The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab the moles of dextrose is this equivalent to is 3.6888 moles.
<h3>What are moles?</h3>
A mole is described as 6.02214076 × 1023 of a few chemical unit, be it atoms, molecules, ions, or others. The mole is a handy unit to apply due to the tremendous variety of atoms, molecules, or others in any substance.
To calculate molar equivalents for every reagent, divide the moles of that reagent through the moles of the restricting reagent. The calculation is follows:
- 655/12 x 6 + 12+ 16 x 6
- = 655/ 180 = 3.6888 moles.
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Answer:
1.) 13 g C₄H₁₀
2.) 41 g CO₂
Explanation:
To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O
48 g O₂ 1 mole 2 moles C₄H₁₀ 58.124 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 13 g C₄H₁₀
48 g O₂ 1 mole 8 moles CO₂ 44.007 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 41 g CO₂
<h2>Answer:</h2>
Rutherford's models
<h2>Explanations:</h2><h2>What is the electron cloud model?</h2>
There are known as the region where electrons are found especially in the nucleus.
According to the five basic atomic models which have contributed to the structure of the atom itself, the Rutherford's models of the atom include a structure that is mostly made of empty space compared to thomson that proposed the plum pudding model of the atom
<span>Boron has a lot of different isotopes, most of which having a very short half life (ranging from 770 milliseconds for Boron-8 down to 150 yoctoseconds for boron-7). But the two isotopes Boron-10 and Boron-11 are stable with about 80.1% of the naturally occurring boron being boron-11 and the remaining 19.9% being boron-10. The weighted average weight of those 2 isotopes has the value of 10.81.
The reason they use the average mass of an element for it's atomic weight is because elements in nature are rarely single isotopes. The weighted average allows us to easily compare relative number of atoms of one element against relative numbers of atoms of another element assuming that the experimenters are getting isotope ratios close to their natural ratios.</span>
