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Mandarinka [93]
2 years ago
7

Draw the ring chain of isomer of propene ?​

Chemistry
1 answer:
aksik [14]2 years ago
4 0

Explanation:

Molecular formula for Propene = C3H6

The isomer of propene is cyclopropane.

(Draw a triangle to show that it is cyclopropane)

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Attempt 1
Vsevolod [243]

Answer:

12.01

Explanation:

(12.00*98.93% + 13*1.07%) /100% = 12.01

3 0
3 years ago
PLEASE HELP <br> I do not understand
Vikentia [17]

Answer:

Absorbing beta particle because the beta is the numbers and are less and the  big numbers are positive and they are the alpha so when you add beta particle it is called Absorbing so the answer is Absorbing beta particle

4 0
2 years ago
Read 2 more answers
0.58 mol of Mg contains how many atoms? please show work
Jet001 [13]

The number of atoms present in 0.58 mole of magnesium, Mg is 3.49×10²³ atoms

<h3>Avogadro's hypothesis </h3>

1 mole of Mg = 6.02×10²³ atoms

<h3>How to determine the atoms in 0.58 mole of Mg </h3>

1 mole of Mg = 6.02×10²³ atoms

Therefore,

0.58 mole of Mg = 0.58 × 6.02×10²³

0.58 mole of Mg = 3.49×10²³ atoms

Thus, 3.49×10²³ atoms are present in 0.58 mole of Mg

Learn more about Avogadro's number:

brainly.com/question/26141731

#SPJ1

3 0
2 years ago
Why is density referred to as a characteristic property of matter?
amm1812

Answer : Density of any particular matter remains the same at any concentration at any weight.


Density does not varies with the concentration. The formula for density is mass / volume.


So if the mass changes the density will change accordingly, also if the volume changes the density will get changed.


because, of the above reasons density is referred as a characteristic property of matter.

6 0
3 years ago
A 275 g sample of a metal requires 10.75 kJ to change its temperature from 21.2 oC to its melting temperature, 327.5 oC. What is
Vlada [557]

Answer:

c=0.127\ J/g^{\circ} C

Explanation:

Given that,

Mass of the sample, m = 275 g

It required 10.75 kJ of heat to change its temperature from 21.2 °C to its melting temperature, 327.5 °C.

We need to find the specific heat of the metal. The heat required by a metal sample is given by :

Q=mc\Delta T

c is specific heat of the metal

c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{10.75\times 10^3\ J}{275\times (327.5 -21.2)}\\\\=0.127\ J/g^{\circ} C

So, the specific heat of metal is 0.127\ J/g^{\circ} C.

4 0
2 years ago
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