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Andreyy89
4 years ago
9

a rough guide to fluid requirements based on body weight is 100mL/kg for the first 10kg of body weight, 50mL/kg for the next 10k

g, and 20mL/kg for weight over 20kg. What volume of fluid per day is needed by a 55kg woman? give answer to 2 significant figures
Chemistry
1 answer:
jeka57 [31]4 years ago
4 0

Given, weight of the woman = 55 kg

For the first 10 kg, fluid required is 100 mL/kg

So for the first 10 kg volume of fluid required is 100 ml/kg  x 10 kg = 1000 ml = 1 L

For the second 10 kg, fluid required is 50 mL/kg

So for the second 10 kg volume of fluid required is 50 ml/kg x 10 kg = 500 ml = 0.5 L

For the remaining (55 kg - 20 kg ) or 35 kg, fluid required is 20mL/kg

Now for the the remaining 35 kg  volume of fluid required is 20 ml/kg x 35 kg = 700 ml = 0.7 L

The total volume of fluid per day required by a 55 kg woman = 1L + 0.5L + 0.7L = 2.2 L

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This means that for every mole of nitrogen consumed, 3 moles of hydrogen are consumed.

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The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc
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Answer: 9.9 grams

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\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

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b) moles of C_2H_4

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1 mole of C_2H_4 combine with 1 mole of H_2

Thus 0.33 mole of C_2H_4 will combine with =\frac{1}{1}\times 0.33=0.33 mole of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product.

As 1 mole of C_2H_4 give =  1 mole of C_2H_6

Thus 0.33 moles of C_2H_4 give =\frac{1}{1}\times 0.33=0.33moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g

Thus theoretical yield (g) of C_2H_6 produced by the reaction is 9.9 grams

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3 years ago
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