A. K+, OH-
B. C6H5CO+, OH-
C. NH4+, Cl-
D. Mg++, 2 NO3-
Everything has 1 except for the Nitrate ion in D, which has 2
<h3>
Answer:</h3>
0.127 mol Au
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 25.0 g Au
[Solve] moles Au
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.126923 mol Au ≈ 0.127 mol Au
Velocity and mass are directly proportional to the quantity of momentum by:
p = mv. Therefore, and increase in either velocity or mass will lead to an increase in momentum and vice versa. Momentum during a reaction is always conserved, meaning that the mass and initial velocity before a reaction will always be equal to the change in mass and velocity produced after the reaction. Kinetic energy after a reaction, however, is not always conserved. For example if a fast moving vehicle collided with a stationary vehicle, and moved together, the overall kinetic energy would be after the reaction, as a heaver mass would be moved by the same velocity causing a decrease in kinetic energy.
I don't know if this is exactly what you are looking for, but in physics this is how it is understood.
Answer:
2.835 moles of carbon
Explanation:
By definition, there are 6.022x10^23 atoms (or compounds) in one mole.
Write and use this as a conversion factor:
(6.022x10^23 atoms)/mole
(1.707 x 10^24 atoms of carbon)/((6.022x10^23 atoms)/mole) = 2.835 moles carbon
A general equation for a combustion reaction would be expressed as follows:
CxHy + (x+y/2)O2 = xCO2 + y/2H2O
Propane would obviously would only have carbon and hydrogen in its structure. Assuming a complete combustion, all of the carbon atoms would go to carbon dioxide and all of the hydrogen atoms to water. To determine the empirical, we determine the number of carbon and hydrogen atoms present.
moles C = 2.461 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.06 mol C
moles H = 1.442 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H ) = 0.16 mol H
Then, we divide the smallest amount to the each mole of the atoms. We do as follows:
C = 0.06 / 0.06 = 1
H = 0.16 / 0.06 = 2.67
Then we multiply a number in order to obtain a whole number ratio between the atoms.
1 CH2.67
2 C2H5.34
3 C3H8 <-------- empirical formula
