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2 years ago

List 3 examples of physical change and 3 examples of chemical change.?

1 answer:

3 0

Examples of chemical change : 1.food scraps 2. gasoline 3. rust.
examples of physical change : 1. juice box. 2. water. 3. rubbing alcohol.

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Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation:

12345 [234]

Answer:

Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J

Explanation:

Given reaction: 2S + 3O₂ → 2 SO₃

Given: The enthalpy of reaction: ΔH = - 792 kJ

Given mass of S: w₂ = 42.8 g, Molar mass of S: m = 32 g/mol

In the given reaction, the number of moles of S reacting: n = 2

As, Number of moles: $n = \frac{mass\: (w_{1})}{molar\: mass\: (m)}$

∴ mass of S in 2 moles of S: $w_{1} = n \times m = 2\: mol \times 32\: g/mol = 64\: g$

Given reaction: 2S + 3O₂ → 2 SO₃

In this reaction, the limiting reagent is S

⇒ 2 moles S produces (- 792 kJ) heat.

or, 64 g of S produces (- 792 kJ) heat.

∴ 42.8 g of S produces (x) amount of heat

⇒ The amount of heat produced by 42.8 g S:

$x = \frac{(- 792\: kJ) \times 42.8\: g}{64\: g} = (-529.65)\: kJ$

$\Rightarrow x = (-5.2965 \times 10^{2})\: kJ = (-5.2965 \times 10^{5})\: J$

$(\because 1 kJ = 10^{3} J)$

Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J

8 0

3 years ago

How many liters are equivalent to 645 mililiters

abruzzese [7]

Answer:

0.645 liters

Explanation:

THE QUESTION IS equivalent 0.645 Liters

6 0

3 years ago

Read 2 more answers

Electric current is the flow of particles with ________. a. Negative charge b. Positive charge c. Both positive and negative cha

Lelechka [254]

I believe Positive & Negative

4 0

2 years ago

Read 2 more answers

The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat

guajiro [1.7K]

First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

= 1 mol * 18 mol/g

= 18 g

and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

= 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

= 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

= 4514 J = 4.514 KJ

∴the total change of enthalpy = q1+q2+q3

= 1.881 KJ +6.01 KJ + 4.514 KJ

= 12.405 KJ

5 0

3 years ago

The solubility of oxygen gas in water at 40 ∘c is 1.0 mmol/l of solution. What is this concentration in units of mole fraction?

juin [17]

The formula for mole fraction is:

$mole fraction of solute = \frac{number of moles of solute}{total number of moles of solution}$ -(1)