Answer:
Different atmospheric pressure. When there is a different atmospheric pressure, air moves from the higher pressure to the lower pressure area which results in what you call <u>WIND</u> but can result in various speeds and pressure.
Hope this helped and if it did, please give my answer a brainliest.
B I hope it’s right I don’t really help a lot but yeah lol
Answer:
B) 12.9 grams.
Explanation:
How many moles of molecules in that 6.30 L of H₂?
The volume of one mole of an ideal gas at STP (0 °C, 1 atm) is 22.4 liters.
(The volume of that one mole of gas at STP will be 22.7 liters if STP is defined as 0 °C and 10⁵ Pa).
.
How many moles of Na will be needed?
The coefficient in front of Na in the equation is twice the coefficient in front of H₂. It takes two moles of Na to produce one mole of H₂.
.
What's the mass of that many Na atoms?
Refer to a modern periodic table. The molar mass of ₁₁Na is 22.990. The mass of one mole of Na atoms is 22.990 gram. The mass of 0.5625 moles of Na atoms will be
.
(2 sig. fig. as in the volume of the H₂ gas.)
Answer:
Embryonic stem cells are pluripotent, meaning they can give rise to every cell type in the fully formed body, but not the placenta and umbilical cord. ... Human embryonic stem cells have been derived primarily from blastocysts created by in vitro fertilization (IVF) for assisted reproduction that were no longer needed
Explanation:
Answer:
68.1% is percent yield of the reaction
Explanation:
The reaction of methane with oxygen is:
CH₄ + 2O₂ → CO₂ + 2H₂O
<em>Where 2 moles of oxygen react per mole of CH₄</em>
<em />
Percent yield is:
Actual yield (28.2g CO₂) / Theoretical yield * 100
To solve this question we need to find theoretical yield finding limiting reactant :
<em>Moles CH₄:</em>
15.1g CH₄ * (1mol / 16.04g) = 0.9414 moles
<em>Moles O₂:</em>
81.2g * (1mol / 32g) = 2.54 moles
For a complete reaction of 0.9414 moles of CH₄ are needed:
0.9414 moles CH₄ * (2 mol O₂ / 1mol CH₄) = 1.88 moles of O₂. As there are 2.54 moles, O₂ is in excess and <em>CH₄ is limiting reactant</em>
In theoretical yield, the moles of methane added = Moles of CO₂ produced. That is 0.9414 moles CO₂. In grams = Theoretical yield:
0.9414 moles CO₂ * (44.01g / mol) = 41.43g CO₂
Percent yield: 28.2g CO₂ / 41.43g CO₂ * 100=
<h3>68.1% is percent yield of the reaction</h3>
