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Sholpan [36]
3 years ago
7

The complete combustion of 0.611 g of a snack bar in a calorimeter (Ccal = 6.15 kJ/°C) raises the temperature of the calorimeter

by 1.29 °C. Calculate the food value (in Cal/g) for the snack bar.
Chemistry
1 answer:
Bogdan [553]3 years ago
4 0
Focus your attention to the units of the given data and the unit of the final measurement you're going to solve. Through dimensional analysis,you know that the working equation would be:

Food Value = CcalΔT/m
Food value = (6.15 kJ/°C)(1.29 °C)/0.611 g = 12.98 kJ/g

Convert kJ to cal (1 kJ = 239 cal)
Food value = 12.98*239 =<em> 3,103.28 cal/g</em>
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A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many m
oksian1 [2.3K]

Answer : The 866.66\mu L must be administered.

Solution :

As we are given that a vial containing radioactive selenium-75 has an activity of 3.0mCi/mL.

As, 3.0 mCi radioactive selenium-75 present in 1 ml

So, 2.6 mCi radioactive selenium-75 present in \frac{2.6mCi}{3.0mCi}\times 1ml=0.86666ml\times 1000=866.66\mu L

Conversion :

(1ml=1000\mu L)

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4 0
3 years ago
How many moles of NH₃ can be produced from 4.81 moles of nitrogen in the following reaction:
makkiz [27]

Answer:

9.62moles of NH₃

Explanation:

Given parameters:

Number of moles of nitrogen  = 4.81moles

Unknown:

Number of moles of  NH₃ = ?

Solution:

To solve this problem, we need to establish a balanced reaction equation:

            N₂  +  3H₂  →  2NH₃

Now, we can solve the problem by working from the known to the unknown.

          1 mole of N₂ produced 2 moles of NH₃

         4.81moles of N₂ will produce 2 x 4.81  = 9.62moles of NH₃

8 0
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Answer:

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6 0
3 years ago
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nasty-shy [4]

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Solution:

Molar mass of CH₄O (Methanol) is 32 g.mol⁻¹.

It means,

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Then,

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Solving for X,

                    X  =  (69 g × 4 g) ÷ 32 g

                    X  =  8.62 g of Hydrogen

6 0
3 years ago
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Answer:2,4&5 A.

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