B. energy level of outer electrons
C. nuclear charge
D. number of valence electrons
THE ANSWER BOI:
27 g
HERE IS DA EXPLANATION FAM:
27 g
M(C6H12O6) = 6*12 + 12*1 + 6*16 = 180 g/mol
100 mL = 0.1 L solution
1.5 M = 1.5 mol/L
1.5 mol/L * 0.1 L = 0.15 mol C6H12O6
0.15 mol * 180 g/1 mol = 27 g C6H12O6
To find the needed grams of glucose for a 1.5 M solution in 100ml, you multiply the molarity by the molecular weight of glucose and the volume of the solution. The calculation is 1.5 mol/l * 180 g/mol * 0.1 l = 27 grams. Therefore, 27 grams of glucose is needed.
The problem in question requires you to make use of the formula M= mass (mol)/ Volume (L). To find the required grams of glucose, C6H12O6, for a 1.5 M solution in 100mL, you would multiply the molarity with the molecular weight of glucose - approximately 180g per mole - and the volume of the solution - 0.1L. Therefore, to solve the problem you would calculate it as: 1.5 mol/l X 180 g/mol x 0.1 l which equals 27g. Hence, you need 27 grams of glucose to make a 100ml, 1.5M solution.
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B) gold.
C) nickel.
D) zinc.
its d zinc if your using usa test this dummy got me wrong trust me sis the answer is d
B. oxygen
C. ore
D. hydrogen
Answer;
- Oxygen
-Ancient cyanobacteria released oxygen, which assisted in creating the atmosphere as we know it today.
Explanation;
-The oxygen atmosphere that we depend on was generated by numerous cyanobacteria during the Archaean and Proterozoic Eras.
-Cyanobacteria are aquatic and photosynthetic, meaning they live in the water, and can manufacture their own food. During the process of photosynthesis, energy from sunlight is used together with carbon (iv) oxide and water to form organic molecules that serve as nutrients for the organisms and also oxygen gas is released to the atmosphere.