Answer:
Plants normally turn dark green and look stunted (both leaves and stalks). Older leaves are first affected and may become violet. Sometimes the brown tips of the leaves remain fragile and their maturity seems to be delayed.
Explanation:
it is what it is booiiii
You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.
For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.
6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2
For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.
13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2
As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.
In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!
Answer:
Sample C is most likely the metal.
Explanation:
The Sample C is the metal, because the properties given in the sample c are all of the metal. As we know that the metals are the lustrous or the shiny elements. They are often good conductor of heat and also electricity. The metals possess high melting point. The density of the metals are heavy for their size. Metals can be easily hammered, and hence are malleable. They can easily be stretched into wires hence are ductile. They remains solid at room temperature but in case of mercury it remains as liquid. Metals are opaque object and cannot be see through it.
K:
m=155g
M=39g/mol
n = 155g / 39g/mol ≈ 3,97mol
KNO₃:
m=122g
M=101g/mol
n = 122g/101g/mol = 1,21mol
2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent
KNO₃ is limiting reagent
Answer:
Number of moles = 0.94 mol
Explanation:
Given data:
Number of moles of sodium chloride = ?
Volume of sodium chloride = 1.25 L
Concentration of solution = 0.750 mol/L
Solution:
Formula:
Concentration = number of moles/ volume in L
By putting values.
0.750 mol/L = number of mole / 1.25 L
Number of moles = 0.750 mol/L×1.25 L
Number of moles = 0.94 mol
