Answer:
1.45 x 10⁻² g CO₂
Explanation:
To find the mass of carbon dioxide, you need to (1) convert grams CH₄ to moles CH₄ (via molar mass), then (2) convert moles CH₄ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass). The final answer should have 3 sig figs to reflect the given value (5.30 x 10⁻³ g).
Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)
Molar Mass (CH₄): 16.043 g/mol
Combustion of Methane:
1 CH₄ + 2 O₂ ---> 2 H₂O + 1 CO₂
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
5.30 x 10⁻³ g CH₄ 1 mole 1 mole CO₂ 44.007 g
--------------------------- x ---------------- x --------------------- x ----------------- =
16.043 g 1 mole CH₄ 1 mole
= 0.0145 g CO₂
= 1.45 x 10⁻² g CO₂
The simplest way to use the periodic table to identify an element is by looking for the element's name or elemental symbol. The periodic table can be used to identify an element by looking for the element's atomic number. The atomic number of an element is the number of protons found within the atoms of that element.
hope it helps
Answer:
1 = strontium
2 = technetium
3 = silver
Explanation:
Strontium is present in 2 group of periodic table. It is alkaline earth metal. It is most reactive element and can not found free in nature. it is insoluble in water but react with water and form strontium hydroxide and hydrogen gas is also produced. It is radioactive element.
It is similar to the calcium and can incorporated into bones. The food rich with strontium is used to avoid the osteoporosis. It is found in leafy vegetables, grains, sea foods etc.
In given series strontium is highest in reactivity then technetium, while the technetium is more reactive than silver. The silver is less reactive as compared to both given elements.
Answer:
Explanation:
To calculate the theoretical yield, determine the number of moles of each reactant, in this case the sole reactant ethanol. Convert the 100 g to moles; the molecular weight of ethanol is 46 g/mole, therefore: Since there is only one reactant, it is also the limiting reagent.
