Answer:
The answer to your question is: 3000 cm
Explanation:
30 meters long to cm
We can use a rule of three to solve it
We know that 1 meter ------------------ 100 cm
30 meters ---------------- x
x = 30(100)/1 = 3000 cm
Ok then! So mitosis is when a cell splits and doesn't lose/gain any chromosomes. In meiosis the chromosomes join and split evenly at the cell's "poles". Chromosomes will be lost evenly through this process.
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6
<em>Nuclear power releases less radiation into the environment than any other major energy source.</em>
Answer:
3.33 tanques de O₂
Explanation:
Basados en la reacción:
2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)
<em>2 moles de acetileno reaccionan con 5 moles de oxígeno produciendo 4 moles de dióxido de carbono y 2 moles de agua</em>
<em />
La ley de Avogadro dice que el volumen de un gas bajo temperatura y presión constantes es proporcional a las moles de este gas. Así, como 2 moles de acetileno reaccionan con 5 moles de oxígeno, los litros de O₂ necesarios para quemar 9340L de acetileno son:
9340 L C₂H₂ × (5 moles O₂ / 2 moles C₂H₂) = <em>23350L de O₂</em>
Si un tanque contiene 7x10³ L de O₂ serán necesarios:
23350L O₂ ₓ (1 tanque / 7x10³L) =<em> 3.33 tanques de O₂</em>
