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Answer: option C.Water will move into the cell
Explanation:
1) Start by analyzing what the statement means in terms of relative concentrations:
------------------------ | inside the cell ------------ | outside the cell |
sugar --------------- | higher ----------------------- | lower ------------- |
water -------------- | lower ------------------------- | higher ------------ |
2) Osmosis is the process where a barrier (the celll membrane) permits the pass of some component and not others.
The component that can pass is that whose particles are smaller. Sugar molecules (the solute) are bigger than water molecules (the solvent), so sugar molecules cannot pass the cell membrane. Only water can.
3) The driviing force for the motion of water molecules is called diffusion. The diffusion occurs from higher concentrations to lower concentrations.
Hence, the water molecules will from outside the outiside the cell, where they have the greater concentration, toward the inside of the cell, where water hasa the lower concentration.
As result, the water will move into the cell, which is the option C.
Answer: 1) Maximum mass of ammonia 198.57g
2) The element that would be completely consumed is the N2
3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g
Explanation:
- In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:
N2(g) + 3H2(g) ⟶2NH3(g)
Both equal amount of atoms side to side.
- Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)
Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3
Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3
- As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
- We proceed calculating the maximum mass of NH3 from the 163.3g of N2.
11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3
- In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:
163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2
That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left
Answer:
Transition Metals
Explanation:
The elements in groups 3-12 are called Transition Metals. These groups contain metals that usually form multiple cations. All other groups on the table (1, 2, 13-18) are called Main Group Elements.
Answer:
The same number of each element present before the reaction takes place must also be present on the product side of the equation. Coefficients are placed in front of a chemical formula to show the number of moles of that substances that are necessary for the reaction to occur.
Explanation:
