Chemistry, 05.11.2019 10:31 jeffyisdrunk

The molecular mass of octane is 114.22 g/mol. what is the mass of 22.05 mol of octane?

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In this item, we are asked to calculate for the mass of 22.05 mole of octane given that its molecular mass is equal to 114.22 g/mol. To answer this item, we simply have to multiply the number of moles with the molecular mass. That is, (114.22 g/mol)(22.05 mol) which is equal to 2518.221 grams.

Answer from: gd9075

In order to compute the mass of a substance, we may simply mulitply the number of moles of the substance by the molar mass of the substance, since molar mass is mass per mole. Therefore, we get: 114.22 * 22.05 = 2,518.55 grams 2,518.55 grams of octane are present.

Answer from: Quest

answer: e. sour taste

explanation: acids are those substances which dissociates in water to give $h^+$ ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

$hcl\rightarrow h^++cl^-$

bases are those substances which dissociates in water to give $oh^-$ ions or accepts $h^+$ ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.

$naoh\rightarrow na^++oh^-$

$nh_3+h^+\rightarrow nh_4^+$

Answer from: Quest

For cyclohexane compounds drawn in a chair conformation: -axial bonds are vertical -equatorial bonds are parallel to two bonds within the ring. thus, as i assume the chair conformation is drawn: -the x group is in an axial position, as are the five other vertical bonds. -three axial bonds are on the top face of the ring -the other three axial bonds are on the bottom face the position of the second substituent on each ring carbon is equatorial. gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other. drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1. click this link and watch some academic videos that may you more understand the topic

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