Answer:
i know you think that they are just related because kg has kilo but they are related because
Explanation:
1000g makes 1 g.You see how much there is a difference in the solution because how is one=one thousand,well that is all i can help you from here.bye panta
Answer:
B. Nuclear fission
Explanation:
Nuclear binding energy is used to determine whether fission or fusion will be a favorable process. The mass defect of a nucleus represents the mass of the energy binding the nucleus, and is the difference between the mass of a nucleus and the sum of the masses of the nucleons of which it is composed.
Answer:
133.74 L
Explanation:
First we <u>convert the given pressures and temperatures into atm and K</u>, respectively:
- 750.0 Torr ⇒ 750/760 = 0.9868 atm
- 20°C ⇒ 20+273.16 = 293.16 K
- 40°C ⇒ 40+273.16 = 313.16 K
Then we<u> use the PV=nRT formula to calculate the number of moles of helium in the balloon</u>, using<em> the data of when it was on the ground</em>:
- 0.9868 atm * 8.50 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293.16 K
Then, knowing the value of n, we <u>use PV=nRT once again, this time to calculate V</u> using <em>the data of when the balloon was high up:</em>
- 0.550 atm * V = 2.866 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 313.16 K
Answer:
Total pressure = 16.42× 10⁻⁹atm
Explanation:
Given data:
Moles of H₂ = 2.50 × 10⁻³ mol
Moles of He = 1.00 × 10⁻³ mol
Mass of Ne = 3 × 10⁻⁴ mol
Volume = 10 L
Temperature = 35°C
Total pressure = ?
Solution:
Pressure of hydrogen:
P = nRT / V
P = 2.50 × 10⁻³ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L
p = 63.22× 10⁻³ atm. L /10 L
P = 6.3 × 10⁻³atm
Pressure of helium:
P = nRT / V
P = 1.00 × 10⁻³ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L
p = 25.29 × 10⁻³ atm. L /10 L
P = 2.53× 10⁻³ atm
Pressure of neon:
P = nRT / V
P = 3 × 10⁻⁴ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L
p = 75.86× 10⁻³ atm. L /10 L
P = 7.59× 10⁻³ atm
Total pressure of mixture:
P(mixture) = pressure of hydrogen + pressure of helium+ pressure of neon
P(mixture) = 6.3 × 10⁻³atm + 2.53× 10⁻³ atm + 7.59× 10⁻³ atm
P(mixture) = 16.42× 10⁻⁹atm
Mid-Ocean Ridges
Two examples of Mid-Ocean Ridges are the Mid-Atlantic Ridge and the East Pacific Rise.
Explanation:
Mid-ocean ridges are mountain-like geographical features on the ocean floor. They usually line along the divergent boundaries of tectonic plates moving away from each other. This is because as the plates move in opposite direction, the vacuum in between the plates is filled by upwelling magma from the mantle. Usually the texture of the rocks that form these mountain ridges has bands. This is because, before the magma cools, the iron in the magma (due to its ferromagnetic property) aligns with the earths magnetic field. The earth magnetic flux flips over several 1000 years hence these bands orient themselves differently depending on the magnetic field orientation at the time they were formed.
Learn More:
For more on mid-ocean ridges check out;
brainly.com/question/7666628
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