Journal making expressing qualitative data and quantitative data.

Help me please!!!!!!!!

ale4655 [162]

Answer:

that's a hard one

Explanation:

6 0

2 years ago

Read 2 more answers

5. A particular anticodon is 5'-AGC-3'. The codon that is a precise complementary base pair match for this is 5'-AGU-3'. However

Veseljchak [2.6K]

Explanation:

Wobble Hypothesis given by Francis Harry Compton Crick states that 3rd base of mRNA codon can base pair with 1st base of a tRNA anticodon undergoing non-Watson-Crick.

The first 2 bases of the mRNA codon form Hydrogen bonds with the corresponding bases on tRNA anticodon in Watson-Crick manner. Through this, they only form the base pairs with the complimentary bases. However, formation of the Hydrogen bonds between 3rd base on codon and 1st base on anticodon can occur potentially in non-Watson-Crick manner.

Thus, the Wobble Hypothesis explains that why the multiple codons can code for single amino acid.

3 0

3 years ago

Can someone please help me with this?

Sergio039 [100]

Amu valie on the top and abundsnce percent in decimals on the bottom for each isotope

7 0

2 years ago

A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.

seropon [69]

Answer: The p-function of $Zn^{2+}$ and $NO_3^{-}$ ions are 2.51 and 2.14 respectively.

Explanation:

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

Concentration of zinc nitrate = 0.0031 M = 0.0031 mol/L

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, $M_1=(2\times 0.0031)=0.0062M$

Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, $M_2=(2\times 0.0042)=0.0084M$

To calculate the concentration of nitrate ions in the solution, we use the equation:

$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$

Putting values in above equation, we get:

$M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M$

Calculating the p-function of zinc ions and nitrate ions in the solution:

For zinc ions:

$\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}]$

$\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51$

For nitrate ions:

$\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}]$

$\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14$

Hence, the p-function of $Zn^{2+}$ and $NO_3^{-}$ ions are 2.51 and 2.14 respectively.

5 0

3 years ago

The combustion of ammonia in the presence of excess oxygen yields no2 and h2o: 4 nh3 (g) + 7 o2 (g) â 4 no2 (g) + 6 h2o (g) the

VLD [36.1K]

The answer would be 118.68 g.
Explanation for this is:4 moles of NH3 give 4 moles of NO2
so 1mole of NH3 will give 1 mole of NO2
43.9 grams of NH3 contains 2.58 moles
so 2.58 moles will be produced of NO2
which is 118.7 grams this the amount of oxygen that is used.

8 0

3 years ago