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Vinil7 [7]
2 years ago
5

As the thermal energy of matter increases, its particles usually spread out, causing

Chemistry
1 answer:
sveta [45]2 years ago
8 0

Answer:

As the thermal energy of matter increases, its particles usually spread out, causing the substance to expand.

Explanation:

<em> i have a book about this stuff</em>

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Why is gold considered matter?
Yakvenalex [24]

Answer:

All matter is made up of substances called elements, which have specific chemical and physical properties and can't be broken down into other substances through ordinary chemical reactions. Gold, in this case, is an element so, therefore, it is considered matter.

Hope it helped :)

3 0
3 years ago
In a reaction of a potential new fuel, it is found that when 2.81 moles of the fuel combusts, 1,612 kJ of energy is released. Wh
Tema [17]

Answer:

-573.67

Explanation:

whenever energy is released in a chemical reaction, we would then expect the delta H of the reaction to be negative because the reaction is an exothermic reaction.

now we have that 2.81 moles of fuel when it combusts would releases 1612kJ of energy

thus, 1 mole will release 1612/2.81 = -573.67kJ of heat

Therefore the delta H of the reaction = -573.67 kJ/mol

3 0
2 years ago
Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have
Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

7 0
3 years ago
What is the final pressure of a system (atm) that has the volume increased from 0.75 l to 1.1 l with an initial pressure of 1.25
lana66690 [7]
Boyle's law states that pressure is inversely proportional to volume of gas at constant temperature 
PV = k
where P - pressure , V - volume and k - constant 
P1V1 = P2V2
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 
substituting these values in the equation 
1.25 atm x 0.75 L = P x 1.1 L
P = 0.85 atm 
final pressure is B) 0.85 atm 
3 0
3 years ago
Calculate the mass percent composition of sulfur in Al2(SO4)3.
Anastaziya [24]
The molar mass of aluminum sulftae is 342.14 g/mol.

Since the subscript shows that there are 3 sulfurs within the substance, the total mass of sulfur is 96.21g/mol

Now take the mass of the sulfur and divide it by the molar mass of aluminum sulfate, then multiply by 100:
(96.21/342.15)(100) = 28.1% mass composition of sulfate
6 0
3 years ago
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