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3 years ago

Helium on the Moon was found to be 0.420% 2He, 2.75% 3He, and 96.83% 4He. What is the average atomic mass of helium on the Moon?

Chemistry

1 answer:

Tasya [4]3 years ago

6 0

Answer:

Average atomic mass = 3.9 amu

Explanation:

Given data:

Percent abundance of He-2 = 0.420%

Percent abundance of He-3 = 2.75%

Percent abundance of He-4 = 96.83%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100

Average atomic mass = (0.420×2)+(2.75×3) +(96.83×4)/100

Average atomic mass = 0.84 + 8.25 +387.32 / 100

Average atomic mass = 396.41 / 100

Average atomic mass = 3.9 amu.

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Which statement justifies that hydrogen peroxide (H2O2) is a polar molecule?

Rudik [331]

To determine whether a compound is polar or nonpolar you have to take into account:

1) formation of dipoles due to the difference in electronegativities of the atoms

2) shape of the molecule to conclude whether there is a net dipole momentum.

You already, likely, know that the electronegativities of H and O are significatively different, being O more electronegative thatn H. So, you can conclude easilty that the electrons are atracted more by O than by H, thus creating two dipoles H→O

Regarding the shape, it may appear that the molecule is symmetrical, which would lead to the cancellation of the two dipoles. But that is not the true. The H2O2 is not symmetrical.

The lewis structure just show this shape

** **
H - O - O - H
** **

which is what may induce to think that the molecule is symmetrical, leading to the misconception that it is nonpolar.

But in a three dimensional arrangement you could see that the hydrogens are placed in non symmetrical positions, which leads to the formation of a net dipole momentum, and thus to a polar molecule.

The fact that H2O2 is a polar compound is the reason why it can be mixed with water and the H2O2 that you buy in the pharmacy is normally a solution in water.

So, the hydrogen peroxide is polar because the hydrogens are not placed symmetrically in the molecule, which result in a net dipole momentum.

4 0

4 years ago

What mass of nitrogen is needed to fill an 855 L tank at STP?

scoundrel [369]

Answer:

It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C or 273.15 °K are used and are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

So, in this case:

P= 1 atm
V= 855 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 273.15 K

Replacing:

1 atm* 855 L= n* 0.082 $\frac{atm*L}{mol*K}$ * 273.15 K

Solving:

$n=\frac{1 atm* 855 L}{0.082\frac{atm*L}{mol*K} *273.15 K }$

n= 38.17 moles

Being the molar mass of nitrogen N2 equal to 28 g / mol, you can apply the following rule of three: if there are 28 grams in 1 mole, how much mass is there in 38.17 moles?

$mass=\frac{38.17 moles*28 grams}{1 mole}$

mass= 1,068.76 grams

<u><em> It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.</em></u>

3 0

4 years ago

"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volum

UkoKoshka [18]

Explanation:

Formula to calculate osmotic pressure is as follows.

Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = $25^{o} C$

= (25 + 273) K

= 298.15 K

Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

0.698 = $C \times 0.082 \times 298.15 K
$

C = 0.0285

This also means that,

$\frac{\text{moles}}{\text{volume (in L)}}$ = 0.0285

So, moles = 0.0285 × volume (in L)

= 0.0285 × 0.100

= $2.85 \times 10^{-3
}$

Now, let us assume that mass of $C_{12}H_{23}O_{5}N$ = x grams

And, mass of $C_{12}H{22}O_{11}$ = (1.00 - x)

So, moles of $C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}$

= $\frac{x}{369}$

Now, moles of $C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}$

= $\frac{x}{369} + \frac{(1.00 - x)}{342}$

= $2.85 \times 10^{-3}$

= x = 0.346

Therefore, we can conclude that amount of $C_{12}H_{23}O_{5}N$ present is 0.346 g and amount of $C_{12}H_{22}O_{11}$ present is (1 - 0.346) g = 0.654 g.

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3 years ago

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Answer:

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3 years ago

Read 2 more answers

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mariarad [96]

$2NaOH+H_2SO_4\Rightarrow\ Na_2SO_4+2H_2O\\
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3 0

3 years ago