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AVprozaik [17]
3 years ago
13

How does a concave lens differ from a convex lens? A) A convex lens is another way to describe a plane mirror; a concave lens is

curved. B) A concave lens us curved inward on both sides; a convex lens is curved on only one side. C) A convex lens has a thick center and thin edges; a concave lens has a thin center and thicker edges. D) A convex lens has a thin center and thick edges; a concave lens has a thick center and thiner edges.
Chemistry
2 answers:
elixir [45]3 years ago
7 0
<h3><u>Answer;</u></h3>

C) A convex lens has a thick center and thin edges; a concave lens has a thin center and thicker edges.

<h3><u>Explanation;</u></h3>
  • Convex lens refers to the lens which merges the light rays at a particular point, that travels through it, while a concave lens can be identified as the lens which disperses the light rays around, that hits the lenses.
  • A convex lens is thicker at the center, as compared to its edges, while a concave lens is thinner at the center as compared to its edges.
  • A concave lens is thicker at the edges than in the middle and spreads light rays apart producing an image smaller than the actual object. A convex lens on the other hand, is thinner at the edges and thicker towards the center, that is they are bent towards a central point.
Tatiana [17]3 years ago
3 0

c. A convex lens has a thick center and thin edges; a concave lens has a thin center and thicker edges.

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earnstyle [38]
Zinc is a metal. At STP, it exists as solid and is stable as it is. It is an important mineral and is used in many applications like in food, metal and drugs. Zinc can be found in the Earth's crust and also it is present in small amounts in some food.
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Murrr4er [49]

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Look at the properties of Oxygen and Silicon - the two most abundant elements in the Earth's crust - by clicking on their symbols on the Periodic Table.

Explanation:

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This is the balances equations. C3H8 + 5O2 → 3CO2 + 4H2O How many moles of oxygen are required to produce 37.15 g CO2
AleksandrR [38]
Molar mass O2 = 31.99 g/mol

Molar mass CO2 = 44.01 g/mol

Moles ratio:

<span>C3H8 + 5 O2 = 3 CO2 + 4 H2O 
</span>
5 x 44.01 g O2 ---------------- 3 x 44.01 g CO2
( mass of O2) ------------------ 37.15 g CO2

mass of O2 = 37.15 x 5 x 44.01/ 3 x 44.01

mass of O2 = 8174.8575 / 132.03

mass of O2 = 61.916 g 

Therefore:

1 mole O2 ----------------- 31.99 g
moles O2 -------------------- 61.916

moles O2 = 61.916 x 1 / 31.99

moles = 61.916 / 31.99 => 1.935 moles of O2
4 0
3 years ago
Help me please I need to pass my exam
Troyanec [42]

Answer:

I think it is 1115 kJ but I don't see the answer

Explanation:

8 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
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