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devlian [24]
3 years ago
11

Which properties can be measured without changing the chemical composition of the matter? _______ properties can be measured wit

hout changing the chemical composition of the matter.
Chemistry
2 answers:
ArbitrLikvidat [17]3 years ago
4 0

Answer:

Physical Properties! :)

Explanation:

evablogger [386]3 years ago
3 0

Answer:

\boxed {\sf Physical}

Explanation:

There are two main kinds of properties: chemical and physical.

Chemical properties, like the name suggests, have to be observed by changing the chemical composition.

That leaves <u>physical properties.</u> They can be measured without any chemical composition changes.

Some examples include: color, odor, mass, density, and volume. All can be measured with just the senses or measuring tools and no composition alterations are needed.

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Answer:

diamond is denser because it is more tightly packed than coal

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At its closest approach to the sun, Pluto is less than 30 AU from the sun. Because Pluto is approaching that point now, which pl
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Neptune: its the 8th planet and since Pluto isn't a planet anymore Neptune is the farthest planet from the sun with a distance of around 2.8 billion miles (4.5 billion km) or (30.07 AU)
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3 years ago
What is the partial pressure (in atm) of CO₂ at 468.2 K in a 25.0 L fuel combustion vessel if it contains 60.0 grams CO₂, 82.1 g
NeTakaya

Answer:

2.09 atm

Explanation:

Step 1: Given and required data

  • Mass of CO₂ (m): 60.0 g
  • Volume of the vessel (V): 25.0 L
  • Temperature (T): 468.2 K

We won't need the data of water and uncombusted fuel, since the partial pressures are independent of each other.

Step 2: Calculate the number of moles (n) corresponding to 60.0 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

60.0 g × 1 mol/44.01 g = 1.36 mol

Step 3: Calculate the partial pressure of CO₂

We will use the ideal gas equation.

P × V = n × R × T

P = n × R × T/ V

P = 1.36 mol × (0.0821 atm.L/mol.K) × 468.2 K/ 25.0 L = 2.09 atm

5 0
3 years ago
Identify each of the following:
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Most metallic: Ge

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7 0
3 years ago
How much heat is transferred when during a reaction, 531 g of water increases from 22.7 ○C to 38.8 ○C?
vazorg [7]

Answer:

8.55 × 10³ cal

Explanation:

Step 1: Given and required data

  • Mass of water (m): 531 g
  • Specific heat of water (c): 1 cal/g.°C
  • Initial temperature: 22.7 °C
  • Final temperature: 38.8 °C

Step 2: Calculate the temperature change (ΔT)

ΔT = Final temperature - Initial temperature = 38.8 °C - 22.7 °C = 16.1 °C

Step 3: Calculate the heat required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 1 cal/g.°C × 531 g × 16.1 °C = 8.55 × 10³ cal

8 0
3 years ago
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