Answer:
4.14 x 10²⁴ molecules CO₂
Explanation:
2 C₄H₁₀ + 13 O₂ --> 8 CO₂ + 10 H₂O
To find the number of CO₂ molecules, you need to start with 100 grams of butane (C₄H₁₀), convert to moles (using the molar mass), convert to moles of CO₂ (using coefficients from equation), then convert to molecules (using Avagadro's number). The molar mass of C₄H₁₀ is calculated using the quantity of each element (subscript) multiplied by the number on the periodic table. The ratios should be arranged in a way that allows for units to be cancelled.
4(12.011g/mol) + 10(1.008 g/mol) = 58.124 g/mol C₄H₁₀
100 grams C₄H₁₀ 1 mol C₄H₁₀ 8 mol CO₂
-------------------------- x ---------------------- x ---------------------
58.124 g 2 mol C₄H₁₀
6.022 x 10²³ molecules
x ------------------------------------ = 4.14 x 10²⁴ molecules CO₂
1 mol CO₂
Im pretty sure the answer is 4 but not 100 percent
Answer:
A) 4.3 × 10²⁴ atoms
Explanation:
Step 1: Given data
Moles of neon: 7.2 moles
Step 2: Calculate the number of atoms present in 7.2 moles of neon
In order to convert moles to toms, we need a conversion factor. In this case, we will use Avogadro's number: there are 6.02 × 10²³ neon atoms in 1 mole of neon atoms.
7.2 mol × 6.02 × 10²³ atoms/mol = 4.3 × 10²⁴ atoms
Answer:
0. 414
Explanation:
Octahedral interstitial lattice sites.
Octahedral interstitial lattice sites are in a plane parallel to the base plane between two compact planes and project to the center of an elementary triangle of the base plane.
The octahedral sites are located halfway between the two planes. They are vertical to the locations of the spheres of a possible plane. There are, therefore, as many octahedral sites as there are atoms in a compact network.
The Octahedral interstitial void ratio range is 0.414 to 0.732. Thus, the minimum cation-to-anion radius ratio for an octahedral interstitial lattice site is 0. 414.
1 result would be the decimation of crops and vegetation, which causes a reduction in farmers income rates which could lead to his workers losing their jobs or even poverty
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