Answer:
121.63 g/mol
Explanation:
Sr(OH) 2 = Strontium Hydroxide
Answer:
The Photosynthesis process
Explanation:
Plants, algae, and some other organisms can transform the sunlight energy into chemical energy. The photosynthesis process occur thanks to the chloroplasts. The chloroplast is an organelle found in all green plants. Inside of the chloroplast you can find the thylakoids which are arranged in stacks named grana, they have membranes with chloropyll a photosynthetic pigment, also you can find the photosystems, they are functional and structural units of protein complexes. The thylakoids capture the light and allow the reactions to transform CO2. The set of reactions that occurs in the chloroplasts are known as the Calvin cycle.
The general equation of photosynthesis is:

6 CO2 + 6 H2O + Energy -> C6H12O6 + 6 O2
Carbon Dioxide + water + Light -> Glucose (sugar) + Oxygen
After, this glucose is transformed into pyruvate, and it allowed the release of denosine triphosphate (ATP) by cellular respiration. The ATP is an organic chemical that is requires for the cell to perform any process (any kind or work).
Answer:
The correct option is: c. phospholipid
Explanation:
Phospholipids, also known as glycerophospholipids, are the derivatives of fatty acids which is a major structural component of the cell membranes.
Phospholipid is the class of lipids that is composed of a <u>glycerol molecule that forms ester bonds with the two long-chain fatty acids and one phosphate group.</u>
<u>Therefore, Molecule A is a </u><u>phospholipid.</u>
Answer:
pH = 4.543
Explanation:
- CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
- pKa = - Log Ka
∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
∴ pKa = 4.87
⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
added 300 mL 0f 0.02 M NaOH:
⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)
⇒ <em>C</em> CH3CH2COOH = 0.048 M
⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M
mass balance:
⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)
charge balance:
⇒ [H3O+] + [Na+] = [CH3CH2COO-]
∴ [Na+] = 0.02 M
⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)
(2) in (1):
⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]
replacing in Ka:
⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])
⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]
⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]
⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0
⇒ [H3O+ ] = 2.867 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = 4.543