Contains DNA, contains ribosomes, lacks a nucleus
Answer: option A. 2
Explanation: in the formula Sr3(PO4)2, the 2 behind (PO4) is affecting both P and O4. It means that we have P2 in the formula
Answer:
a. 3; b. 5; c. 10; d. 12
Explanation:
pH is defined as the negative log of the hydronium concentration:
pH = -log[H₃O⁺] (hydronium concentration)
For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.
For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:
pOH = -log[OH⁻] (hydroxide concentration)
Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.
Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.
(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)
A net ionic equation simply means to cancel out any ions which appear on both sides of the chemical equation that are not involved in the reaction - they're called spectator ions.
We'll first write out the full ionic equation, showing all ions and compounds formed, then rewrite and not include spectator ions.
2FeBr3(aq) + 3Na2S(Aq) --> Fe2S3(s) + 6NaBr(aq) [original eqation]
2Fe3+(aq) + 6Br-(aq) + 3Na+(aq) + 3S2-(aq)--> Fe2S3(s)+6Na+(aq) + 6Br-(aq)
[full ionic equation]
2Fe3+(aq) + 3S2-(aq)--> Fe2S3(s) [net ionic equation]
notice that Br- and Na+ appear unreacted on both sides of the full ionic equation, so they cancel out and do not appear in the net ionic.
*Please give me a 'brainliest' if you can! Thanks!
3.46 g Cu 1 mol Cu 6.02*10^23 atoms Cu =
63.55 g Cu 1 mol Cu
3.28*10^22 atoms Cu
