a weak acid undergoes _ ionization in water

Answer:

0.80 M

Explanation:

Step 1: Write the generic neutralization reaction

HA + NaOH ⇒ NaA + H₂O

Step 2: Calculate the reacting moles of NaOH

20.05 mL of 1.000 M NaOH react.

0.02005 L × 1.000 mol/L = 0.02005 mol

Step 3: Calculate the reacting moles of HA

The molar ratio of NaOH to HA is 1:1. The reacting moles of HA is 1/1 × 0.02005 mol = 0.02005 mol.

Step 4: Calculate the concentration of HA

0.02005 moles of HA are in 25 mL.

[HA] = 0.02005 mol/0.025 L = 0.80 M

Answer:

6.88 mg

Explanation:

Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄

The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.

175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P

Step 2: Calculate the rate constant for the decay of ³²P

The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.

k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹

Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days

For first-order kinetics, we will use the following expression.

ln P = ln P₀ - k × t

ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d

P = 6.88 mg

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:

the equilibrium concentration of HF is 2.85 M

Option a) 2.85 M is the correct answer.

Explanation:

Given the data in the question;

     H₂         +      F₂      ⇄     2HF

I    1.69 M        1.69 M           0

C    -x                 -x               +2x

E    1.69-x         1.69-x          +2x

given that Kc = 115        

Kc = [ HF ]² / [H₂][F₂]

we substitute

115 = [ 2x ]² / [ 1.69-x  ][ 1.69-x ]

lets find the square root of both sides

10.7238 = 2x / [ 1.69-x  ]

10.7238[ 1.69-x  ] = 2x

18.123222 - 10.7238x = 2x

2x + 10.7238x = 18.123222

12.7238x = 18.123222

x = 18.123222 / 12.7238

x = 1.424356

Hence, equilibrium concentration of HF = 2x

that is;

HF = 2 × 1.424356

HF = 2.8487 ≈ 2.85 M

Therefore, the equilibrium concentration of HF is 2.85 M

Option a) 2.85 M is the correct answer.

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.

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.

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:

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:

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:

10764 J

Explanation:

Remember the equation for specific heat::

q = mcΔT

q = 100 x 0.897 x 120

q = 10764

Answer: C

Explanation:

Answer:

compressable

Explanation:

as liquid nitrogen came out from container the force exerted on it which changes it into liquid ends making it gas

liquid nitrogen boils at 77 K or -196° C

Answer:

see explanation

Explanation:

SO₃(g) + 395.77 Kj/mole => S°(s) + 3/2O₂(g)

The standard heat of formation for SO₃(g) is given by the following rxn:

S°(s) + 3/2O₂(g) => SO₃(g) + 395.77 kJ/mole. Reversing this reaction is the decomposition of SO₃(g) into its basic elements in their standard state (25°C, 1atm) and is endothermic with +295.77Kj/mole.

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:

the correct answer is c

Explanation:

becuase i had the same question

Answer:

C

Explanation:

KE at B is max and PE is 0

KE at C is 0 and PE is max

so when student swings from C to B

its KE increases

and PE decreases

Answer:

The theoretical yield potassium permanganate, KMnO₄ when starting with 50.0 g MnO₂ is 90.8 g

Explanation:

Molar mass of MnO₂ = (55 + 2 × 16) = 87.0 g/mol

Molar mass of KMnO₄ = (39 + 55 + 4 × 16) = 158 g/mol

Moles of MnO₂ in 50 g = reacting mass / molar mass

where reacting mass = 50 g

Moles of MnO₂ in 50 g = 50 g /87 g/mol = 0.575 moles

The equation for the production of potassium permanganate is as follows:

2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2

From the equation of the reaction above, 2 moles of MnO₂ produces 2 moles of KmNO₄. The mole ratio of MnO₂ to KMnO₄ is 1 : 1

Therefore, 0.575 moles of MnO₂ will produce theoretically 0.575 moles of KMnO₄

Mass of 0.575 moles of KMnO₄ = number of moles × molar mass

Mass of 0.575 moles of KMnO₄ = 0.575 moles × 158 g/mol = 90.8 g of KMnO₄

Therefore, the theoretical yield potassium permanganate when starting with 50.0 g MnO₂ is 90.8 g

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: The correct option is C (One of the fatty acid salts was unsaturated, and it completely isomerized under the reaction conditions).

Explanation:

Fats and oils belongs to a general group of compounds known as lipids. Fatty acids are weak acid and are divided into two:

--> Saturated fatty acids: These have NO double bonds in their hydrocarbon chain, and

--> Unsaturated fatty acids: These have one or more double bonds in their hydrocarbon chain.

SAPONIFICATION is defined as the process by which fats and oil is hydrolyzed with caustic alkali to yield propane-1,2,3-triol and the corresponding sodium salt of the component fatty acids. During this process, One hydroxide ion is required to hydrolyze one ester linkage of a triacylglycerol molecule. Because there are three ester linkages in a triacylglycerol, three equivalents of sodium hydroxide will be needed to completely saponify the triacylglycerol. This explains the reason why saponification of the triacylglycerol iresulted in four different fatty acid salts.

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%.

BEST ANSWER IS

have a great summer

1.01 x 10^-3 moles

n = 0.20 mol

N

N = n × 6.02 × 10²³ atoms/mol

N = 0.2 mol × 6.02 × 10²³ atoms/mol

N = 1.20 × 10²³ atoms

Therefore, there are 1.20 × 10²³ gold atoms in 0.2 mol of a gold sample.

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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:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

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.

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:

A

Explanation:

Answer:

.04 cg/L

Explanation:

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