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zhuklara [117]
3 years ago
6

Elements in which the d-sublevel is being filled have the properties of

Chemistry
2 answers:
andre [41]3 years ago
7 0

Elements in which the d-sublevel is being filled have the properties of;

<em><u>transition metals. </u></em>

In quantum theory the sublevel is an energy level. Now, the d-sublevel in atomic configuration has 5 orbitals and as such can contain a maximum of 10 electrons.

Now, this d-sublevel is used primarily by elements called "transition elements" as they are also called d-block elements.

However, these transition elements are also called transition metals because all the transition elements possess properties of metals generally.

In conclusion, elements in which the d-sublevel is being filled have the properties of transition metals.

Read more at; brainly.com/question/3948476

Nadya [2.5K]3 years ago
4 0

Answer:

Elements in which the d-sublevel is being filled have the properties of metals

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If the aluminum block is initially at 25 ∘C∘C, what is the final temperature of the block after the evaporation of the alcohol?
Effectus [21]

Answer:

Final temperature of aluminum block = 12.1°C

<em>Note: The question is not complete. The complete question is given below:</em>

<em>If the aluminum block is initially at 25 ∘C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 ∘C. Heat of vaporization of the alcohol at 25 ∘C is 45.4 kJ/mol Suppose that 1.12 g of rubbing alcohol (C3H8O) evaporates from a 73.0 g aluminum block.</em>

Explanation:

Heat lost by aluminum block = heat required for vaporization of alcohol

Heat required to vaporize ethanol, H = mass of alcohol * heat of vaporization of alcohol

Mass of alcohol = 1.12 g; molar mass of rubbing alcohol = 60 g/mol

Heat of vaporization = (45.4 kJ/mol)/ 60 g/mol =  0.75666 kJ/g

H = 1.12 g × 0.7566 kJ/g

H = 0.8474 kJ = 847.4 J

Heat lost by aluminum block, Q = -(mass × specific heat capacity × temperature change)

Mass of aluminum block = 73.0 g

Specific heat capacity of aluminum = 0.900 J/g

Temperature change = (Final temperature, T - 25)

Q = -73.0 g × 0.900 J/g × (T - 25)

Q = -65.7 J × (T - 25°C)

Since, Heat lost by aluminum block = heat for vaporization of alcohol

-65.7 J × (T - 25°C) = 847.4 J

T - 25 = 847.4/-65.7

T - 25 = -12.9

T = -12.9 + 25

T = 12.1°C

5 0
3 years ago
How many moles of NaCl are contained in 100.0 mL of a 0.20 M solution?​
AURORKA [14]

Answer:

0.02 mol.

Explanation:

8 0
3 years ago
What is the relationship between concentration gradient and diffusion?
Juli2301 [7.4K]
A concentration gradient happens when there is a change of the concentration in solution for period of time. Diffusion is the movement of molecules or ions. The relationship between the two of them is molecules diffuse down from an area with a higher concentration to a lower one. An example of this is putting dye in water.
4 0
4 years ago
Given that Kp = 3.5 x 10-4 for the reaction 2 CO(g) &lt;=&gt; C(graphite) + CO2(g), what is the partial pressure of CO2(g) at eq
spin [16.1K]

<u>Answer:</u> The partial pressure of carbon dioxide at equilibrium is 0.0056 atm

<u>Explanation:</u>

The given chemical equation follows:

                     2CO(g)\rightleftharpoons C\text{ (graphite)}+CO_2(g)

<u>Initial:</u>             4.00

<u>At eqllm:</u>       4.00-2x        x                 x

The expression of K_p for above reaction follows:

K_p=\frac{p_{CO_2}}{(p_{CO})^2}

The partial pressure of pure solids and liquids are taken as 1 in the equilibrium constant expression.

We are given:

K_p=3.5\times 10^{-4}

Putting values in above expression, we get:

3.5\times 10^{-4}=\frac{x}{(4-2x)^2}\\\\x=0.0056,718.28

Neglecting the value of x = 718.28 because equilibrium pressure cannot be greater than initial pressure

Partial pressure of CO_2 = 0.0056 atm

Hence, the partial pressure of carbon dioxide at equilibrium is 0.0056 atm

4 0
3 years ago
Write a net ionic equation for the reaction that occurs when excess hydroiodic acid and nickel(II) sulfide are combined.
tiny-mole [99]

Explanation:

Step 1: Write the balanced "molecular" equation:

2HI(aq) + NIS (s) → NiI₂(aq) + H₂S(g)

Step 2: H₂S is written in molecular form. Consult the solubility and net ionic equation rules on the information page to determine which of the other substances will dissociate:

2HI(aq) + NIS (s) → NiI₂(aq) + H₂S(g)

yes no yes

Step 3: Dissociate all soluble salts, strong acids, and strong bases (except calcium hydroxide). Leave together all "not soluble" salts and weak acids or bases:

{2H₃O}^{ + }  (aq) +  {2I}^{ - }  (aq) + NiS(s)→  { Ni}^{2 + }(aq) +  {2I}^{ - }(aq) + H₂S(g) + 2H₂O(l)

Step 4: Cross out "spectator ions" that appear on both sides of the reaction (these ions do not participate in the chemistry) and rewrite the "net" reaction using the smallest possible coefficients:

{2H₃O}^{ + } (aq) + NiS(s) →   { Ni}^{2 + } (aq) + H₂S(g) + 2H₂O(l)

4 0
3 years ago
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