The ER takes up a lot of space in some cells<span>. The endoplasmic reticulum may be “rough” or “smooth.” ER that has no attached ribosomes is called smooth endoplasmic reticulum. </span>
Answer:
Explanation:
so if the reaction is C2H4+3O2-->2H2O+2CO2
if there were 7 moles of C2H4:
(see the attachment)
Answer:
28.93 g/mol
Explanation:
This is an extension of Graham's Law of Effusion where 
We're only talking about molar mass and time (t) here so we'll just concentrate on 
. Notice how the molar mass and time are on the same position, recall effusion is when gas escapes from a container through a small hole. The time it takes it to leave depends on the molar mass. If the gas is heavy, like Xe, it would take a longer time (4.83 minutes). If it was light it would leave in less time, that gives us somewhat an idea what our element could be, we know that it's atleast an element before Xenon.
Let's plug everything in and solve for M2. I chose M2 to be the unknown here because it's easier to have it basically as a whole number already.
The square root is easier to deal with if you take it out in the first step, so let's remove it by squaring each side by 2, the opposite of square root essentially.
M2= 0.22 x 131
M2= 28.93 g/mol
Answer:
4.12 mol
Explanation:
Given data:
Moles of LiOH required = ?
Volume of solution = 4.2 L
Molarity of solution = 0.98 M
Solution:
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
we will calculate the moles from above given formula.
0.98 M = number of moles / 4.2 L
0.98 M × 4.2 L = number of moles
Number of moles = 0.98 M × 4.2 L
Number of moles = 4.12 mol (M = mol/L)
Answer:
Mass = 2.12 g
Explanation:
Given data:
Volume of KMnO₄ = 255 mL (255/1000 = 0.255 L)
Molarity = 0.0525 M
Mass in gram = ?
Solution:
First of all we will calculate the number of moles.
<em>Molarity = number of moles of solute / volume in litter</em>
0.0525 M = number of moles of solute / 0.255 L
Number of moles of solute = 0.0525 M ×0.255 L
Number of moles of solute = 0.0134 mol
Mass in gram:
<em>Number of moles = mass/ molar mass</em>
Mass = moles × molar mass
Mass = 0.0134 mol × 158.04 g/mol
Mass = 2.12 g
