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nekit [7.7K]
2 years ago
3

A positively charged rod is brought close to one end of a neutral metallic plate as seen below. What type of charge is induced o

n the right side of the plate?
A. Positive
B. Negative
C. Neutral
D. Mixed charges
Chemistry
1 answer:
katrin2010 [14]2 years ago
4 0
<h3><u>Answer;</u></h3>

A. Positive

<h3><u>Explanation;</u></h3>
  • When a rod with positive charge is placed near a neutral piece of metal, positive charges in the metal are repelled by the positive charge on the rod.
  • The negative electrons move towards the positively charged rod.
  • The metal now has more negative charge at the end near the rod and more positive charge at the end away from the rod.
  • The positively charged rod attracts the nearer negative charge more strongly than it repels the farther positive charge and an attractive force results.
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How many moles of lithium chloride will be formed by the reaction of chlorine with 0.046 mol of lithium bromide in the following
Feliz [49]

Answer:

The answer is 0.046 mol.

Explanation:

By looking at the balanced equation, you can form a ratio of lithium chloride and lithium bromide using the coefficient value :

ratio of lithium bromide <u>(</u>2LiBr)

= 2

ratio of lithium chloride (2LiCl)

= 2

So the ratio is 2 : 2 then simplify into 1 : 1 . Which means that 1 mol of lithium bromide is equal to 1 mole of lithium chloride.

In this case, 0.046 mol of lithium bromide will form <u>0</u><u>.</u><u>0</u><u>4</u><u>6</u><u> </u><u>m</u><u>o</u><u>l</u> of lithium chloride.

8 0
2 years 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 <u>5.44</u>

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

<em><u>The acid dissociation constant </u></em>(K_{a})<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>

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}

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

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

Initial:                     0.1 M                    x         x            

Change:                   -x                      +x       +x

Equilibrium:           0.1 - x                    x         x

<u>The acid dissociation constant: </u>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}

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

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

<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>

5 0
2 years ago
The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s
denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0
2 years ago
A student adds solid KCl to water in a flask. The flask is sealed with a stopper and thoroughly shaken until no more solid KCl d
AlladinOne [14]

Answer:

Option (A) saturated and is at equilibrium with the solid KCl

Explanation:

A saturated solution is a solution which can not dissolve more solute in the solution.

From the question given above, we can see that the solution is saturated as it can not further dissolve any more KCl as some KCl is still visible in the flask.

Equilibrium is attained in a chemical reaction when there is no observable change in the reaction system with time. Now, observing the question given we can see that there is no change in flask as some KCl is still visible even after thorough shaking. This simply implies that the solution is in equilibrium with the KCl solid as no further dissolution occurs.

4 0
2 years ago
Ka, the acid dissociation constant, for an acid is 9 × 10^{−4} at room temperature. At this temperature, what is the approximate
devlian [24]
Dissociation=k×no of moles
percentage of dissociation=9.0×10^-4×1×100
knowing that x%=x/100,we then say;
x/100=9.0×10^-4×1×100
therefore, x=100×100×9×10^-4×1
x=9
x percentage of dissociation=9%
8 0
2 years ago
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