Answer:
We need 10.14 grams of sodium bromide to make a 0.730 M solution
Explanation:
Step 1: Data given
Molarity of the sodium bromide (NaBr) = 0.730 M
Volume of the sodium bromide solution = 135 mL = 0.135 L
Molar mass sodium bromide (NaBr) = 102.89 g/mol
Step 2: Calculate moles NaBr
Moles NaBr = Molarity NaBr * volume NaBr
Moles NaBr = 0.730 M * 0.135 L
Moles NaBr = 0.09855 moles
Step 3: Calculate mass of NaBr
Mass NaBr = 0.09855 moles * 102.89 g/mol
Mass NaBr = 10.14 grams
We need 10.14 grams of sodium bromide to make a 0.730 M solution
D. a and b because same-charged particles repel each other
The coefficient in a chemical formula represents the amount of each chemical present. The amount of a substance is measured in moles.
Answer: potassium iodide is the basic test for starch,and the positive test is blue-black coloration, any other test substance which is not starch will give a negative results.
Explanation:
Starch is an example of polysaccharide and since the standard test for it is potassium iodide solution, it gives a positive test.
Diasaccharides e.g maltose are reducing sugars.their standard test is BENEDICT test .
Therefore, in the hydrolysis; starch should give a positve test, while Diasaccharides should give negative rest.
Answer:
Correct choice are C and D (they are both, the same).
Explanation:
Cathode is the positive(+) electrode where a reduction occurs.
Reduction is the chemical reaction where the oxidation state is reduced.
2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag+(aq) + H2O (l)
A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-
B. 2Ag (s) → 2Ag+ (aq) + 2e-
C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)
D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)
C or D, are ok. They are the same equation.
Oxygen from ground state reduce the oxidation state from 0 to -2
