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Sati [7]
2 years ago
14

What is similar between seeing an object and seeing a shadows

Chemistry
1 answer:
Naya [18.7K]2 years ago
3 0

The similarity between seen an object and seeing a shadow is their shape.

The shape of shadow of an object is always similar with object.

If object is in standing position, its image also is in standing position. The size of the shadow sometime greater than the object or less than the object. It never same.

If object is in sitting position, its image also is in sitting position.

If the source of light is on the back side, the shadows formed is in our front side and always act opposite to our activity.

For example, if we raise your left hand, the shadow appear to be raise right hand.

If the source of light is on the front side, the shadows formed is in back side and always act similar like us.

For example: if we raise your left hand, the shadow also appear to be raise left hand.

Everything changes at different position of source of light but shape never changes.

Thus, we concluded that similarity between seen an object and seeing a shadow is their shape.

learn more about images:

brainly.com/question/12629638

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Because the chemicals are different
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In just thomsons experiments with electricity he showed that an electrical current can be
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4 years ago
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What was the original pressure in a rigid container if the new pressure is 82.5atm and the temperature was raised from 44C to 67
zhenek [66]

Answer: 54 atm

Explanation:

I did 67/82.5 then got 0.8121212121. I them divided 44 by 0.81212121 and got 54.1791044776

4 0
3 years ago
Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The
love history [14]

Answer:

(a) V_B=11.68L

(b) x_{He}=0.533

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol

Thus, since the final pressure is 3.60 bar, we can write:

P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar

The moles of helium could be computed via solver as:

n_{He}=2.358mol

Or algebraically:

3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol

In such a way, the volume of the compartment B is:

V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\  \\V_B=11.68L

Finally, he mole fraction of He is:

x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533

Regards.

8 0
3 years ago
Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to
valina [46]

5.451 X 10³ kg of sodium carbonate must be added to neutralize 5.04×103 kg of sulfuric acid solution.

<u>Explanation</u>:

  • Sodium carbonate is used to neutralized sulfuric acid, H₂SO₄. Sodium carbonate is the salt of a strong base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:

                        Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃

  • From a balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of  H₂SO₄.
  • Molar mass of Na₂CO₃= 106 g/mol = 0.106 kg/mol and                                Molar mass of H₂SO₄= 98 g/mol = 0.098 kg/mol.
  • To neutralize 0.098 kg of H₂SO₄ amount of Na₂CO₃ required is 0.106 kg, so, To neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ required is =  5.451 X 10³ kg.

 

5 0
3 years ago
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