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2 years ago

Which of the following reactions shows a chemical chand

Chemistry

2 answers:

dexar [7]2 years ago

8 0

There is no picture or any thing that could help me help you

sladkih [1.3K]2 years ago

6 0

Answer:

where is it?

Explanation:

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Is carbon tetrafluoride a polar molecule?

atroni [7]

It is non-polar molecule.
CF₄ - dipole moment = 0D

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3 years ago

Iodine 131, cesium 137, and strontium 90 are all considered isotopes. What are isotopes?

Nitella [24]

Answer:

Isotopes are variants of an element, where the number of neutrons is different but its number of protons stay the same.

Explanation:

If the number of protons had changed then it would just be a different element.

Electron number changing simply adds charge to the atom, and doesn't change it or anything.

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2 years ago

Read 2 more answers

Energy is released when the nucleus of an atom splits and two smaller atoms are formed. What is the name of this process?

Svetllana [295]

Explanation:

It's (D), nuclear fission................

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2 years ago

What is the pH of a 0.300 M NH₃ solution that has Kb = 1.8 × 10⁻⁵ ? The equation for the dissociation of NH₃ is: NH₃ (aq) + H₂O

nalin [4]

Answer:

11.4

Explanation:

Step 1: Given data

Concentration of the base (Cb): 0.300 M

Basic dissociation constant (Kb): 1.8 × 10⁻⁵

Step 2: Write the dissociation equation

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

Step 3: Calculate the concentration of OH⁻

We will use the following expression.

$[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8 \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M$

Step 4: Calculate the pOH

We will use the following expression.

$pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6$

Step 5: Calculate the pH

We will use the following expression.

$pH+pOH=14\\pH = 14-pOH = 14-2.6 = 11.4$

8 0

3 years ago

A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0

bija089 [108]

Answer:

E = 0.062 V

Explanation:

(a) See the attached file for the answer

(b)

Calculating the voltage (E) using the formula;

E = - (2.303RT/nf)log Cathode/Anode

Where,

R = 8.314 J/K/mol

T = 35°C = 308 K

F- Faraday's constant = 96500 C/mol,

n = number of moles of electron = 2

Substituting, we have

E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)

= -0.031 * -2

= 0.062V

Therefore, the voltmeter will show a voltage of 0.062 V

5 0

2 years ago