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3 years ago

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Bombardment of cobalt-59 with a neutron produces a manganese-56 atom and another particle. What is this particle?A. an alpha par

ticleB. a beta particle (electron)C. a positronD. a gamma rayE. a neutron

1 answer:

Aleksandr-060686 [28]3 years ago

4 0

Answer:A

Explanation:

A nuclear reaction is balanced by ensuring that the Masses and charges of te reactants and products are exactly balanced on the left and right hand side of the reaction equation. If there are 60 mass units on the LHS and manganese has only 56 mass units then four mass units are left. If there is no charge on a neutron and there is a charge of 27 on the cobalt, then two charges are left. Four mass units and a charge of +2 corresponds to a helium which is actually an alpha particle.

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Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe

Illusion [34]

Answer:

a. Kp=1.4

$P_{A}=0.2215 atm$

$P_{B}=0.556 atm$

b.Kp=2.0 * 10^-4

$P_{A}=0.495atm$

$P_{B}=0.00995 atm$

c.Kp=2.0 * 10^5

$P_{A}=5*10^{-6}atm$

$P_{B}=0.9999 atm$

Explanation:

For the reaction

A(g)⇌2B(g)

Kp is defined as:

$Kp=\frac{(P_{B})^{2}}{P_{A}}$

The conditions in the system are:

A B

initial 0 1 atm

equilibrium x 1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.

Replacing these values in the expression for Kp we get:

$Kp=\frac{(1-2x)^{2}}{x}$

Working with this equation:

$x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0$

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

$x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a}$ (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

$x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128$

$x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215$

With x1 we get a partial pressure of:

$P_{A}=1.128 atm$

$P_{B}=1-2*1.128 = -1.256 atm$

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

$P_{A}=0.2215 atm$

$P_{B}=1-2*0.2215 = 0.556 atm$

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

$P_{A}=0.505atm$

$P_{B}=1-2*0.505 = -0.01005 atm$

Not sense.

With x2

$P_{A}=0.495atm$

$P_{B}=1-2*0.495 = 0.00995 atm$

X2 is the solution for this equation.

Kp=2*10^5

X1=50001

$X2=5*10^{-6}$

With x1

$P_{A}=50001atm$

$P_{B}=1-2*50001=-100001atm$

Not sense.

With x2

$P_{A}= 5*10^{-6}atm$

$P_{B}=1-2*5*10^{-6}= 0.9999 atm$

X2 is the solution for this equation.

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How many moles are equal to 3.0x10^23 atoms of germanium? PLS ANSWER WITH WORK!

horrorfan [7]

1 mole=6.02 x 10^23 atoms so how many moles are there in 3.0 x 10^23 we will cross multiply, 1 x 3.0 x 10^23 / 6.02 x 10 ^23. Which will give us 0.498 moles.
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3 years ago