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Tpy6a [65]
4 years ago
10

Mature elephants consume an average of 600 pounds of food per day. If an elephant

Chemistry
1 answer:
MAXImum [283]4 years ago
5 0

Answer:

\large \boxed{\text{4 600 000 kg or 4600 Mg}}

Explanation:

1. Calculate the adult life

If the elephant is an adult from its 10th birthday until the day before its 56th birthday, its  

Adult life = 55 - 10 + 1 = 46 yr

2. Convert years to days

\text{Adult life} = \text{46 yr} \times \dfrac{\text{365.25 da}}{\text{1 yr}} = \text{16 800 da}

3. Convert days to pounds of feed

\text{Feed} = \text{16 800 da} \times \dfrac{\text{600 lb}}{\text{1 da}} = 1.01 \times 10^{7} \text{ lb}

4. Convert pounds to kilograms and megagrams

\text{Feed} =1.01 \times 10^{7} \text{ lb} \times \dfrac{\text{1 kg}}{\text{2.20 lb}} =\textbf{4 600 000 kg} = \textbf{4600 Mg}\\\\\text{The elephant will eat $\large \boxed{\textbf{4 600 000 kg or 4600 Mg}}$ of food.}

Note: The answer can have only two significant figures because that is all you gave for age of the elephant.

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I shall assume the following equations are the ones of interest as none was provided:

(a) 63 Li + 56 28Ni →?

(b) 40 20Ca + 248 96Cm→147 62Sm + ?

(c) 88 38Sr + 84 36Kr→116 46Pd + ?

(d) 40 20Ca + 238 92U→70 30Zn + 4 10 n + 2?

I shall also assume in the following calculations that the mass number of the unknown products is 'a' and the atomic number of the unknown products is 'b'.

In balancing nuclear equations, we observe the following rules:

I. The total number of protons and neutrons in the products and reactants must be the same ( conservation of mass number, A ).

II. The total number of nuclear charges in the products and reactants must be the same ( conservation of atomic number, Z ).

For (a),

63 Li + 56 28Ni→?

Add mass number, A, of reactants = 6+56 = 62

Add atomic number, Z, of reactants = 3+28 = 31

A of reactants = A of unknown product = 62

and, Z of reactants = Z of unknown product = 31

Hence, unknown product is the element with Z as 31 = Ga

Balanced equation becomes:

63 Li + 56 28Ni → 62 31Ga

For (b),

40 20Ca + 248 96Cm→147 62Sm + ?

A of reactants = 40+248 = 288

A of products = 147 + 'a'

But A of reactants = A of products

288 = 147 + 'a'

‘a’ = 288 – 147 = 141

Also,

Z of reactants = 20 + 96 = 116

Z of products = 62 + 'b'

Z of reactants = Z of products

116 = 62 + 'b'

'b' = 116 – 62 = 54

So, unknown product is the element with Z = 54 = Xe

Balanced equation is:

40 20Ca + 248 96Cm→147 62Sm + 141 54Xe

For (c),

88 38Sr + 84 36Kr→116 46Pd + ?

A of reactants = 88 + 84 = 172

A of products = 116 + 'a'

But A of reactants = A of products

172 = 116 + 'a'

‘a’ = 172 – 116 = 56

Also,

Z of reactants = 38 + 36 = 74

Z of products = 46 + 'b'

Z of reactants = Z of products

74 = 46 + 'b'

'b' = 74 – 46 = 28

So, unknown product is the element with Z = 28 = Ni

Balanced equation is:

88 38Sr + 84 36Kr→116 46Pd + 56 28Ni

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40 20Ca + 238 92U→70 30Zn + 4 10 n + 2?

A of reactants = 40 + 238 = 278

A of products = 70 + (4 x 1) + (2 x 'a') = 70 + 4 + 2a = 74 + 2a

But, A of reactants = A of products

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Z of reactants = 20 + 92 = 112

Z of products = 30 + (4 x 0) + (2 x b) = 30 + 0 + 2b = 30 + 2b

Z of reactants = Z of products

112 = 30 + 2b

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Therefore, unknown product is element with Z = 41 =Nb

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40 20Ca + 238 92U→70 30Zn + 4 10 n + 2 102 41Nb

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