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2 years ago

How many grams of nitrogen are in a sample of Al(NO3)3 that contains 364.0 g of aluminum? Answer: ________ grams of nitrogen.

Chemistry

1 answer:

saw5 [17]2 years ago

7 0

Answer: The mass of nitrogen that is present for given amount of aluminium is 566.22 g.

Explanation:

We are given:

Mass of aluminium = 364 grams

The chemical formula of aluminium nitrate is $Al(NO_3)_3$

Molar mass of nitrogen = 14 g/mol

Molar mass of aluminium = 27 g/mol

In 1 mole of aluminium nitrate, 27 grams of aluminium is combining with 42 grams of nitrogen.

So, 364 grams of aluminium will combine with = $\frac{42}{27}\times 364=566.22g$ of nitrogen.

Hence, the mass of nitrogen that is present for given amount of aluminium is 566.22 g.

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Explanation:

Formula for work done is as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

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Now, we will put the given values into the above formula and calculate work done as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

= $\frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}$

= $3.68 \times 10^{-18} J$

Thus, we can conclude that work required to increase the separation of the two ions to an infinite distance is $3.68 \times 10^{-18} J$ .

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2 years ago

Approximately ____ days have passed when 20% of Cobalt (Co) remains? A) 19 B) 20 C) 21 D) 29

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2 years ago

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Answer:

The volume is 19.7 mL

Explanation:

Step 1: Given data

Pressure at sea level = 1.00 atm

Pressure at 50 ft = 2.47535 atm

kH for N2 in water at 25°C is 7.0 × 10−4 mol/L·atm

Molarity (M) = kH x P

Step 2: Calculate molarity

M at sea level:

M = 7.0*10^-4 * (1.00atm * 0.78) = 5.46*10^-4 mol/L

M at 50ft:

M = 7.0*10^-4 * (2.47535atm * 0.78) = 13.5*10^-4 mol/L

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Step 3: Calculate volume

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with P = 1.00 atm

with V = TO BE DETERMINED

with n = 8.04*10^-4 mol/L *1L = 8.04*10^-4

with R= 0.0821 atm * L/ mol *K

with T = 25 °C = 273+25 = 298 Kelvin

To find the volume, we re-organize the formula to: V=nRT/P

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The volume is 19.7 mL

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