Answer:
Because of the ground above and the molten iron core.
Explanation:
As you go deeper into the earth you have more and more of the earth above you that applies pressure on you. The heat near the very top of the plate is relatively low and it decreases by a little as you go down. But once you start to go down into the earth by hundreds of meters then the temperature starts to rise and it rises pretty quickly due to the heat being transferred from the molten iron outer core of the earth.
A weak Bronsted-Lowry base is a weak proton acceptor, where the proton is in the form of H+, so the conjugate acid formed contains one more H atom and an extra positive charge.
Hope this helps!
To make 1 Molar solution of hemoglobin ; 1600 grams of hemoglobin will be dissolved in 1 liter of water
The molecular weight of Hemoglobin is approximately 16,000 Daltons, when hemoglobin is converted to mM
16000 Dalton = 16000 ( g/mol )
given that 1 Dalton = 1 g/mol
To make 1 molar solution of hemoglobin using 1 liter of water
1 liter = 1000 grams
16000 Dalton = 16000 g/mol
Hence 16,000 grams of Hemoglobin is required to make 1 Molar solution of hemoglobin using 1 liter of water.
learn more : brainly.com/question/23517096
Answer:
grams H₂O produced = 8.7 grams
Explanation:
Given 2C₂H₆(g) + 7O₂(g) => 4CO₂(g) + 6H₂O(l)
7g 18g ?g
Plan => Convert gms to moles => determine Limiting reactant => solve for moles water => convert moles water to grams water
Moles Reactants
moles C₂H₆ = 7g/30g/mol = 0.233mol
moles O₂ = 18g/32g/mol = 0.563mol
Limiting Reactant => (Test for Limiting Reactant) Divide mole value by respective coefficient of balanced equation; the smaller number is the limiting reactant.
moles C₂H₆/2 = 0.233/2 = 0.12
moles O₂/7 = 0.08
<u><em>Limiting Reactant is O₂</em></u>
Moles and Grams of H₂O:
Use Limiting Reactant moles (not division value) to calculate moles of H₂O.
moles H₂O = 6/7(moles O₂) = 6/7(0.562) moles H₂O = 0.482 mole H₂O yield
grams H₂O = (0.482mol)(18g·mol⁻¹) = 8.7 grams H₂O
Answer: X2Y3
Explanation: I had the same question ;) (and got it right)
