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Phantasy [73]
3 years ago
11

A magnifying glass is an example of a _______ lens.

Chemistry
2 answers:
Svetllana [295]3 years ago
4 0
The correct answer is B. The magnifying lens is a converging lens. It focuses the incoming light to one single point. It can be proved when you use the magnifying glass to burn a piece of paper. You can observe that the light will concentrate at a single point. 
Kryger [21]3 years ago
4 0

Penn Foster Students:  converging

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A gravel can be broken down into different kinds of substances by physical processes. No chemical reactions are needed to separa
butalik [34]

answer:

heterogenous mixture

Explanation:

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3 years ago
Read 2 more answers
Write the correct chemical formula for the following:
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I believe its

A. AlF

B.2S5O

C.HBr


5 0
3 years ago
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
graduated cylinder is filled with water to a volume of 6.2 ML. an irregular shaped plastic object weighing 1.2 g is placed in th
Aleks04 [339]

Answer:

A. Density of object = 0.86 g/mL

B. The object will float in the water.

Explanation:

The following data were obtained from the question:

Volume of water = 6.2 mL

Mass (m) of object = 1.2 g

Volume of water + Object = 7.59 mL

Density of object =?

Density of water = 1 g/mL

Next, we shall determine the volume of the object. This can be obtained as follow:

Volume of water = 6.2 mL

Volume of water + Object = 7.59 mL

Volume of object =?

Volume of object = (Volume of water + Object) – (Volume of water)

Volume of object = 7.59 – 6.2

Volume of object = 1.39 mL

Therefore, the volume of the object is 1.39 mL

A. Determination of the density of the object.

Mass (m) of object = 1.2 g

Volume (V) of object = 1.39 mL

Density (D) of object =?

Density = mass /volume

Density = 1.2/1.39

Density of object = 0.86 g/mL

B. Determination of whether the object will float or sink.

Density of object = 0.86 g/mL

Density of water = 1 g/mL

From the above, we can see that the density of water is greater than that of the object. This implies that the object is lighter than water. Therefore, the object will float in the water.

8 0
3 years ago
A rolling ball will eventually stop rolling. Which describes one reason why this happens
kogti [31]
Friction .......................................................................................
5 0
4 years ago
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