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babunello [35]
3 years ago
10

What is the movement of air from high-pressure areas to low-pressure areas? is the movement of air from high- pressure areas to

low pressure areas.
Chemistry
1 answer:
mezya [45]3 years ago
7 0

Answer:

The "pressure gradient" causes the movement of air from high-pressure areas to low-pressure areas.

Explanation:

The pressure gradient is the physical entity which tells us about the direction of the air in which it will move. It also tells us about the rate at which it will increase as well as decrease in the specific location. The unit of the pressure gradient used is pascals per metre. The difference of this quantity creates uneven increase in the temperature of the surface of the Earth, only when the solar radiation focuses over the equator. They are also the reason behind the beginning movement of the air.

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Distinguish between sugar and non-sugar with examples.
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Answer:

Reducing sugars are sugars where the anomeric carbon has an OH group attached that can reduce other compounds. Non-reducing sugars do not have an OH group attached to the anomeric carbon so they cannot reduce other compounds. ... Maltose and lactose are reducing sugars, while sucrose is a non-reducing sugar

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What is the molarity of a solution that contain 1.1 miles of lithium in 0.5 liters of solution
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4 0
3 years ago
2C2H6 + 7O2 ------&gt; 4CO2 + 6H2O
Romashka-Z-Leto [24]

We convert the masses of our reactants to moles and use the stoichiometric coefficients to determine which one of our reactants will be limiting.

Dividing the mass of each reactant by its molar mass:

(10 g C2H6)(30.069 g/mol) = 0.3326 mol C2H6

(10 g O2)(31.999 g/mol) = 0.3125 mol O2.

Every 2 moles of C2H6 react with 7 moles of O2. So the number of moles of O2 needed to react completely with 0.3326 mol C2H6 would be (0.3326)(7/2) = 1.164 mol O2. That is far more than the number of moles of O2 that we are given: 0.3125 moles. Thus, O2 is our limiting reactant.

Since O2 is the limiting reactant, its quantity will determine how much of each product is formed. We are asked to find the number of grams (the mass) of H2O produced. The molar ratio between H2O and O2 per the balanced equation is 6:7. That is, for every 6 moles of H2O that is produced, 7 moles of O2 is used up (intuitively, then, the number of moles of H2O produced should be less than the number of moles of O2 consumed).

So, the number of moles of H2O produced would be (0.3125 mol O2)(6 mol H2O/7 mol O2) = 0.2679 mol H2O. We multiply by the molar mass of H2O to convert moles to mass: (0.2679 mol H2O)(18.0153 g/mol) = 4.826 g H2O.

Given 10 grams of C2H6 and 10 grams of O2, 4.826 g of H2O are produced.

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