Answer: A comprehensive legal term for uniform standards ascribed to the quantity, capacity, volume, or dimensions of anything. Legislation that adopts and mandates the use of a uniform system of weights and measures is a valid exercise of Police Power, and such laws are constitutional.
Explanation:
The common substance among the product(s) of the first equation and among the reactant(s) in the second equation is H2O(g). We can eliminate that as an intermediate. The overall chemical equation will thus be:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l),
which is the first answer choice.
In essence, all you’re doing here is swapping water vapor for liquid water.
Answer : The molar mass of unknown substance is, 39.7 g/mol
Explanation : Given,
Mass of unknown substance = 9.56 g
Volume of solution = 100.0 mL
Molarity = 2.41 M
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :

Now put all the given values in this formula, we get:


Therefore, the molar mass of unknown substance is, 39.7 g/mol
Answer:

Explanation:
The molar mass of uranium-235 is 235 g/mol. So one mole of uranium-235 has a mass of 235 g. Put differently 6.022×10^23 atoms of uranium-235 have a mass of 235 g. Knowing that, how can we use that to find the mass of one atom?
mass of one atom = 
Answer:
Reagent O₂ will be consumed first.
Explanation:
The balanced reaction between O₂ and C₄H₁₀ is:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles
- O₂: 13 moles
- CO₂: 8 moles
- H₂O: 10 moles
Being:
- C: 12 g/mole
- H: 1 g/mole
- O: 16 g/mole
The molar mass of the compounds that participate in the reaction is:
- C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
- H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles* 58 g/mole= 116 g
- O₂: 13 moles* 32 g/mole= 416 g
- CO₂: 8 moles* 44 g/mole= 352 g
- H₂O: 10 moles* 18 g/mole= 180 g
If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

mass of O₂= 223.78 grams
But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>