Answer:
Option 2= Glucose
Explanation:
Cell membrane is made up of two phospholipid layers and each contain phosphate head and fatty acid or lipid tails. the head is present between the outer and inner boundaries and tail is present in between. The small non- polar molecules can pass the membrane through simple diffusion. This lipid tail restrict the passage of polar molecules including water soluble substances like glucose. However, transmembranes are present that allow the molecules to inter that are blocked by the tails.
Facilitated diffusion:
it is a type of diffusion in which caries protein without using the cellular energy shuttle the molecules to the cell membrane. Glucose is bind on the carrier protein ,change the shape and transport it from one to another side of membrane. In order to absorb the glucose red blood cells use this kind of diffusion.
Primary active transport:
The cells that are present along small intestine use this type of transport to pump the glucose inside the cell. The primary active transport require energy to transport the glucose inside.
Secondary active transport:
It is another method of transport of glucose into the cell. This method can not use ATP but it is based on concentration gradient of the sodium that provide electro chemical energy for the glucose transport.
The mass of carbon dioxide that would be made by reacting 30 grams C2H6 with 320 grams O2 will be 80 grams
From the balanced equation of the reaction:
The mole ratio of C2H6 to O2 is 2:7.
- Mole of 30 grams C2H6 = mass/molar mass
= 30/30
= 1 mole
- Mole of 320 grams O2 = 320/32
= 10 moles
Thus, C2H6 is the limiting reactant.
Mole ratio of C2H6 to CO2 according to the equation = 1:2
Since the mole of C2H6 is 1, the equivalent mole of CO2 would, therefore, be 2.
Mass of 2 moles CO2 = mole x molar mass
= 2 x 44
= 88 grams
More on stoichiometric calculations can be found here: brainly.com/question/8062886?referrer=searchResults
Answer:
8.8g of Al are necessaries
Explanation:
Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.
To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:
<em>Moles H2:</em>
PV = nRT; PV/RT = n
<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>
Replacing:
1atm*11L/0.082atmL/molK*273.15K = n
n = 0.491 moles of H2 must be produced
<em />
<em>Moles Al:</em>
0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required
<em />
<em>Mass Al -Molar mass: 26.98g/mol-:</em>
0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries
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