[tex]65.8 L \times 16.00\\= 1052.8 L[/tex]

Now, final pressure is as follows.

[tex]\frac{2.07}{16.00}\\= 0.129375 atm[/tex]

Initially the product of pressure and volume is as follows.

[tex]PV = 2.07 \times 65.8\\= 136.206[/tex]

Hence, if volume of the vessel increased by a factor of 16.00 and pressure is decreased by a factor of 16.00 then its product is as follows.

[tex]PV = 0.129375 \times 1052.8\\= 136.206[/tex]

Here, product of pressure and volume remains the same.

Thus, we can conclude that pressure of the gas is 0.129375 atm when the volume of the vessel increased by a factor of 16.00.

Answer:

A) 1.3 × 10⁻⁵ mol/L

Explanation:

Step 1: Write the balanced equation for the solution of AgCl

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

Step 2: Make an ICE Chart

        AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

I                           0             0

C                        +S            +S

E                          S              S

If we replace the solubility (S) in the Ksp expression, we get,

Ksp = [Ag⁺] [Cl⁻] = S × S = S²

S = √Ksp = √1.8 × 10⁻¹⁰ = 1.3 × 10⁻⁵ mol/L

Answer:

It is always less than the theoretical yield.

Answer:

The system is not at equilibrium and the reaction will proceed to the left.

Explanation:

Step 1: Write the balanced equation

H₂(g) + CO₂(g) ⇄ CO(g) + H₂O(g)

Step 2: Calculate the reaction quotient (Q)

The reaction is calculated in the same way as the equilibrium constant (Kc) but it uses the concentrations at any time.

Q = [CO] × [H₂O] / [H₂] × [CO₂]

Q = 0.610 × 0.695 / 0.425 × 0.500 = 2.00

Since Q ≠ Kc, the reaction is not at equilibrium.

Since Q > Kc, the reaction will proceed to the left.

Answer:

0.375mL of solution of nalorphine must be injected

Explanation:

A solution of 0.40% (w/v) contains 0.40g of solute (In this case, nalorphine), in 100mL of solution. To obtain 1.5mg of nalorphine = 1.5x10⁻³g of nalorphine are needed:

1.5x10⁻³g * (100mL / 0.40g) =

Answer:

Explanation:

Continental tropical air masses are extremely hot and dry. Arctic, Antarctic, and polar air masses are cold. The qualities of arctic air are developed over ice and snow-covered ground. Arctic air is deeply cold, colder than polar air masses.

Answer:

1.46 × 10⁻⁸ g

Explanation:

Step 1: Given data

Molecules of CO₂: 2.00 × 10¹⁴ molecules

Step 2: Convert molecules to moles

We need a conversion factor: Avogadro's number. There are 6.02 × 10²³ molecules in 1 mole of molecules.

2.00 × 10¹⁴ molecules × 1 mol/6.02 × 10²³ = 3.32 × 10⁻¹⁰ mol

Step 3: Convert moles to mass

We need a conversion factor: the molar mass. The molar mass of CO₂is 44.01 g/mol.

3.32 × 10⁻¹⁰ mol × 44.01 g/mol = 1.46 × 10⁻⁸ g

Answer: The %v/v of the given KCl solution is 7.6%.

Explanation:

Given: Volume of solute = [tex]4.6 \times 10^{1} ml[/tex]

Volume of solution = [tex]6.0 \times 10^{-1} g/ml[/tex]

Formula used to calculate %v/v is as follows.

[tex]\frac{volume of solute}{volume of solution} \times 100[/tex]

Substitute the values into above formula as follows.

[tex]\frac{volume of solute}{volume of solution} \times 100\\\frac{4.6 \times 10^{1}}{6.0 \times 10^{-1}} \times 100\\= 7.6[/tex]

Thus. we can conclude that the %v/v of the given KCl solution is 7.6%.

Answer:

Explanation:

When an electron moves from a lower energy level to a higher energy level, energy is absorbed by the atom. When an electron moves from a higher to a lower energy level, energy is released and photon is emitted.

this emitted photon is depicted as a small wave-packet being expelled by the atom in a well-defined direction.

Actual yield over theoretical yield, then multiply by 100

The question is incomplete, the complete question is;

Groups of the periodic table correspond to elements with a. the same color b. the same atomic number c. similar chemical properties d. similar numbers of neutrons

Answer:

similar chemical properties

Explanation:

In the periodic classification of elements, elements are divided into groups and periods. Elements in the same group of the periodic table have the same number of outermost electrons and share very similar chemical properties.

Elements in the same period have the same number of shells and the same maximum energy level of the outermost electron. Chemical properties carry markedly across a period.