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lisov135 [29]
3 years ago
12

Drag the correct symbol to the equation. Not all symbols will be used. A plutonium atom undergoes nuclear fission. Identify the

missing element in the nuclear equation. 239Pu + 1n ---> ____ + 134Xe + 3 1n 94 0 54 0 Options: 101Zr 40 134Xe 54 105Zr 40 103Zr 40 102Xe 54
Chemistry
1 answer:
Naya [18.7K]3 years ago
5 0

Answer:

_{40}^{103}Zr

Explanation:

For any nuclear equation, we should utilize the law of mass conservation and the law of charge conservation. The sum of the masses on the left-hand side of the arrow should be equal to the sum of the masses on the right-hand side of the arrow (those are the superscripts for each nucleus). Similarly, the sums of charges should be equal (this is the law of charge conservation).

Let's say that the missing species is X with a mass of 'M' and charge of 'Z':

_{94}^{239}Pu+_0^1n\rightarrow _Z^MX+_{54}^{134}Xe+3_0^1n

Find mass applying the mass balance law:

239+1=M+134+3\cdot1\\240 = M+137\\M=240 -137\\M=103

This means our particle X has a mass of 103. Let's find the atomic number (the charge) same way:

94+0=Z+54+3\cdot0\\94=Z+54\\Z=94-54=40

The atomic number of our nucleus is 40. That said, we have:

_{40}^{103}X

Find the element in the periodic table with Z = 40. This is Zr. Meaning we can now identify it fully:

_{40}^{103}Zr

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Which of the following sets of elements could replace the generic "X" symbol with their own to produce a valid dot notation for
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Symbol  "X" represents Nitrogen, phosphorus, and arsenic.as they have five valence electrons in their valance shell.

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6 0
2 years ago
A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )
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Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g)   (1)

The reaction rate of equation (1) is given by:

rate = k*[O_{3}][NO]     (2)

<u>We have:</u>

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Hence, to find the inital reacion rate we will use equation (2):

rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s  

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

\frac{\Delta[NO_{2}]}{\Delta t} = rate

\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!                                

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