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adoni [48]
3 years ago
7

The illuminance on a surface is 6 lux and the surface is 4 meters from the light source. What is the intensity of the source?

Chemistry
2 answers:
Elza [17]3 years ago
7 0

The correct answer is:

D.  96 candelas


Ghella [55]3 years ago
3 0

Answer: D. 96 candelas

Explanation: Illuminance is defined as the quantity of light incident on a surface per unit of area. It depends on the light’s intensity and its distance from the surface.

E=  illuminance on the surface (in lux)= 6 lux

I =  the intensity of the light from the source (in candelas) = ?

S= distance from the light source to the surface (in meters) = 4 m

E=\frac{I}{S^2}

I=E\times S^2

I=6\times (4)^2=96candela

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3 years ago
A 3.00-L flask is filled with gaseous ammonia, NH3. The gas pressure measured at 27.0 ∘C is 2.55 atm . Assuming ideal gas behavi
Whitepunk [10]

Answer : The mass of ammonia present in the flask in three significant figures are, 5.28 grams.

Solution :

Using ideal gas equation,

PV=nRT\\\\PV=\frac{w}{M}\times RT

where,

n = number of moles of gas

w = mass of ammonia gas  = ?

P = pressure of the ammonia gas = 2.55 atm

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V = volume of ammonia gas = 3.00 L

Now put all the given values in the above equation, we get the mass of ammonia gas.

(2.55atm)\times (3.00L)=\frac{w}{17g/mole}\times (0.0821L.atm/mole.K)\times (300K)

w=5.28g

Therefore, the mass of ammonia present in the flask in three significant figures are, 5.28 grams.

8 0
3 years ago
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3 years ago
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In code 250.52(B)(3) it is clearly specified that the bonding grid and reinforcing steel that is related to a pool should not be used as grounding electrodes.

This is essential because when a metal that lies beneath a swimming pool is used as a grounding electrode, current from nearby electrical systems can be introduced into the pool.

This could cause the electrocution of anybody in the swimming pool at that time.

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