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arlik [135]
3 years ago
8

Write 2 typical properties that are only common to transition metals​

Chemistry
2 answers:
butalik [34]3 years ago
8 0

Answer:

high density and oxidation states

Vinil7 [7]3 years ago
6 0

Answer:

Properties of transition elements

they are all metals and that most of them are hard, strong, and lustrous, have high melting and boiling points, and are good conductors of heat and electricity.

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A gas has a volume of 25.0 mL when under a pressure of 525 mmHg. What is the new pressure when the volume has been increased to
Fudgin [204]

Answer:

152.26 mmHg

Explanation:

pv=p'v'

525× 25=p'×86.2

p'=525×25÷ 86.2

p'=152.26

5 0
3 years ago
A solution
timurjin [86]
The answer to your question is  C.  A solution is a homogeneous mixture composed of two or more substances, so it couldn't have been A and D. Since a solution can't have its substances separated by a chemical means because they are chemically bonded, thus they are able to be separated by physical means 
3 0
3 years ago
What is the density of an 84.7 g sample of an unknown substance if the sample occupies 49.6cm cubed
Ahat [919]

Answer:

             1.70 g.cm⁻³

Solution:

Data Given;

                   Mass  =  84.7 g

                   Volume  =  49.6 cm³

                   Density  =  ?

Formula Used;

                   Density  =  Mass ÷ Volume

Putting values,

                   Density  =  84.7 g ÷ 49.6 cm³ 

                   Density  =  1.70 g.cm⁻³

7 0
3 years ago
Write 0.000443200 to two significant figures
Tju [1.3M]
0.00044


Zeros to the right of the decimal place are not significant UNLESS they are found in between or after a non-zero number, therefore, we take the 3200 away because those ARE significant so then after you round your answer (if needed) you're left with only two numbers that are significant.
6 0
3 years ago
What is the pH of pure water at 40.0°C if the Kw at this temperature is 2.92 - 10-147 OA) 6.767 O B) 0 465 O c) 7.000 OD) 7 233
Sergeeva-Olga [200]

Answer:

PH= 6.767     (answer is the A option)

Explanation:

first we need to correct the value in Kw at this temperature is 2.92*10^-14

so, in this case we have that:

Kw=2.92*10^-14 M²

[ H3O^+] [ H3O^+]

[H_{3}O^{+}  ] [OH^{-}  ] = Kw = 2.92*10^{-14} M^{2}   \\\\

at 40ºC

[H_{3}O^{+}  ] = [OH^{-}  ]

[H_{3}O^{+}  ]^{2} = 2.92*10^{-14} M^{2}

[H_{3}O^{+}  ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M

PH= -log10[H_{3}O^{+}  ] = -log10(1.71*10^{-7} ) = 6.767

7 0
3 years ago
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