Answer:
C-Rb20
Explanation:
The chemical formula for Rubidium oxide is Rb20.
Hope this helps!
5.00111 moles in 45.05g of br
<span>the balanced equation for the reaction is as follows
2C</span>₄H₁₀ + 13O₂ ---> 8 CO₂ + 10H₂<span>O
stoichiometry of C</span>₄H₁₀ to O₂ <span>is 2:13
stoichiometry applies to the molar ratio of reactants and products. Avagadros law states that volume of gas is directly proportional to number of moles of gas when pressure and temperature are constant.
Therefore volume ratio of reactants is equal to molar ratio, volume ratio of C</span>₄H₁₀ to O₂<span> is 2:13
2 L of </span>C₄H₁₀ reacts with 13 L of O₂<span>
then 100 L of </span>C₄H₁₀<span> reacts with 13/2 x 100 = 650 L
therefore 650 L of O</span>₂<span> are required </span>
<u>Answer:</u> The number of moles of ammonium nitrate is 0.004 moles.
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
We are given:
Molarity of 
solution = 0.125 M
Volume of solution = 32.5 mL
Putting values in above equation, we get:
Hence, the number of moles of ammonium nitrate is 0.004 moles.
Answer: hre
Explanation:
N2(g) + 3H2-> 2NH3(g) This is the balanced equation
Note the mole ratio between N2, H2 and NH3. It is 1 : 3 : 2 This will be important.
moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present
moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present
Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2.
moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced
grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced
NOTE: The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.
