How many mm are equal to 21 L?

1 L = 1000000 mm

Convert:-

1000000 × 21 = 21000000

21 l = 21000000 mm

How many mg are equal to 9 g?

1 g = 1000 mg

Convert:-

9 × 1000 = 9000

9 g = 9000 mg

8 0

3 years ago

Give two test substances that would react with each other to produce salt and water

Mademuasel [1]

Answer:

NaOH +HCl==>Nacl+H2O

KOH+HCl==>KOH+H2O

6 0

3 years ago

Mutations that neither benefit nor harm the organism have (blank) effect on the organisms survival

Daniel [21]

Answer:

I would say that the mutation has no effect on the organism, as it doesn't help or harm it.

hope this helps :)

Explanation:

3 0

3 years ago

Read 2 more answers

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

3 years ago