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zysi [14]
4 years ago
8

Will chlorine gas effuse more quickly than ammonia gas

Chemistry
2 answers:
beks73 [17]4 years ago
6 0
This statement is false due to the fact that the ammonia gas has the lower molar mass.
SVEN [57.7K]4 years ago
6 0
No ammonia has lower molar mass . So this is not true
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A 8 kg cat is running 4 m/s. How much kinetic energy does it have
Debora [2.8K]
<span>KE = 1/2mv^2
KE = 1/2(8)4 m/s^2
KE = 4*4
KE = 16 Joules

Kinetic energy would equal 16 J 
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3 years ago
diamond has a density of 3.26 g/cm3. what is the mass of a diamond that has a volume of 0.313 cm cubed
DaniilM [7]
D = m / V

3.26 = m / 0.313

m = 3.26 x 0.313

m = 1.02038 g

hope this helps!
6 0
3 years ago
How would you arrange the objects below from greatest to least volume?
ikadub [295]

Answer:

A balloon

A bottle of soda

A rock

A water bottle

Explanation:

4 0
3 years ago
Read 2 more answers
The longest day, and shortest night, of the year occurs on the
BaLLatris [955]

Answer:

This is the first answer,

winter solstice

The winter solstice occurs during the hemisphere's winter. In the Northern Hemisphere, this is the December solstice (usually December 21 or 22) and in the Southern Hemisphere, this is the June solstice (usually June 20 or 21).

and this is the second one,

the tilt of the Earth on its axis

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5 0
3 years ago
Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t
Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

A=\log \frac{I_o}{I}

\log \frac{I_o}{I}=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution = 3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}

l = path length = 2.5 mm = 0.25 cm

I_o = incident light

I = transmitted light

\epsilon = molar absorptivity coefficient = 855dm^3mol^{-1}cm^{-1}

Now put all the given values in the above formula, we get:

\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)

\log \frac{I_o}{I}=0.6947

\frac{I_o}{I}=10^{0.6947}=4.951

If we consider I_o = 100

then, I=\frac{100}{4.951}=20.198

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light I_A = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100

\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%

Therefore, the percentage reduction in intensity is 79.80 %

3 0
4 years ago
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