Answer:
Explanation:
1) A fulcrum is a pivot point that plays a central role (not necessarily located at the center) in a lever. The fulcrum of the attached picture has been circled (in blue).
2) The object placed on this lever's measurement tray is balanced by placing it at the center of the tray. This is the standard way of placing objects on any balance.
CxHy + O2 --> x CO2 + y/2 H2O
Find the moles of CO2 : 18.9g / 44 g/mol = .430 mol CO2 = .430 mol of C in compound
Find the moles of H2O: 5.79g / 18 g/mol = .322 mol H2O = .166 mol of H in compound
Find the mass of C and H in the compound:
.430mol x 12 = 5.16 g C
.166mol x 1g = .166g H
When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.
Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).
In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).
Explanation:
When an acid is dissolved in water it forms hydrogen ions (H+) that combine with water to form the hydronium ion (H3O+) .
Answer:
a) The volume is 5.236x10⁻¹³L
b) The molarity of a single virus is 1.91x10¹² mol/L
c) The molarity for a 100 virus particles is 1.91x10¹⁴ mol/L
Explanation:
a) Given:
D = diameter of the cell = 10 μm
r = radius = 10/2 = 5 μm
The volume of the spherical cell is equal:
If 1 μm³ = 1x10⁻¹⁵L, then 523.6 μm³ = 5.236x10⁻¹³L
b) The molarity is:
For a single virus within the cell
c) For a 100 virus particles the molarity is:
Answer:
The position of equilibrium would not be appreciably affected by changes in the volume of the container for NiO(s) + CO(g) ⇌ Ni(s) + CO2(g).
Correct Answer : Option A
Explanation:
The equilibrium position tends to change with increase or decrease in pressure or volume, or both of them. This happens because considering change in volume, when the volume of the container increases, the reactant molecule increases i.e. mole of gases and thus the position of equilibrium shifts towards the right side. Similarly in case of decreasing volume, reactant molecule decreases and the equilibrium position shifts left side.
And in case, when the mole of gases on both the sides of the equation i.e. reactant side and product side are equal, it will not have any effect on the equilibrium position on increasing or decreasing volume, or pressure.
