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3 years ago

How did qualitative chemical and enzyme tests help avery identify dna as the transforming principle?

2 answers:

7 0

Qualitative chemical and enzyme tests helped Avery identify DNA as the transforming principle by conducting a series of tests to find out if the transforming principle was DNA or a protein and the result is that no protein was present but that DNA was present. Hope this answers your question.

kvv77 [185]3 years ago

5 0

Answer:

Qualitative chemical and enzyme tests help Avery to identify DNA as the transforming principle as they clarify the chemical nature of genes, that they are form of nucleic acids.

Explanation:

One of Avery’s qualitative chemical test result was that the transforming principle could precipitate with alcohol, so this demonstrate it was not a carbohydrate.

The enzyme test demonstrated that proteases (enzymes that digest protein) and lipases (enzymes that digest lipids) did not destroy the transforming principle, so the scientist concluded that it was neither a protein nor a lipid.

The bacteriologist found out that the transforming principle was formed of nucleic acids so they tested the sample with ribonuclease (enzyme that digest RNA) but it did not inactivate the substance. And finally he tested it with deoxyribonuclease and it was enzymatically destroyed. So he concluded that the isolated transforming principle was DNA.

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What is the mass in grams of 9.45*10^24 molecules of methanol (CH3OH)?

Angelina_Jolie [31]

Number of moles:

1 mole ---------- 6.02x10²³ molecules
? moles --------- 9.45x10²⁴ molecules

1 x ( 9.45x10²⁴) / 6.02x10²³ =

9.45x10²⁴ / 6.02x10²³ => 15.69 moles of CH3OH

Therefore:

Molar mass CH3OH = 32.04 g/mol

1 mole ------------ 32.04 g
15.69 moles ----- mass methanol

Mass methanol = 15.69 x 32.04 / 1 => 502.7076 g

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For all of the following questions 20.00 mL of 0.200 M HBr is titrated with 0.200 M KOH.

HACTEHA [7]

Answer :

The concentration of $H^+$ before any titrant added to our starting material is 0.200 M.

The pH based on this $H^+$ ion concentration is 0.698

Explanation :

First we have to calculate the concentration of $H^+$ before any titrant is added to our starting material.

As we are given:

Concentration of HBr = 0.200 M

As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion $H^+$ and bromide ion $Br^-$ .

As, 1 M of HBr dissociates to give 1 M of $H^+$

So, 0.200 M of HBr dissociates to give 0.200 M of $H^+$

Thus, the concentration of $H^+$ before any titrant added to our starting material is 0.200 M.

Now we have to calculate the pH based on this $H^+$ ion concentration.

pH : It is defined as the negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH=-\log (0.200)$

$pH=0.698$

Thus, the pH based on this $H^+$ ion concentration is 0.698

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2 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$