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2 years ago

1. Calculate the [H] in a solution that has a pH of 9.88.

Chemistry

2 answers:

Olenka [21]2 years ago

5 0

The answer to the question is D.

Aleksandr-060686 [28]2 years ago

4 0

Answer:

The answer to your question is the letter D) [H⁺] = 1.3 x 10⁻¹⁰ M

Explanation:

Data

[H⁺] = ?

pH = 9.88

pH measures the [H⁺] in a solution. It has a range from 0 to 14. Acids have a pH from 0 to 6.9, a neutral solution has a pH of 7.0 and alkalis have a pH from 7.1 to 14.

Formula

pH = -log [H⁺]

Substitution

9.88 = -log[H⁺]

Simplification

antilog -9.88 = [H⁺]

-Result

[H⁺] = 1.3 x 10⁻¹⁰ M

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A gas-filled weather balloon with a volume of 65.0 L is released at sea level where conditions are745 torr and 25 ºC. The balloo

Blababa [14]

Answer:

It will reach to its maximum volume.

Explanation:

Using Ideal gas equation for same mole of gas as :-

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 65.0 mL

V₂ = ?

P₁ = 745 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 745 / 760 atm = 0.9803 atm

P₁ = 0.9803 atm

P₂ = 0.066 atm

T₁ = 25 ºC

T₂ = -5 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (25 + 273.15) K = 298.15 K

T₂ = (-5 + 273.15) K = 268.15 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac{{0.9803}\times {65.0}}{298.15}=\frac{{0.066}\times {V_2}}{268.15}$

Solving for V₂ as:-

$V_2=\frac{0.9803\times \:65.0\times 268.15}{298.15\times 0.066}$

$V_2=\frac{17086.38392}{19.6779}$

V₂ = 868 L

Given that:- V max = 835 L

<u>Thus, it will reach to its maximum volume.</u>

3 0

2 years ago

What is the equilibrium constant expression for the reaction below?<br> H2 (g) + I2 (g) → 2 HI(g)

brilliants [131]

<h3>Further explanation</h3>

Given

Reaction

H₂ (g) + I₂ (g) → 2HI(g)

Required

The equilibrium constant

Solution

The equilibrium constant is the value of the product in the equilibrium state of the substance in the right (product) divided by the substance in the left (reactant) with the exponents of each reaction coefficient

The equilibrium constant for reaction

pA + qB ⇒ mC + nD

$\large {\boxed {\bold {K ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}$