The asthenosphere is directly below the lithosphere so the answer to your question is the asthenosphere, because the outer core is towards the center of Earth.
Answer:
Explanation:
Hello,
In this case, it is convenient to compute the car's mileage in km/L as follows:
In such a way, since the distance is measured to be 250 km, the volume requirement is:
Regards.
Answer:
Dark matter makes up 85% of the mass of the universe. Dark matter is not directly observable because it doesn't interact with any electromagnetic wave. In the development of the universe, without dark matter, the universe will not function, move or rotate as it does now (this speculation led to the quest to find the anomaly of mass and energy in the known universe, eventually leading to the idealization of dark matter) and will not have enough gravitational force to hold it together. After the big bang,<em> the presence of dark matter and energy ensured that the newly formed universe didn't just float away, rather, it provided enough gravitational force to hold the universe while still allowing it to expand sufficiently</em>.
The development of the universe would have been different without the universe in the sense that the young universe won't have enough mass to hold it together, and the universe would have simply floated apart. The behavior of the universe would have been different from what we observe now, and some physical laws that applies now will not apply to the universe.
Answer:
51207 torr is the new pressure of the gas
Explanation:
We can solve this question using combined gas law that states:
P1V1T2 = P2V2T1
<em>Where P is pressure, V volume and T absolute temperature of 1, initial state and 2, final state of the gas</em>
<em> </em>
Computing the values of the problem:
P1 = 710torr
V1 = 5.0x10²mL
T1 = 273.15 + 30°C = 303.15K
P2 = ?
V2 = 25mL
T2 = 273.15 + 820°C = 1093.15K
Replacing:
710torr*5.0x10²mL*1093.15K = P2*25mL*303.15K
3.881x10⁸torr*mL*K = P2 * 7.579x10³mL*K
P2 = 51207 torr is the new pressure of the gas
First, we have to get:
1- The heat required to increase T of ice from -50 to 0 °C:
according to q formula:
q1 = m*C*ΔT
when m is the mass of ice = mol * molar mass
= 1 mol * 18 mol/g
= 18 g
and C is the specific heat capacity of ice = 2.09 J/g-K
and ΔT change in temperature = 0- (-50) = 50°C
by substitution:
∴q1 = 18 g * 2.09 J/g-K *50°C
= 1881 J = 1.881 KJ
2- the heat required to melt this mass of ice is :
q2 = n*ΔHfus
when n is the number of moles of ice = 1 mol
and ΔHfus = 6.01 KJ/mol
by substitution:
q2 = 1 mol * 6.01 KJ/mol
= 6.01 KJ
3- the heat required to increase the water temperature from 0°C to 60 °C is:
q3 = m*C*ΔT
when m is the mass of water = 18 g
C is the specific heat capacity of water = 4.18 J/g-K
ΔT is the change of Temperature of water = 60°C - 0°C = 60°C
by substitution:
∴q3 = 18 g * 4.18 J/g-K * 60°C
= 4514 J = 4.514 KJ
∴the total change of enthalpy = q1+q2+q3
= 1.881 KJ +6.01 KJ + 4.514 KJ
= 12.405 KJ
