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HACTEHA [7]
4 years ago
9

What are five primary signs of a chemical change

Chemistry
1 answer:
MissTica4 years ago
5 0
1) size 
2) shape 
3)odor 
4)state  
5)form 
This is from my knowledge please make sure to check it before learning it. 
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A very old tree limb contains an amount of carbon-14 that is approximately 1/8 of the current atmospheric 14C levels.. Calculate
Tomtit [17]

A.       The radioactive decay equation is N = N0e^{-ln(2)*t/T }

where T is the half-life (5730 years), N0 is the number of atoms at time t = 0 and N is the number at time t.

Rewriting this as:

(N/N0) = e^{-ln(2)*t/T }

Since N = (1/8) N0 and substituting known values:

1/8 = e^{-ln(2)*t/5730}

Taking ln of both sides:

ln(1/8)= -ln(2)*t/5730

t = - 5730 * ln(1/8) / ln (2)

t = 17,190 years

The tree was cut down 17,190 years ago.

B.   N0 = 1,500,000 carbon-14 atoms

Since N = (1/8) N0

N = 187,500 carbon atoms left

3 0
3 years ago
During respiration _____.
Alex73 [517]

Answer:

Shawty's like a melody in my head

That I can't keep out, got me singin' like

Na-na-na-na, everyday

It's like my iPod stuck on replay, replay-ay-ay-ay

Shawty's like a melody in my head

That I can't keep out, got me singin' like (ayy!)

Na-na-na-na, everyday

It's like my iPod stuck on replay (J-J-J-JR), replay

Explanation:

Shawty's like a melody in my head

That I can't keep out, got me singin' like

Na-na-na-na, everyday

It's like my iPod stuck on replay, replay-ay-ay-ay

Shawty's like a melody in my head

That I can't keep out, got me singin' like (ayy!)

Na-na-na-na, everyday

It's like my iPod stuck on replay (J-J-J-JR), replay

5 0
4 years ago
Read 2 more answers
Which of the following represents an alpha particle?<br> O A. je<br> B. He<br> O C. Y<br> O D.B
-Dominant- [34]
I think it’s D hope that helped
4 0
3 years ago
Read 2 more answers
Un recipiente cerrado, de 4,25 L, con tapa móvil, contiene H2S(g) a 740 Torr y 50,0°C. Se introduce en ese recipiente N2(g) a te
yulyashka [42]

Answer:

n_{N_2}=6.41mol

Explanation:

¡Hola!

En este caso, teniendo en cuenta la información dada por el problema, inferimos que primero se debe usar la ecuación del gas ideal con el fin de calcular las moles de gas que se encuentran al inicio del experimento:

PV=nRT\\\\n=\frac{RT}{PV} \\\\n=\frac{0.08206\frac{atm*L}{mol*K}*(50.0+273.15)K}{740/760atm*4.25L}\\\\n=6.41mol

Seguidamente, usamos la ley de Avogadro para calcular las moles finales, teniendo el cuenta que el volumen final es el doble del inicial (8.50 L):

n_2=\frac{6.41mol*8.50L}{4.25L}\\\\n_2=12.82mol

Quiere decir que las moles de N2(g) que se agregaron son:

n_{N_2}=12.81mol-6.41mol\\\\n_{N_2}=6.41mol

¡Saludos!

8 0
3 years ago
Draw the Lewis structure (including all lone pair electrons and any formal charges) for one of the four possible isomers of C3H9
Alexus [3.1K]

Answer and Explanation

The isomer picked is the N-Propylamine.

It has a lone pair of electron available on the electron rich Nitrogen and no formal charge.

Since it will be hard to draw the Lewis structure in this answer format, I'll attach a picture of the Lewis structure to this answer.

The lone pair of electron is shown by the two dots on the Nitrogen atom.

6 0
3 years ago
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