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bagirrra123 [75]
3 years ago
12

Classify the following alcohols as primary secondary and tertiary alcohols?​

Chemistry
1 answer:
morpeh [17]3 years ago
4 0
Is this all you need or more ?

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How are strong acids and strong bases similar?
Lady_Fox [76]

Answer:

An acid or base which strongly conducts electricity contains a large number of ions and is called a strong acid or base and an acid or base which conducts electricity only weakly contains only a few ions and is called a weak acid or base.

5 0
4 years ago
What type of bond is this?
satela [25.4K]
Is what? It depends on what your talking about.
4 0
3 years ago
Read 2 more answers
What is the m of a solution where 0.500 moles of a salt are dissolved in 100.0 ml of solution? 25.0m 5.00m 50.0m o.500m 2.50m?
My name is Ann [436]
Molarity=moles/litre
molarity=0.5/0.1
molarity=5.00m
3 0
3 years ago
Question 6 (1 point)
Mnenie [13.5K]

Answer:

The correct option is;

d 4400

Explanation:

The given parameters are;

The mass of the ice = 55 g

The Heat of Fusion = 80 cal/g

The Heat of Vaporization = 540 cal/g

The specific heat capacity of water = 1 cal/g

The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice

The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal

The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change

The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal

The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice

The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal

The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal

However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.

The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal

3 0
3 years ago
The combustion method used to analyze for carbon and hydrogen can be adapted to give percentage N by collecting the nitrogen fro
12345 [234]

Answer:

The percentage of N in the compound is 0.5088

Explanation:

Mass of compound = 8.75 mg = 8.75×1000 = 8750 g

Mass of N2 = number of moles of N2 × MW of N2 = 1.59 × 28 = 44.52 g

% of N in the compound = (mass of N2/mass of compound) × 100 = (44.52/8750) × 100 = 5.088×10^-3 × 100 = 0.5088

6 0
3 years ago
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