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Ilya [14]
3 years ago
9

What is a vocab word for Friction?

Chemistry
1 answer:
Leona [35]3 years ago
7 0

Answer: Friction is the resistance to motion of one object moving relative to another. It is not a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.

Explanation:

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The answer is solute
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If the alum reaction was done starting with 12.000 g of al, how many moles of hydrogen gas would be produced according to the ba
mylen [45]

0.67 moles of hydrogen gas.

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3 years ago
A substance is considered to have a smaller surface area when...
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Answer: C you have a small amount of substance
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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
4 years ago
2.5 piece of lithium is dropped into a 500 g sample of water at 22.1 C.ten temperature of the water increases to 23.5 C. How muc
Dominik [7]

Answer:

2.6 kJ  

Explanation:

The formula for the amount of heat (q) absorbed by the water is

q = mCΔT

1. Calculate ΔT

ΔT = 23.5 °C - 22.1 °C = 1.4 °C

2. Calculate q

q₂ = mCΔT = 500 g × 4.184 J·°C⁻¹g⁻¹ × 1.4 °C = 2900 J = 2.9 kJ

3 0
4 years ago
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