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3 years ago

Consider the model of the nitrogen atom.

Chemistry

2 answers:

ch4aika [34]3 years ago

8 0

Answer:

the answer is B

Explanation:

teach gave me the answer

frutty [35]3 years ago

3 0

Answer:

prob B not too sure but its better than nothin

Explanation:

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A water bath in a physical chemistry lab is 1.85m long, 0.810m wide and 0.740m deep. If it is filled to within 2.57 in from the

jenyasd209 [6]

Answer:

The volume of water in water bath is 1,011 Liters.

Explanation:

Length of the water bath, L = 1.85 m

Width of the water bath, W= 0.810 m

Height of the water bath ,H= 0.740 m

Height of the water in water bath, h= 0.740 m - 2.57 inches

1 m = 39.37 inch

$= 0.740 m - \frac{2.57}{39.37} m = 0.6747 m$

Volume of the water in bath = L × W × h

$=1.85 m\times 0.810 m\times 0.6747 m=1.011 m^3$

$1 m^3=1000 L$

$1.011 m^3=1.011\times 1000 L=1,011 L$

The volume of water in water bath is 1,011 Liters.

6 0

3 years ago

Explain how the concept of the “survival of the fittest” relates to the theory of evolution.

Vinvika [58]

The fittest species for a certain biome will adapt and change so as to keep living. This process is evolution and an example can be giraffes. If they had smaller necks, they couldn't reach the tree tops where food is, so only the fittest one with large necks survived, and they evolved by having the entire species grow up with large necks.

8 0

3 years ago

Read 2 more answers

How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr

pickupchik [31]

Answer: There are $1.469 \times 10^{23}$ molecules present in 7.62 L of $CH_4$ at $87.5^{o}C$ and 722 torr.

Explanation:

Given : Volume = 7.62 L

Temperature = $87.5^{o}C = (87.5 + 273) K = 360.5 K$

Pressure = 722 torr

1 torr = 0.00131579

Converting torr into atm as follows.

$722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm$

Therefore, using the ideal gas equation the number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

$0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}$

Thus, we can conclude that there are $1.469 \times 10^{23}$ molecules present in 7.62 L of $CH_4$ at $87.5^{o}C$ and 722 torr.

5 0

3 years ago

Un móvil se mueve con movimiento acelerado. En los segundo 2 y 3 los espacios recorridos son 90 y 120 m, Calcula la velocidad in

faust18 [17]

Answer:

La velocidad inicial es 55 $\frac{m}{s}$ y su aceleración es -10 $\frac{m}{s^{2} }$

Explanation:

Un movimiento es rectilíneo uniformemente variado, cuando la trayectoria del móvil es una línea recta y su velocidad varia la misma cantidad en cada unidad de tiempo . Dicho de otra manera, este movimiento se caracteriza por una trayectoria que es una línea recta y la velocidad cambia su módulo de manera uniforme: aumenta o disminuye en la misma cantidad por cada unidad de tiempo. Y la aceleración es constante y no nula (diferente de cero).

En este caso la posición del objeto esta dada por la expresión:

$x=x0+v0*t+\frac{1}{2} *a*t^{2}$

donde x es la posición del cuerpo en un instante dado, x0 la posición en el instante inicial, v0 la velocidad inicial y a la aceleración.

En este caso, por un lado podes considerar:

x= 90 m
x0= 0 m
v0= ?
t= 2
a= ?

Reemplazando obtenes:

$90=v0*2+\frac{1}{2} *a*2^{2}$

$90=v0*2+\frac{1}{2} *a*4$

$90=v0*2+2*a$

Y por otro lado tenes:

x= 120 m
x0= 0
v0= ?
t= 3
a= ?

Reemplazando obtenes:

$120=v0*3+\frac{1}{2} *a*3^{2}$

$120=v0*3+\frac{1}{2} *a*9$

$120=v0*3+\frac{9}{2} *a$

Por lo que tenes el siguiente sistema de ecuaciones:

$\left \{ {{2*v0+2*a=90} \atop {3*v0+\frac{9}{2} *a=120}} \right.$

Resolviendo por el método de sustitución, que consiste en aislar en una ecuación una de las dos incógnitas para sustituirla en la otra ecuación, obtenes:

Despejando v0 de la primera ecuación:

$v0= \frac{90-2*a}{2}$