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Morgarella [4.7K]
2 years ago
7

Compare the properties of elements in groups 17 and group 18. Make sure to cover the following points:

Chemistry
2 answers:
Andre45 [30]2 years ago
8 0

Answer:

<em>Letter:</em><em> </em><em>C</em>

explanation:

:)

sdas [7]2 years ago
3 0

Answer:

a) Explain why:

(i) All the elements of a group have similar chemical properties.

(ii) All the elements of a period have different chemical properties.

(b) The atomic radii of three elements X, Y and Z of a period of the periodic table are 186 pm, 104 pm and 143 pm respectively. Giving a reason arrange these elements in the increasing order of atomic number in the period.

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Which elements from the list would be likely to form an ionic bond with chlorine, Cl?
monitta

Answer:

metal calcium or (Ca).

Explanation:

For example, the metal calcium (Ca) and the nonmetal chlorine (Cl) form the ionic compound calcium chloride (CaCl2). In this compound, there are two negative chloride ions for each positive calcium ion

8 0
2 years ago
12. Identify the following as a chemical or physical change:
Reika [66]
I think the answer isD,A
3 0
3 years ago
Explain the environmental benefits of recyling paper products.
jenyasd209 [6]

Answer:

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Explanation:

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5 0
3 years ago
The potential difference across the terminals of a battery of e.m.f. 12 V and internal resistance 2 ohm drops to 10 V when it is
lesantik [10]

Explanation:

The given data is as follows.

      E.m.f = 12 V,        Voltage = 10 V,        Resistance = 2 ohm

Hence, calculate the current as follows.

          I = \frac{E - V_{t}}{r_{int}}

Putting the given values into the above formula as follows.

         I = \frac{E - V_{t}}{r_{int}}

           = \frac{12 - 10}{2}

           = 1 A

Atomic weight of copper is 63.54 g/mol. Therefore, equivalent weight of copper is \frac{M}{2}.

That is,          \frac{M}{2}

                = \frac{63.54 g/mol}{2}

Hence, electrochemical equivalent of copper is as follows.

                   Z = (\frac{E}{96500}) g/C

                       = (\frac{63.54 g/mol}{2 \times 96500}) g/C

                       = 3.29 \times 10^{-4} g/C

Therefore, charge delivered from the battery in half-hour is calculated as follows.

                     It = Q

                       = 1 \times \frac{1}{2} \times 60 times 60  

                       = 1800 C

So, copper deposited at the cathode in half-an-hour is as follows.

                     M = ZQ

                = 3.29 \times 10^{-4} g/C \times 1800 C

                = 0.5927 g

Thus, we can conclude that 0.5927 g of copper is deposited at the cathode in half an hour.

8 0
3 years ago
A beaker contains a sample of water with a mass of 325 grams. The water in the beaker
alexira [117]

Answer: a

Explanation:

Bc I just did it and yeah

5 0
3 years ago
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