Answer:
It has 2 eyes
Explanation:
Quantitive observation means that it is facts based of numbers, therefore you can count how many eyes it has, but not that it has red eyes
1Al2(SO4)3 + 3ZnCl2 → 2AlCl3 + 3ZnSO4
The coefficients represents moles. There is 1 mole of Aluminum Sulfate, 3 moles of Zinc(II) Chloride, 2 moles of Aluminum Chloride, and 3 moles of Zinc(II) Sulfate.
Now add all the coefficients/moles.
9 is the sum of all the coefficients.
The flow of stimuli information in the body is from the receptors to sensory neuron (afferent neurons) to the interneurons then to the central nervous system (brain and spinal cord) then is carried of by the motor neurons (efferent neurons) to the mucles involve.
Sensation to transduction to perception. Sensation is the ambiguous information which is received by the receptors or sensory organ henceforth, transduction occurs in the nerve cell protruding to the brain and is now called perception as the vague stimuli is interpreted and processed to be understood then is responded with the proper and apt response or reaction.
Answer:
concave lens
Explanation:
it's concave lens because it diverges the ray/beam of light.
Answer:
CH₂ ; 67.1 %
Explanation:
To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation
Assume 100 grams of the compound.
# mol C = 85.7 g / 12.01 g/mol = 7.14 mol
# mol H = 14.3 g / 1.008 g/mol = 14.19 mol
The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C
So the empirical formula is CH₂
For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.
We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )
1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol
Theoretical yield from balanced chemical equation:
0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆
Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g
% yield = 0.295 g/ 0.440 g x 100 = 67.1 %
