Answer: The correct answer is Option 1.
Explanation:
Control group is the group used in the experiment by the researchers in which no change in the variable is done. It is then set as a benchmark for the groups which are being tested.
We are given 4 groups which are being experimented for the flotation of an egg in water. As in Cup 1, there is no addition of salt and hence there is no change in the variables. So, this is set as a benchmark fro the cups which are further used in the experiment conducted.
Hence, the correct answer is Option 1.
Answer:
0.18 moles
Explanation:
Applying,
PV = nRT................... Equation 1
Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = temperature.
make n the subject of the equation
n = PV/RT............... Equation 2
Given: V = 5.3 L, T = 22 °C = (22+272) K = 295 K, P = 632 mmHg = (0.00131579×632) = 0.8316 atm, R = 0.083 L.atm/K.mol
Substitute these values into equation 2
n = (0.8316×5.3)/(0.083×295)
n = 0.18 moles
Answer:
2.17 e -14
Explanation:
A strong acid like HCl ionize 100 % in water so [H3O+] = 0.46 M
[OH-] = Kw / [H3O+]
= 1.0 e -14 / 0.46
= 2.17 e -14
3.01 Ă— 10^24 Ă— (12/5) hydrogen atoms
Looking at the formula for the molecule, the ratio of carbon to hydrogen atoms is 5:12, so if we divide the number of carbon atoms by 5 and then multiply by 12, we can find the number of hydrogen atoms. Let's look at the available options and see what makes sense.
3.01 Ă— 10^24 Ă— (12/5) hydrogen atoms
* This is exactly correct.
(3.01 Ă— 10^24 / 5) hydrogen atoms
* Nope. This will tell you how many pentane MOLECULES you have, but not the number of hydrogen atoms.
3.01 Ă— 10^24 Ă— (5/12) hydrogen atoms
* Close, but the ratio (5/12) will tell you the number of carbon atoms you have if you give it the number of hydrogen atoms. So this choice is wrong.
3.01 Ă— 10^24 Ă— 12 hydrogen atoms description
* This would tell you the number of hydrogen atoms you have if you know the number of pentane molecules you have. So this choice is also wrong.
Answer: El carbono, que en estado sólido, puede adoptar muchas formas alotrópicas, siendo las más comunes el diamante (red tridimensional) y el grafito (láminas), aunque también puede formar nanoestructuras en forma de balón de fútbol (fullerenos) o tubos diminutos (nanotubos de carbono), entre otras posibilidades.
Explanation:
