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3 years ago

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There are two naturally occurring isotopes of boron. 10 B has a mass of 10.0129 u. 11 B has a mass of 11.0093 u. Determine the a

bundance of each isotope

1 answer:

Vanyuwa [196]3 years ago

5 0

<h2>Natural Abundance for 10B is 19.60%</h2>

Explanation:

The natural isotopic abundance of 10B is 19.60%.

The natural isotopic abundance of 11B is 80.40%.

The isotopic masses of boron are 10.0129 u and 11.009 u respectively.

For calculation of abundance of both the isotopes -

Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.

Determining it as an equation -

10x + 11y= 10.8

x+y=1 (ratio)

10x + 10y = 10

By taking the denominator away from the numerator

we get;

y = 0.8

x + y = 1

∴ x = 0.2

To get percentages we need to multiply it by 100

So, the calculated abundance is 80% for 11 B and 20% 10 B.

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Elements occur in a number of isotopic forms. In this problem, you will learn about the notation used to distinguish different i

alina1380 [7]

Answer:

<em>Number of protons, Z = 14</em>
<em>Number of neutrons, N = 14</em>

Explanation:

<u>1) About isotopes:</u>

<em>Isotopes</em> are different kind of atoms of the same element. Hence, they have the atomic number (Z), which is the number of <em>protons</em>, the same number of electrons (talking about to neutral atoms, not ions), and different <em>number of neutrons N).</em>

This is, it is the number of neutrons what distinguish different isotopes of an element.

<u>2) About the notation used to distinguish different isotopes:</u>

A superscript and a subscript, both to the left of the chemical symbol of the element, are used to <em>distinguish different isotopes</em>:

A ←------------- This superscript tells the mass number of the isotope

X ←--------- This is the chemical symbol of the element

Z ←-------------- This subscript is the atomic number of isotope

In our case, the notiation for the isotope of silicon is: ²⁸₁₄ Si

So, we have:

28 is the mass number (A)
14 is the atomic number (Z)
Si is the chemical symbol.

Now, you can answer the questions of the <em>part A</em>:

Number of protons: Z = 14

Number of neutrons N:

mass number = number of protons + number of neutrons

A = Z + N

⇒ N = A - Z = 28 - 14 = 14

In <u>conclusion</u>:

Number of protons, Z = 14
Number of neutrons, N = 14

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3 years ago

A chemistry student weighs out of hypochlorous acid into a volumetric flask and dilutes to the mark with distilled water. He pla

riadik2000 [5.3K]

Answer:

<u><em>Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.</em></u>

A chemistry student weighs out 0.0941 g of hypochlorous acid (HClo) into a 250. ml. volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits mL.

Explanation:

1 mole HClO = 74.44g

0.0941g = $\frac{0.0941}{74.44}$ = 0.00126 moles

Concentration = no. of moles/volume in L

Hence, Concentration of HClO = 0.00126/ 0.250L

= 0.005M.

C1V1 =C2V2

0.005 × 250 mL = 0.2 × V2

<u><em>Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.</em></u>

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Answer:

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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:

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

Reemplazando en la segunda ecuación:

$120=\frac{90-2*a}{2} *3+\frac{9}{2} *a$

Resolviendo:

$120=(90-2*a)*\frac{3}{2} +\frac{9}{2} *a$

$120=135-3*a +\frac{9}{2} *a$

$120-135=-3*a +\frac{9}{2} *a$

$-15=\frac{3}{2} *a$

$\frac{-15}{\frac{3}{2} } =a$

-10=a

Reemplazando el valor de a en la expresión despejada anteriormente obtenes:

$v0= \frac{90-2*(-10)}{2}$

Resolviendo:

$v0= \frac{90+20}{2}$

$v0= \frac{110}{2}$

v0=55

<u><em>La velocidad inicial es 55 </em></u> $\frac{m}{s}$ <u><em>y su aceleración es -10 </em></u> $\frac{m}{s^{2} }$ <u><em></em></u>

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3 years ago