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Nina [5.8K]
3 years ago
10

What are the characteristics found in transition elements ?​

Chemistry
2 answers:
Dafna1 [17]3 years ago
7 0

Answer:

they are coloured

they have more than one oxidation state

they have high density

have high melting and boiling points.

vampirchik [111]3 years ago
5 0

Answer:

Properties of transition elements include:

have large charge/radius ratio;

are hard and have high densities;

have high melting and boiling points;

form compounds which are often paramagnetic;

show variable oxidation states;

form coloured ions and compounds;

form compounds with profound catalytic activity;

Explanation:

<h2><u><em>hope this helps !</em></u></h2>
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If more solute can be dissolved in a solvent, the solution is
Mekhanik [1.2K]

Answer: B. Unsaturated

Explanation: in a saturated/super saturated solution, more solute will not be able to dissolve.

8 0
3 years ago
A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure
Lesechka [4]

Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure P_{ext}:

P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}

P_{ext} = 1.03 \ atm

The workdone W = P_{ext}V

The change in volume ΔV= \dfrac{W}{P_{ext}}

ΔV = \dfrac{130.1 \ J  \times \dfrac{1 \ L  \ atm}{ 101.325 \ J}  }{1.03 \ atm }

ΔV = \dfrac{1.28398717 }{1.03  }

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial  volume = 61.5 mL

The change in volume V is \Delta V = V_2 -V_1

-  V_2= -  \Delta V  -V_1

multiply through by (-), we have:

V_2=   \Delta V+V_1

V_2 =  1250 mL + 61.5 mL

V_2 = 1311.5 mL

∴ the final volume of the gas is V_2 = 1311.5 mL

5 0
3 years ago
If a nanometer is one billionth of a meter (0.000 000 00 1 m), how many nanometers are there in one meter?
Anit [1.1K]

Answer:

One billion

Explanation:

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6 0
3 years ago
How many grams are 3.01 × 1023 molecules of CuSO4?
zvonat [6]
Answer is: 79.8 grams of copper(II) sulfate.
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n(CuSO₄) = N(CuSO₄) ÷ Na.
n(CuSO₄) = 3.01·10²³ ÷ 6.02·10²³ 1/mol.
n(CuSO₄) = 0.5 mol; amount of substance.
m(CuSO₄) = n(CuSO₄) · M(CuSO₄).
m(CuSO₄) = 0.5 mol · 159.6 g/mol.
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3 years ago
Ionization refers to the process of
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Answer:

  B

Explanation:

3 0
3 years ago
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