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2 years ago

HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.

Chemistry

1 answer:

Lostsunrise [7]2 years ago

7 0

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, $C_A$ = 6.0mol/dm³

volume of HA, $V_A$ = 25.0cm³, = 0.025dm³

Concentration of HB, $C_B$ = 3.0mol/dm³

volume of HB, $V_B$ = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

$C_iVi = C_fV_f$

where;

$C_i$ is initial concentration

$V_i$ is initial volume

$C_f$ is final concentration of the solution

$V_f$ is final volume of the solution

$C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3$

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

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Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)

gtnhenbr [62]

Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

$Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}$ , where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr 7.96 torr 0 0 initial partial pressure

-x -x +x +x react

105.1-x 7.96-x x x equilibrium

Then:

$Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}$

$2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}$

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

$x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}$

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

8 0

2 years ago

Read 2 more answers

Calculate the mass of 3.75 x 10^23 molecules of CaSO4 Please HelpIn a hurry Ill give brainliest

SOVA2 [1]

Answer:

Mass = 84.82 g

Explanation:

Given data:

Number of molecules of CaSO₄ = 3.75× 10²³

Mass in gram = ?

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

3.75× 10²³ molecule × 1 mol / 6.022 × 10²³ molecules

0.623 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.623 mol × 136.14 g/mol

Mass = 84.82 g

7 0

1 year ago

A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c

Nata [24]

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

8 0

2 years ago

A 36 gram sample of water has an initial temperature of 22 degrees Celsius after the sample absorbs 1200joules of heat energy th

UkoKoshka [18]

We can use the heat equation,
Q = mcΔT

Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

Let's assume that the finale temperature is T.

Q = 1200 J

m = 36 g
c = 4.186 J/g °C

ΔT = (T - 22)

By applying the formula,

1200 J = 36 g x 4.186 J/g °C x (T - 22)

(T - 22) = 1200 J / (36 g x 4.186 J/g °C)

(T - 22) = 7.96 °C

T = (7.96 + 22) °C = 29.96 °C

T = 30 °C

Hence, the final temperature is 30 °C.

8 0

2 years ago

Read 2 more answers

A voltaic cell consists of a nickel electrode in a solution containing Ni2+ ions, and a copper electrode in a solution containin

Marysya12 [62]

First, we write the half equations for the reduction of the chemical species present:

Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V

In order to determine the potential of the cell, we find the difference between the two values. For this:

E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V

The second option is correct. (The difference in values is due to different values in literature, and it is negligible)

4 0

1 year ago