What will happen when a piece of magnesium metal is dropped into a beaker containing a l M solution of copper(1) chloride?

True or false: In a combustion reaction, matter is destroyed.

sp2606 [1]

Matter is never destroyed, nor created. So, this is false.

6 0

1 year ago

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

1 year ago

Use coulomb's law to calculate the ionization energy in kj/mol of an atom composed of a proton and an electron separated by 185.

Tems11 [23]

Coulomb's law mathematically is:
F = kQ₁Q₂/r²
we integrate this with respect to distance to obtain the expression for energy:
E = kQ₁Q₂/r; where k is the Coulomb's constant = 9 x 10⁹; Q are the charges, r is the seperation
Charge on proton = charge on electron = 1.6 x 10⁻¹⁹ C
E = (9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹) / (185 x 10⁻¹²)
E = 1.24 x 10⁻¹⁸ Joules per proton/electron pair
Number of pairs in one mole = 6.02 x 10²³
Energy = 6.02 x 10²³ x 1.24 x 10⁻¹⁸
= 746.5 kJ

5 0

1 year ago

Read 2 more answers

By which process is a precipitate most easily separated from the liquid in which it is suspended

Gelneren [198K]

Filtration, if its a precipitate that means its insoluble.

7 0

1 year ago

Read 2 more answers

In an experiment, 0.42 mol of co and 0.42 mol of h2 were placed in a 1.00-l reaction vessel. at equilibrium, there were 0.29 mol

Likurg_2 [28]

To determine the Keq, we need the chemical reaction in the system. In this case it would be:

CO + 2H2 = CH3OH

The Keq is the ration of the amount of the product and the reactant. We use the ICE table for this. We do as follows:

CO H2 CH3OH
I .42 .42 0
C -0.13 -2(0.13) 0.13
-----------------------------------------------
E = .29 0.16 0.13

Therefore,

Keq = [CH3OH] / [CO2] [H2]^2 = 0.13 / 0.29 (0.16^2)
Keq = 17.51

4 0

2 years ago