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DaniilM [7]
3 years ago
10

How many significant figures are in 28.6 g

Chemistry
2 answers:
Alex787 [66]3 years ago
8 0
There are no zeros, so all three digits<span> are significant.</span>
<span>b. 3440.</span>
rewona [7]3 years ago
5 0
There are 3 significant figures in 28.6g :)

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Calculate the pH and pOH of 0.010 mol OH- per liter
Sati [7]
We can use a variety of formulas to determine our answers here.

Our formula for pOH is -log(mol), and we can plug it in as -log(0.010). Take note that OH- is a base, not an acid. 

So, the pOH of OH- is 2.

To find pH we can set up this simple equation:
pH + pOH = 14
All we need to do is subtract 2 from 14, therefore the pH is 12.

This makes sense since acids range in the pH of 1-6, and we are dealing with a base. Hope I could help!
7 0
3 years ago
How many moles are 3.21×10^28 molecules of N2?
kow [346]

Answer:5 to the 7th power

Explanation:

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8 0
3 years ago
Sodium metal (Na) reacts with chlorine gas ( Cl2 ) to produce solid sodium chloride (NaCl).
Fynjy0 [20]

The unbalanced equations are the equations with different atomic numbers on the sides of the reaction. The unbalanced reaction is Na + Cl₂ → NaCl

<h3>What are balanced equations?</h3>

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An unbalanced equation between sodium metal and chloride can be shown as:

Na + Cl₂ → NaCl

The equation is unbalanced as the number of chloride ions is more on the reactant side than the product side.

The balanced reaction will be:

2 Na + Cl₂  2NaCl

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3 0
2 years ago
A solution contains 50.0g of heptane (C7H16)and 50.0g of octane (C8H18) at 25 degrees C.The vapor pressures of pure heptane and
AleksandrR [38]

Answer:

a)Pheptane = 24.3 torr          

Poctane = 5.12 torr    

b)Ptotal vapor = 29.42 torr

c)  81 % heptane

    19 % octane

d) See explanation below

Explanation:

The partial pressure is given by Raoult´s law as:

Pa = Xa Pºa where Pa = partial pressure of component A

                               Xa = mole fraction of A

                               Pºa = vapor pressure of pure A

For a binary solution what we have to do is compute the partial  vapor pressure of each component and then add them together to get total vapor pressure.

In order to calculate the composition of the vapor  in part b), we will first calculate the mole fraction of each component in the vapor which is given by the relationship:

          Xa = Pa/Pt where Xa = mol fraction of  in the vapor

                                       Pa = partial pressure of A as calculated above

                                        Pt = total vapor pressure

Once we have mole fractions we can calculate the masses of the components for part c)    

a)                  

 MW heptane = 100.21 g/mol

 MW octane = 114.23 g/mol

mol heptane = 50.0 g / 100.21 g/mol = 0.50 mol

mol octane = 50.0 g/ 114.23 g/mol = 0.44 mol

mol total = 0.94 ⇒ Xa= 0.50/0.94 = 0.53 and

                             Xb= 0.44/0.94 = 0.47

Pheptane = 0.53 x 45.8 torr = 24.3 torr

Poctane = 0.47 x 10.9 torr = 5.12 torr

b) Ptotal = 24.3 torr +5.12 torr = 29.42 torr

c) We will call Y the mole fraction in the vapor to differentiate it from the mole fraction in solution

Y heptane (in the vapor) = 24.3 torr/ 29.42 torr = 0.83

Y octane (in the vapor) = 5.12 torr/ 29.42 torr = 0.17

d) To solve this part   we will assume that since the molecular weights are similar then having a mole fraction for heptane of 0.82, we could say that for every mole of mixture we have 0.82 mol heptane and 0.17 mol octane  and then we can calculate the masses:

0.82 mol x 100.21  g/mol = 82.2 g

0.17 mol x 114.23 g/mol =  19.4 g

total mass = 101.6

% heptane = 82.2 g/101.6g x 100 = 81 %

% octane = 19 %

There is another way to do this more exactly by calculating the average molecular weight of the mixture:

average MW = 0.83 (100.21 g/mol)  + 0.17 ( 114.23 g/mol ) = 102. 6 g/mol

and then  having a mol fraction of 0.83  means in 1 mol of mixture we have 0.83 mol heptane and 0.17 mol octane then the masses are:

mass heptane = 0.83 x 100.21 g/mol = 83.2 g

mass octane = 0.17 x  114.23 g/mol = 19.4 g

mass of mixture = 1 mol x MW mixture = 1 mol x 102.6 g/mol 102.6 g

% heptane = (83.2 g/ 102.6 g ) x 100 g = 81 %

% octane = 100 - 81 = 19 %

d)The composition of the vapor is different from the composition of the solution because the vapor is going to be richer in the more volatile compound in the solution which in this case is heptane ( 45.8  vs 10.9 torr).

4 0
3 years ago
Can u help me out its <br>a quick question<br>​
tia_tia [17]

The second one is the way to go.

5 0
3 years ago
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