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2 years ago

Compare the type of change that occurs when an iron bar rusts and when a substance freezes.

Chemistry

1 answer:

Ivanshal [37]2 years ago

5 0

Explanation:

when an iron bar rust is an example of a chemical change in which a new substance is formed and the change is not easily reversible.for iron to rust moisture and air must be present.while when a substance freezes,it can be easily reversed through melting and no new substance is formed.this change is termed a physical change.

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A sample of three mixed gases is at 632.0 mmhg. if the partial pressure of co2 is 124.3 mmhg and the partial pressure of n2 is 4

goldfiish [28.3K]

According to Dalton's Law of Partial pressures, the sum of the partial pressures of the individuals gas molecules that occupy a specific volume allows us to find the total pressure or find one of the partial pressures if the total pressure is known.

Partial pressure O2 = 632 - 124.3 - 461.9 = 45.8 mm Hg.

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2 years ago

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A piece of gold jewelry weighs 11.54 g and has a volume of 0.675 cm3. The jewelry contains only gold (density = 19.3 g/cm3) and

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Answer:

Percentage of gold = 84.6%

Explanation:

Suppose the jewelry has x g of gold and y g of silver

The mass of both is the total mass of the jewelry then:

x + y = 11.54

And as the problem says the total volume is the sum of both volumes.

volume = mass / density

$\frac{x}{19.3} + \frac{y}{10.5} = 0.675$

Solving the system you get:

x = 9.764 g

y = 1.776 g

The percentage of gold in the jewelry is:

$X_{gold} = \frac{9.764}{11.54} * 100 = 84.6\%$

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2 years ago

(Immediate) Which structural formula represents a nonpolar molecule?

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Answer:

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Explanation:

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A weather balloon at Earth’s surface has a

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The temperature : 263.016 K

<h3>Further explanation</h3>

Combined with Boyle's law and Gay Lussac's law

$\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}$

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V1 = initial gas volume (m³)

P2 = gas end pressure

V2 = the final volume of gas

T1 = initial gas temperature (K)

T2 = gas end temperature

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V1= 4 L

T1 = 275 K

P2=704 mmHg

V1=4.13 L

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