Answer:
749 grams CO₂
Explanation:
To find the amount of carbon dioxide produced, you need to (1) convert grams C₃H₈ to moles C₃H₈ (via molar mass from periodic table), then (2) convert moles C₃H₈ to moles CO₂ (via mole-to-mole ratio via reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass from periodic table). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The desired unit should be in the numerator. The final answer should have 3 significant figures because the given value (250. grams) has 3 sig figs.
Molar Mass (C₃H₈): 3(12.01 g/mol) + 8(1.008 g/mol)
Molar Mass (C₃H₈): 44.094 g/mol
1 C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O
Molar Mass (CO₂): 12.01 g/mol + 2(16.00 g/mol)
Molar Mass (CO₂): 44.01 g/mol
250. g C₃H₈ 1 mole C₃H₈ 3 moles CO₂ 44.01 g
------------------ x ---------------------- x ---------------------- x -------------------- =
44.094 g 1 mole C₃H₈ 1 mole CO₂
= 749 grams CO₂
Answer : Linear
Explanation : Hydrogen Cyanide (HCN) when drwan in the Lewis diagram shows carbon atom at the center with no lone electron pairs.
The carbon and nitrogen atoms are bonded through a triple bond which counts as "one electron pair".
The molecule has two electron pairs in all and appears to be linear.
Also, according to the VSEPR theory; the electron clouds on atoms around the carbon will try to repel each other.
They will get pushed apart, which gives HCN molecule a linear molecular geometry or shape.
The bond angle that is developed will be 180 degrees since it has a linear molecular geometry of HCN. The hybridisation observed in this molecule is SP.
Some meringues are different and don't require the same ingredients just follow the recipe as said!!! Good luck
<span><span>Percent =mass of the elementx100</span>molar mass of compound</span><span> Find the mass percents for the elements present in H2SO4.molar mass of H2SO4<span> = 2(1) + 32 + 4(16) = 98 g</span><span>mass of H = 2 mol H x 1.008g/mol = 2.016 gmass of S = 1 mol S x 32.00g /mol = 32.00 gmass of O = 4 mol O x 16.00g/mol = 64.00 gmass % H = 2.016g/98g x 100 = 2.06%mass % S = 32.00g/98g x 100 = 32.6%mass % O = 64.00g/98g x 100 = 65.3% </span>Calculate the mass % of the elements present in glucose, C6H12O6. </span>
Answer:
2 is most likely instructions.
Explanation:
sorry if its not I'm not that good with the method
