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netineya [11]
3 years ago
7

Various pieces of safety equipment are used in the lab to provide protection against injury. What piece of lab safety equipment

is least likely to be used if proper eye goggles are being worn?
Chemistry
2 answers:
Elina [12.6K]3 years ago
6 0
Various pieces of safety equipment are used in the lab to provide protection against injury. However, the lab safety equipment which is least likely to be used when proper eye goggles are being worn is the eye wash station. If this question has the same choices as the ones posted before, the answer is letter C. eye wash station.
julsineya [31]3 years ago
6 0

Answer:

Eye wash station.

Explanation:

Hello,

Due to the fact that the security googles provide the proper eye protection, one could say that eye wash station becomes needless since the associated risk is lowered due the coverage the goggles provide.

Best regards.

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2 years ago
What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?
mars1129 [50]

The concentration of solution in M or mol/L can be calculated using the following formula:

C=\frac{n}{V} .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is K_{2}SO_{4} thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass

Putting the values,

n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol

Thus, number of moles of  K_{2}SO_{4} will be 0.1061\times 0.5=0.053 mol.

The volume of solution is 225 mL, converting this into L,

1 mL=10^{-3}L

Thus,

225 mL=0.225 L

Putting the values in equation (1),

C=\frac{(0.053 mol}{0.225 L}=0.236 M

Therefore, concentration of potassium sulfate solution is 0.236 M.


4 0
3 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
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