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olga_2 [115]
3 years ago
5

Does the antifreeze you put in a car radiator have a lower or higher freezing point than water

Chemistry
1 answer:
saveliy_v [14]3 years ago
7 0
Hello!
Antifreeze says it in the name. This means that it helps prevent freezing. The liquid that would freeze if you wouldnt use the antifreeze is water. This means that the antifreeze has to have a higher freezing point then water, because otherwise it would only freeze faster!

Hope this helped you!
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Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter? 
NARA [144]

Answer:

ΔH = q_{p}

Explanation:

In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.

The heat transfer is represented by

q_{com} = q_{p}

where

q_{p} = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.

q_{com}  = the heat of combustion

Also, we know that the total heat change of the any system is

ΔH = ΔQ + ΔW

where

ΔH = the total heat absorbed by the system

ΔQ = the internal heat absorbed by the system which in this case is q_{p}

ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0

substituting into the heat change equation

ΔH = q_{p} + 0

==> ΔH = q_{p}

5 0
4 years ago
If 6.0g of carbon is heated in air the mass of the product obtained could be either 22.0g or 14.0g depending on the amount of ai
Gnom [1K]

In accordance with Dalton's Law of multiple proportions

<h3>Further explanation</h3>

Given

6.0g of carbon

22.0g or 14.0g of product

Required

related laws

Solution

the amount of air present ⇒ as an excess or limiting reactant

  • air(O₂) as a limiting reactant(product=14 g)

C+0.5O₂⇒CO

6 + 8 = 14 g

mol O₂=8 g : 32 g/mol=0.25

mol C = 6 g : 12 g/mol = 0.5(2 x mol O₂)

mol CO= 2 x mol O₂ = 0.5 mol = 0.5 x 28 g/mol = 14 g

  • air(O₂) as an excess reactant(product=22 g) an C as a limiting reactant

C+O₂⇒CO₂

6 + 16 = 22 g

mol C = 6 g : 12 g/mol = 0.5

mol O₂ = 16 g : 32 g/mol=0.5

mol CO₂ = 22 g : 44 g/mol = 0.5

if the mass firs element (C) constant, then the mass of the second element(O) in the two compounds will have a ratio as a simple integer.

CO = 6 : 8

CO₂ = 6 : 16

the ratio O = 8 : 16 = 1 : 2

In accordance with Dalton's Law of multiple proportions

4 0
3 years ago
Calculate the fraction of atoms in a sample of argon gas at 400 K that have an energy of 12.5 kJ or greater. Report your answer
emmainna [20.7K]

Answer:

0.023

Explanation:

The Arrhenius' equation states that:

k = A*e^{\frac{-Ea}{RT}

Where k is the velocity constant of the reaction, A is the constant of the collisions, Ea is the activation energy (the energy necessary to the molecules have so the reaction will happen), R is the gas constant (8.314 J/molK) and T is the temperature.

This equation is derivated of:

k = pZf

Where

p=fraction of collisions that occur with reactant molecules properly oriented

f=fraction of collisions having energy greater than the activation energy

Z=frequency of collisions

Thus, p*Z = A, and

f = e^{\frac{-Ea}{RT} }

So, if the energy of the molecules is 12.5 kJ/mol = 12500 J/mol, thus the fraction will be:

f = e^{\frac{-12500}{8.314*400} }

f = 0.023

6 0
3 years ago
Which of the following can be used to neutralize a sodium hydroxide (NaOH) solution? NaCl NH3 HCl Ca(OH)2
Andru [333]
NaOH is a base and hence it can only be neutralized by an acid.

So it is by HCl, which is an acid.
4 0
3 years ago
Read 2 more answers
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user100 [1]

Tides are the rise and fall of the oceans. They are caused by the gravity, or pull, of the Moon and Sun. The Moon's gravity is the main force in causing tides. It makes the oceans bulge out toward it. Another bulge occurs on the opposite side, because Earth is being pulled toward the Moon and away from the water. The water on the side farthest away from the Moon is least affected by its gravity.

3 0
3 years ago
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