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ki77a [65]
3 years ago
14

Identify the group of ONLY metalloids.

Chemistry
1 answer:
Alinara [238K]3 years ago
5 0

Answer:

C boron, germanium, arsenic, silicon

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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
Does anyone know the answer?????​
Juli2301 [7.4K]

Answer:

In chemistry, pH (/piːˈeɪtʃ/) (abbr. power of hydrogen or potential for hydrogen) is a scale used to specify how acidic or basic a water-based solution is. Acidic solutions have a lower pH, while basic solutions have a higher pH.

Explanation:

that should answer ur question

3 0
3 years ago
1. Look at the DNA molecule shown at right. What does it look like?
jarptica [38.1K]

Answer:

i dont see it but i hope it looks like a dna molecule or something would be wrong

Explanation:

there is no pic

8 0
2 years ago
Read 2 more answers
Design a test to determine whether thorium-234 also emits particles. First, explain how Rutherford’s experiment measured positiv
liubo4ka [24]

The characteristics of the α and β particles allow to find  the design of an experiment to measure the ²³⁴Th particles is:

  • On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.
  • The neutrons cannot be detected in this experiment because they have no electrical charge.

In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.

The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.

The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.

Thorium has several isotopes, with different rates and types of emission:

  • ²³²Th emits α particles, it is the most abundant 99.9%
  • ²³⁴Th emits β particles, exists in small traces.

In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.

Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.

In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:

  • On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.
  • The neutrons cannot be detected in this experiment because they have no electrical charge.

Learn more about radioactive emission here: brainly.com/question/15176980

7 0
3 years ago
The nuclear binding energy of one nitrogen-14 atom is (BLANK) x 10A J. Round to 3
d1i1m1o1n [39]

mass defect = mass of constituents - mass of atom

N has 7p and 9n

proton mass ~ 1.00728 amu

neutron mass ~ 1.00866 amu

electron mass ~ 0.000549 amu

Nitrogen mass ~ 14.003074 amu

mass defect = (7*1.00728)-(7*1.00866)-(7*0.000549) - 14.003074

= 0.11235amu

convert to energy, the binding energy = 1.68x10^-11 J


5 0
3 years ago
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