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2 years ago

What is the best definition of a eukaryotic cell?

Chemistry

1 answer:

Novosadov [1.4K]2 years ago

3 0

Answer:

A. has an internal structure consisting of a membrane–bound nucleus and membrane–bound organelles

Explanation:

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BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

5 0

2 years ago

When the volume of a gas is

Montano1993 [528]

Answer:

$\boxed {\boxed {\sf 232.9 \textdegree C}}$

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

$\frac {V_1}{T_1}= \frac{V_2}{T_2}$

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

$\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}$

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

$\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

$250 \ mL * T_2 = 137 \textdegree C * 425 \ mL$