Bronsted lowry bases
NO2- amd OH-
Answer:
B.) Magma
Explanation:
Lava - Magma that reaches Earth's through a volcanic vent
Magma - Composed of molten mixture of rock-forming substances, gases and water
Batholith - Very large igneous intrusion extending deep in the earth's crust
Slurry - Semiliquid mixture, typically of fine particles of manure, cement, or coal suspended in water
Answer:
2.6%
Explanation:
As, 1 ounce (oz) = 0.0625 pounds (lb)
Therefore, weight of baby at discharge = 7 lb,1 oz = 7+0.0625 lb = 7.0625 lb
Since, 1 oz = 0.0625 lb
⇒ 4 oz = 4×0.0625 = 0.25 lb
Therefore, weight of baby at birth = 7 lb,4 oz = 7+0.25 lb = 7.25 lb
The <u>amount of weight lost</u> is equal to the difference of weight of the baby at birth and discharge.
Therefore, <u>weight lost</u> = 7.25 lb - 7.0625 lb = <u>0.1875 lb</u>
Now, the <u>percentage of weight lost</u> by the baby is given by the amount of weight lost divided by the weight of the baby at birth.
Therefore, <u>the percentage of weight los</u>t = weight lost ÷ weight at birth = 0.1875 lb ÷ 7.25 lb × 100 = <u>2.6% </u>
The full question is shown in the image attached to this answer
Answer:
6.2 * 10^-3
Explanation:
The reaction at equilibrium is as follows;
TiCl4(l) → Ti(s)+ 2Cl2(g)
We have to obtain the concentration of chlorine gas as follows;
[Cl2] = 1.08/70.91 g/mol × 5.2 L
[Cl2] = 0.079 M
Kc = [Cl2]^2
Kc = [ 0.079]^2
Kc = 6.2 * 10^-3
The balanced equation for the above reaction is as follows
C₆H₁₂O₆(s) + 6O₂(g) --> 6H₂O(g) + 6CO₂<span>(g)
the limiting reactant in the equation is glucose as the whole amount of glucose is used up in the reaction.
the amount of </span>C₆H₁₂O₆ used up - 13.2 g
the number of moles reacted - 13.2 g/ 180 g/mol = 0.073 mol
stoichiometry of glucose to CO₂ - 1:6
then number of CO₂ moles are - 0.073 mol x 6 = 0.44 mol
As mentioned this reaction takes place at standard temperature and pressure conditions,
At STP 1 mol of any gas occupies 22.4 L
Therefore 0.44 mol of CO₂ occupies 22.4 L/mol x 0.44 mol = 9.8 rounded off - 10.0 L
Answer is B) 10.0 L CO₂
