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erastova [34]
3 years ago
12

ANSWER THIS QUICKLY Which of the following is an example of natural selection?

Chemistry
2 answers:
mojhsa [17]3 years ago
7 0

B is a case of natural selection.

kaheart [24]3 years ago
3 0

A  

blah blah blah it wants 20 characters

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Describe how you would prepare a supersaturated solution.
Anastasy [175]
To make a supersaturated solution<span>, make a saturated </span>solution<span> of sugar by adding 360 grams of sugar to 100 mL of water at 80 degrees Celsius. When the water cools back down to 25 degrees, that 360 grams of sugar will still be dissolved even though the water </span>should<span> only dissolve 210 grams of sugar.</span>
4 0
3 years ago
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When the concentration of reactant molecules is increased
a_sh-v [17]

Complete question is;

When the concentrations of reactant molecules are increased, the rate of reaction increases. The best explanation for this phenomenon is that as the reactant concentration increases,

A)the average kinetic energy of molecules increases.

B)the frequency of molecular collisions increases.

C)the rate constant increases.

D)the activation energy increases.

E)the order of reaction increases.

Answer:

B) The frequency of molecular collisions increases.

Explanation:

When we increase number of reactant molecules, the effective collision between the reactant molecules will form a product which also increases. As a result, the overall rate of the reaction will also increase which means the frequency of the molecular collision will also increase as well.

Thus, the correct answer is Option B

6 0
3 years ago
A sample of carbon dioxide gas occuples a volume of 2.5L at standard temperature and pressure (STP). What will be the volume of
MArishka [77]

Answer:

The correct answer is 5.0 L

Explanation:

STP are defined as T=273 K and P= 1 atm

By using the ideal gas equation, we can calculate the number of moles (n) of the gas at a volume V=2.5 L:

PV= nRT

⇒n= (PV)/(RT) =(1 atm x 2,5 L)/(0.082 L.atm/K.mol x 273 K)= 0.112 mol

For a sample of argon gas, with the same number of moles (0.112 mol) but twice the temperature (T = 273 K x 2= 546 K):

V= (nRT)/P = (0.112 mol x 0.082 L.atm/K.mol x 546 K)/1 atm = 5.0 L

That is consistent with the fact that when a gas is heated, it expanses. So, if the temperature increases twice, the volume also increases twice.

5 0
3 years ago
an element has atomic number 10 and an atomic mass of 20. how many neutrons are in the atom of this element
Flura [38]
Hey there!

n = A - Z

n =  20 - 10

n = 10 
6 0
3 years ago
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One mole of oxygen gas is at a pressure of 6.00 atm and a temperature of 27.0°C. (a) If the gas is heated at constant volume unt
Umnica [9.8K]

Answer: a) 900 K

b) 1200 K

Explanation:

According to the ideal gas equation:

PV=nRT

P = Pressure of the gas = 6.00 atm

V= Volume of the gas = ?

T= Temperature of the gas = 27°C = 300 K    0^00C=273K

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas  = 1

V=\frac{nRT}{P}=\frac{1\times 0.0821\times 300}{6.00}=4.10L

a) To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=6.00atm\\T_1=300K\\P_2=3\times 6.00=18.0atm\\T_2=?

Putting values in above equation, we get:

\frac{6.00}{300K}=\frac{18.0}{T_2}\\\\T_2=900K

The final temperature is 900 K

b) The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 6.00 atm

P_2 = final pressure of gas = 2\times 6.00atm=12.0atm

V_1 = initial volume of gas = 4.10 L

V_2 = final volume of gas =  2\times 4.10 L=8.20L

T_1 = initial temperature of gas = 300K

T_2 = final temperature of gas =?

Now put all the given values in the above equation, we get:

\frac{6.00\times 4.10}{300}=\frac{12.0\times 8.20}{T_2}

T_2=1200K

The final temperature is 1200 K

5 0
3 years ago
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