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uranmaximum [27]
2 years ago
9

1. How many mg of lithium phosphide are in 18.0 mL of a 1.25 M solution?

Chemistry
1 answer:
Temka [501]2 years ago
6 0

Answer:

1160mg

Explanation:

Molarity = number of moles ÷ volume

According to the information in the question, molarity = 1.25 M, volume = 18.0 mL = 18/1000 = 0.018L

M = n/V

n = M × V

n = 1.25 × 0.018

n = 0.0225moles.

Using mole = mass/molar mass, to find the mass of lithium phosphide (Li3P)

Molar mass of Li3P = 6.9(3) + 31 = 51.7g/mol

mole = mass/molar mass

0.0225 = mass/51.7

mass = 1.16grams.

In milligrams (mg), mass of Li3P = 1.16 × 1000 = 1160mg

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What nuclide undergoes fission to form molybdenum-103, atomic number 42, tin-131, atomic number 50, and two neutrons?
FrozenT [24]

Answer:

The correct answer is Pu, 234.

Explanation:

In the given case, let us consider the reactant as X. Now the mass number (balanced) on both the sides will be,

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M = 1 * 103 + 1 * 131 + 2 * 0

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Now the atomic number (balanced) on both the sides,

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A = 1*42 + 1*50 + 2*1

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Answer:

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Explanation:

Hello,

In this case, the combustion of methane is shown below:

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Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

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Best regards.

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