____N2 + ____H2  ___

DNA bases are attached to _____.

melomori [17]

Answer:

both sugar and phosphate molecules

5 0

3 years ago

Read 2 more answers

Lithium has two stable isotopes with masses of 6.01512 amu and 7.01600 amu. The average molar mass of Li is 6.941 amu. What is t

Dvinal [7]

Answer : The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % respectively.

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'

For Li isotope-1 :

Mass of Li isotope-1 = 6.01512 amu

Fractional abundance of Li isotope-1 = x

For Li isotope-2 :

Mass of Li isotope-2 = 7.01600 amu

Fractional abundance of Li isotope-2 = 100-x

Average atomic mass of Li = 6.941 amu

Putting values in equation 1, we get:

$6.941=[(6.01512\times x)+(7.01600\times (100-x))]$

By solving the term 'x', we get:

$x=694.048$

Percent abundance of Li isotope-1 = $\frac{694.048}{100}=6.94\%$

Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %

6 0

3 years ago

The mole fraction of CO2 in a certain solution with H2O as the solvent is 3.6 × 10−4. What is the approximate molality of CO2 in

nataly862011 [7]

Answer: C) 0.020 m

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molality=\frac{n\times 1000}{W_s}$

where,

n = moles of solute

$W_s$ = weight of solvent in g

Mole fraction of $CO_2$ is = $3.6\times 10^{-4}$ i.e. $3.6\times 10^{-4}$ moles of $CO_2$ is present in 1 mole of solution.

Moles of solute $(CO_2)$ = $3.6\times 10^{-4}$

moles of solvent (water) = 1 - $3.6\times 10^{-4}$ = 0.99

weight of solvent = $moles\times {\text {Molar mass}}=0.99\times 18=17.82g$

Molality = $\frac{3.6\times 10^{-4}\times 1000}{17.82g}=0.020$

Thus approximate molality of $CO_2$ in this solution is 0.020 m

5 0

3 years ago

In 1911, Ernest Rutherford tested the atomic model existing at the time by shooting a beam of alpha particles (42He, helium nucl

STatiana [176]

Answer:

At the time of Rutherford's experiment, the accepted model for the atom was the Thomson plum-pudding model of the atom, in which the atom consists of a "sphere" of positive charge distributed all over the sphere, with tiny negative particles (the electrons) inside this sphere.

In his experiment, Rutherford shot alpha particles towards a very thin sheet of gold foil. He observed the following things:

1- Most of the alpha particles went undeflected, but

2- Some of them were scattered at very large angles

3- A few of them were even reflected back to their original directions

Observations 2) and 3) were incompatible with Thomson model of the atom: in fact, if this model was true, all the alpha particle should have gone undeflected, or scattered at very small angles. Instead, due to observations 2) and 3), it was clear that:

- The positive charge of the atom was all concentred in a tiny nucleus

- Most of the mass of the atom was also concentrated in the nucleus

So, Rutherford experiment lead to a change in the atomic model of the atom, as it was clear that the plum-pudding model was no longer adequate to describe the results of Rutherford's experiment.

5 0

3 years ago

What is the volume of 1g of ice in cm3?

Amanda [17]

Answer:

What is the volume of 1g of ice in cm3?

Explanation:

3 0

3 years ago

____N2 + ____H2  ____NH3

N2 + H2  NH3