Answer:
C
Explanation:
The answer is C because machines such as tractors and sprinklers are counted as technology, and are better examples than pesticides.
Answer:they will out live us himans
Explanation:
Answer:
Assuming that all of the oxygen is used up, 1.53×4111.53×411 or 0.556 moles of C2H3Br3 are required. Because there are only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.
Limiting Reagent What is the limiting reagent if 76.4 grams of C2H3Br3 were reacted with 49.1 grams of O2? C2H3Br3 + 11O2 → 8CO2 + 6H2O + 6Br2 SOLUTION Using Approach 1: A. 76.4g × (1 mol/ 266.72 g) = 0.286 moles C2H3Br3 49.1g × (1 mole/ 32 g) = 1.53 moles O2 B.
Explanation:
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https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/08%3A_Quantities_in_Chemical_Reactions/8.04%3A_Limiting_Reactant_and_Theoretical_Yield
Answer: 1.2044 × 10^24
Explanation:
1 mole of calcium is 40 gram
Based on Avogadro's law:
1 mole of any substance has 6.02 x 10^23 atoms, also 1 mole of calcium is 40gram
So, 1 mole of calcium = 6.02 x 10^23 atoms
Also, 1 mole of calcium is 40gram
40 grams of calcium = 6.02 x 10^23 atoms
80.156 grams of calcium = Z atoms
To get the value of Z, cross multiply:
Z atoms x 40 grams = 6.02 x 10^23 atoms x 80.156 grams
40Z = 482.54 x 10^23
Z = (482.54 x 10^23/40)
Z = 12.06 x 10^23
Put result in standard form
Z = 1.206 x 10^24 atoms
Thus, 1.206 x 10^24 atoms of Ca are present in 80.156 grams of Ca
Answer:
N(CO2) = n(CO2) x Avogadros constant
n(CO2) = mass(CO2)/Molar mass(CO2) = 112/44 = 2.5454….
N(CO2) = 2.545454… x (6.02 x 10^23) = 1.5323636 x 10^24
Therefore, the number of CO2 molecules in 112 g of CO2 = 1.53 x 10^24 molecules
