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Harlamova29_29 [7]
3 years ago
10

What is a prediction?

Chemistry
2 answers:
Arisa [49]3 years ago
4 0
Something foretold, for example: Jacob's prediction proved true.<span>
</span>
aalyn [17]3 years ago
3 0
A prediction is a guess of something that could happen in the future

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How many moles are in 4.85 x 10^25 atoms of carbon?
Serjik [45]

Answer:

4.85 x 10^25

Explanation:

thats what i was told by a calculator

6 0
3 years ago
Pure chlorobenzene (C6H5Cl) has a normal boiling point of 131.00 °C. A solution of 32.5 g of 2,8-dibromodibenzofuran (C12H6Br2O)
vichka [17]

Answer:

Kb →  1.56 °C / m

Explanation:

This is all about boiling point elevation, the colligative property that shows that boiling point for a solution is higher than boiling point of pure solvent.

This is the formula: ΔT = Kb . m . i

where i is the Van't Hoff factor (ions dissolved in solution). As these are organic compounds, we assume they are non electrolytic,

m is molality (mol of solute / 1kg of solvent)

Kb is our unknown. The value for ebulloscopic constant, it is specific for each solvent.

ΔT = T° boiling from solution - T° boiling from solute

First of all, let's determine the moles of solute.

Mass / Molar mass → 32.5 g/ 113.45 g/mol = 0.286 mol

Molality is mol of solute/ 1 kg of solvent

We must convert the mass from g to kg

195g . 1kg /1000 = 0.195 kg

Molality = 0.286 mol / 0.195 kg = 1.47 m

Let's replace the values in the formula

133.30 °C - 131°C = Kb . 1.47m .1

2.30°C / 1.47 m =  Kb →  1.56 °C / m

3 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION!!
Ksenya-84 [330]
2 I think not 100% tho
5 0
3 years ago
6. A piece of solid gold was heated from 274K to 314K. 35.73 of energy was needed to raise the temperature.
ivanzaharov [21]

From Q = mcΔT, we can rearrange the equation to solve for mass, m = Q/cΔT. The specific heat capacity, c, of solid gold is 0.129 J/g °C. I'm assuming that the energy is given in joules, as it's not specified in the question as written.

m = Q/cΔT = (35.73 J)/(0.129 J/g °C)(40.85 °C - 0.85°C)

m = 6.92 g of gold was present  

5 0
3 years ago
Silver (Ag) has two stable isotopes: 107Ag, 106.90 amu, and 109Ag, 108.90 amu. If the average atomic mass of silver is 107.87 am
inysia [295]

Answer:

  • The abundance of 107Ag is 51.5%.
  • The abundance of 109Ag is 48.5%.

Explanation:

The <em>average atomic mass</em> of silver can be expressed as:

107.87 = 106.90 * A1 + 108.90 * A2

Where A1 is the abundance of 107Ag and A2 of 109Ag.

Assuming those two isotopes are the only one stables, we can use the equation:

A1 + A2 = 1.0

So now we have a system of two equations with two unknowns, and what's left is algebra.

First we<u> use the second equation to express A1 in terms of A2</u>:

A1 = 1.0 - A2

We <u>replace A1 in the first equation</u>:

107.87 = 106.90 * A1 + 108.90 * A2

107.87 = 106.90 * (1.0-A2) + 108.90 * A2

107.87 = 106.90 - 106.90*A2 + 108.90*A2

107.87 = 106.90 + 2*A2

2*A2 = 0.97

A2 = 0.485

So the abundance of 109Ag is (0.485*100%) 48.5%.

We <u>use the value of A2 to calculate A1 in the second equation</u>:

A1 + A2 = 1.0

A1 + 0.485 = 1.0

A1 = 0.515

So the abundance of 107Ag is 51.5%.

3 0
3 years ago
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