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Airida [17]
3 years ago
8

What term refers to the period of life from puberty to the end of the teenage years?

Chemistry
1 answer:
Daniel [21]3 years ago
7 0
A - adolescence.
 
According to google definitions..

Adolescencenounthe period following the onset of puberty during which a young person develops from a child into an adult.
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I need help plz..
Anuta_ua [19.1K]

Answer:

1. a simulation 2. saurology

Explanation:

4 0
2 years ago
Read 2 more answers
Which example illustrates a chemical change
Svetllana [295]
Burning wood because heat indicates a chemical change
7 0
3 years ago
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El fluoruro de hidrógeno HF que se utiliza en
Blizzard [7]

Answer:

25.6g de HF son producidos

Explanation:

<em>...¿Cuánto HF es producido?</em>

Para resolver este problema debemos convertir la masa de cada reactivo a moles usando su masa molar. Como la reacción es 1:1, el reactivo con menor número de moles es el reactivo limitante. Con las moles del reactivo limitante podemos obtener las moles de HF y su masa así:

<em>Moles CaF2:</em>

Masa molar:

1Ca = 40g/mol

2F = 19*2 = 38g/mol

40+38 = 78g/mol

50g CaF2 * (1mol/78g) = 0.641 moles CaF2

<em>Moles H2SO4:</em>

Masa molar:

2H = 2g/mol

1S = 32g/mol

4O = 64g/mol

98g/mol

100g H2SO4 * (1mol / 98g) = 1.02 moles H2SO4

Como las moles de CaF2 < Moles H2SO4: CaF2 es reactivo limitante.

<em>Moles HF usando la reacción:</em>

0.641 moles CaF2 * (2mol HF / 1mol CaF2) = 1.282 moles HF

<em>Masa HF:</em>

Masa molar:

1g/mol + 19g/mol = 20g/mol

1.282 moles HF * (20g/mol) =

<h3>25.6g de HF son producidos</h3>
8 0
2 years ago
Calculate the pH of the solution that results when 20 mL of 0.2M HCOOH is mixed with 25mL of 0.2M NaOH solution.(post-equivalenc
Paul [167]

The pH of the solution : 12

<h3>Further explanation</h3>

Reaction

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

mol HCOOH =

\tt 20~ml\times 0.2~M=4~mlmol

mol NaOH =

\tt 25~ml\times 0.2~M=5~mlmol

Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]

ICE method :

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

4                          5

4                          4                     4                   4

0                          1                      1                    1

Concentration of [OH⁻] from NaOH :

\tt \dfrac{1~mlmol}{20+25~ml}=0.02

pOH=-log[OH⁻]

pOH=-log 10⁻²=2

pH+pOH=14

pH=14-2=12

3 0
3 years ago
A sample of oxygen occupied 626 mL when the pressure increased to 874.27 mm Hg. At constant temperature, what volume did the gas
Usimov [2.4K]

Answer:

709 (With sig figs)

Explanation:

Pressure(1) * volume(1) = Pressure(2) * volume(2)

771.75 mm Hg * unknown = 874.27 mm Hg * 626 mL

771.75 mm Hg * unknown = 547,293.02 mm Hg*mL

   Divide both sides by 771.75 mm Hg

Unknown = 709 (With sig figs)....709.158432 (without sig figs)

4 0
1 year ago
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