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erica [24]
3 years ago
14

How does the amount of matter change when water changes state

Chemistry
2 answers:
LiRa [457]3 years ago
6 0
The amount of matter can not change because Matter can not be created or destroyed.

OlgaM077 [116]3 years ago
4 0
When its frozen it gets bigger, when it evaporates, it gets smaller. It has something to do with the atoms

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Show me the periodic table
yawa3891 [41]
I think this is what you wanted, so good luck!

8 0
3 years ago
What is the mass in grams of 2.6 moles of angelic acid?
Dennis_Churaev [7]

Answer:

260.34g

Explanation:

First, you need to know what angelic acid is comprised of. It is written as C₅H₈O₂.

In order to solve for the mass of 2.6 moles of angelic acid, you need the mass of 1 mole of angelic acid. This can be found by adding the masses from the periodic table, like shown below:

5 carbon atoms = (5)(12.01g) = 60.05g

8 hydrogen atoms = (8)(1.01) = 8.08g

2 oxygen atoms = (2)(16) = 32g

angelic acid = 60.05 + 8.08 + 32 = 100.13g

Then, set up a basic stoichiometric equation and solve. The units should cancel out.

(\frac{2.6mol}{1} )(\frac{100.13g}{1mol} )=260.34g

5 0
3 years ago
Plz give me short note on isotopes in agriculture industries and medicine
Mumz [18]

Uses of isotopes:

In agriculture,

C-14 is used to trace the path of photosynthesis.

In medicine,

Iodine -131 is used in the treatment of

goiter

8 0
3 years ago
is the atomic mass of chlorine represented by the mass of the most common naturally occurring isotope in chlorine?
IrinaVladis [17]
Combine the number of its Protons and Neutrons and you will have its atomic mass.
8 0
3 years ago
The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2
pentagon [3]

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

3 0
3 years ago
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