With the given formula, we can calculate the amount of CO₂ using the balance equation but we first need the moles of CH₄
1) to find the moles of CH₄, we need to use the ideal gas formula (PV= nRT). if we solve for n, we solve for the moles of CH₄, and then we can convert to CO₂. Remember that the units put in this formula depending on the R value units. I remember 0.0821 which means pressure (P) has to be in atm, volume (V) in liters, the amount (n) in moles, and temperature (T) in kelvin.
PV= nRT
P= 1.00 atm
V= 32.0 Liters
n= ?
R= 0.0821 atm L/mol K
T= 25 C= 298 K
let plug the values into the formula.
(1.00 x 32.0 L)= n x 0.0821 x 298K
n= (1.00 x 32.0 L )/ (0.0821 x 298)= 1.31 moles CH₄
2) now let's convert the mole of CH₄ to moles to CO₂ using the balance equation
1.31 mol CH₄ (1 mol CO₂/ 1 mol CH₄)= 1.31 mol CO₂
3) Now let's convert from moles to grams using the molar mass of CO₂ (find the mass of each atom in the periodic table and add them)
molar mass CO₂= 12.00 + (2 x 16.0)= 44.0 g/mol
1.31 mol CO₂ ( 44.0 g/ 1 mol)= 57.6 g CO₂
Note: let me know if you any question.
<span>It must generate hydroxide ions in the water
hope i helped</span>
Answer:
A synthesis reaction occurs when two or more reactants combine to form a single product. ... Replacement reactions occur when elements switch places in compounds. In a single replacement reaction, one element takes the place of another in a single compound. In a double replacement reaction, two compounds exchange elements.
Answer:
There are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution
Explanation:
Molarity of the solution = 2.20 M

Therefore, there are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution