Answer : Both solutions contain
molecules.
Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain
molecules.
Avogadro's Number is
=
which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.
Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.
Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.
Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.
We can calculate the number of molecules for each;
Number of molecules =
;
∴ Number of molecules =
which will be = 
Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.
Answer:
A. cycas
1. it is thick and scaly
2. it grow relatively slowly and have a large, terminal rosette of leaves.
B. bamboo
1. it is very durable
2. it is both flexible and elastic
C. lemna
1. it grows as simple free-floating thalli on or just beneath the water surface
2. it are small, not exceeding 5 mm in length
D. paddy
1. it is variety purity
2. it degree of purity
E. sugarcane
1. it is bear long sword-shaped leaves
2. it's stalks are composed of many segments, and in each joint there is a bud
hey mate here is ur answer
solution
mass{m}=3 gram
=3/1000
volume{v}=16cm
=16/100
density=m/v
=3/1000÷16/100
=3/160
=0.01875kg/m3
Explanation:
<h3 /><h2>
<em><u>H2 </u></em><em><u>+</u></em><em><u> </u></em><em><u>O2 </u></em><em><u>=</u></em><em><u> </u></em><em><u>H2O</u></em></h2>
<h2>
<em><u>Hydrogen</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>Oxygen</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>Water</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em></h2>
<em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em>