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topjm [15]
3 years ago
10

The experiment where the factors being tested are not applied. It allows scientists to set a baseline to compare the experimenta

l groups that do have factors applied to them.
Test variable (independent variable):
Outcome variable (dependent variable):
Control group:
Chemistry
1 answer:
olasank [31]3 years ago
5 0

Answer:

An experiment is a research method performed within a completely controlled environment that uses groups of individuals ( subjects), one of which the factors being tested are not spplied on, an independent factor of the hypothesys ( to be tested ), for example the resilience of a group of teenagers) and a dependant factor that is applied to check how it varyes depending on the tested factor ( for example, the most challanging experience of their generation)

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A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent
Degger [83]

Answer :  The % of (+) limonene isomer = 79%


                The % of (-) limonene isomer = 0%


                The % of enantiomeric excess = 58%


Explanation :   Enantiomeric excess (ee) is the measurement of purity used for chiral substances.


Given,


% of pure limonene enantiomer = The % of (+) limonene isomer = 79%


Therefore, The % of (-) limonene isomer = 0%


Formula used :  

\%(+)\text{ isomer}=\frac{ee}{2}+50\%


Where,         ee → enantiomeric excess


Now, put all the values in above formula, we get the value of enantiomeric excess (ee).


     {ee}=\frac{\%(+)-50\%}{2}


            =\frac{79\%-50\%}{2}


              = 58%



7 0
3 years ago
Read 2 more answers
Addition of 50. J to a 10.0-g sample of a metal will cause the temperature of a metal to rise from 25ºC to 35ºC. The specific he
Snowcat [4.5K]

Answer:

b) C = 0.50 J/(g°C)

Explanation:

  • Q = mCΔT

∴ Q = 50 J

∴ m = 10.0 g

∴ ΔT = 35 - 25 = 10 °C

specific heat (C) :

⇒ C = Q / mΔT

⇒ C = 50 J / (10.0 g)(10 °C)

⇒ C = 0.50 J/(g°C)

7 0
3 years ago
The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is __________
PIT_PIT [208]

Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.

<span>
Moles of HCl = 5.5 g HCl ( 1 mol HCl / 36.46 g HCl ) = 0.1509 mol HCl</span>

<span>
Molality = 0.1509 mol HCl / 200 g C2H6O ( 1 kg / 1000 g )
Molality = 0.7543 mol / kg</span>


<span>The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is 0.7453 molal.</span>
6 0
3 years ago
The two naturally occuring isotopes of antimony are 121Sb (57.21%) and 123Sb (42.79%), with isotopic masses of 120.904 and 122.9
emmasim [6.3K]

Answer:

The average atomic weight = 121.7598 amu

Explanation:

The average atomic weight of natural occurring antimony can be calculated as follows :

To calculate the average atomic mass the percentage abundance must be converted to decimal.

121 Sb has a percentage abundance of 57.21%, the decimal format will be

57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .

123 Sb has a percentage abundance of 42.79%, the decimal format will be

42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .

Next step is multiplying the fractional abundance to it masses

121 Sb = 0.5721 × 120.904 = 69.169178400

123 Sb = 0.4279 × 122.904 = 52.590621600

The final step is adding the value to get the average atomic weight.

69.169178400 + 52.590621600 = 121.7598 amu

5 0
3 years ago
The solubility of N2 in blood at 37°C and a partial pressure of 0.80 atm is 5.6 ✕ 10−4 mol·L−1. A deep-sea diver breathes compre
marysya [2.9K]

Answer:

0.0126 moles are released

Explanation:

Using Henry's law, where the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid:

S = k×P

<em>Where S is solubility (5.6x10⁻⁴molL⁻¹), k is Henry's constant and P is partial pressure (0.80atm)</em>

Replacing:

<em>5.6x10⁻⁴molL⁻¹ / 0.80atm = 7x10⁻⁴molL⁻¹atm⁻¹</em>

Thus, with Henry's constant, solubility of N₂ when partial pressure is 3.8atm is:

S = 7x10⁻⁴molL⁻¹atm⁻¹ × 3.8atm

S = 2.66x10⁻³molL⁻¹

Thus, when the deepd-sea diver has a pressure of 3.8amt, moles dissolved are:

6.0L × 2.66x10⁻³molL⁻¹ = <em>0.01596 moles of N₂</em>

At the surface, pressure is 0.80atm and solubility is 5.6x10⁻⁴molL⁻¹, moles dissolved are:

6.0L × 5.6x10⁻⁴molL⁻¹ = <em>3.36x10⁻³mol</em>

Thus, released moles are:

0.01596mol - 3.36x10⁻³mol = <em>0.0126 moles are released</em>

8 0
3 years ago
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