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vagabundo [1.1K]
4 years ago
9

A sample of gas with an initial volume of 28.4 Liters at a pressure of 725 mmHg and a temperature of 305 K is compressed to a vo

lume of 14.8 Liters and warmed to a temperature of 375 Kelvin. What is the final pressure of the gas?
Chemistry
1 answer:
Lerok [7]4 years ago
6 0

Answer:

The final pressure is 2.25 atm or 1710 mm Hg

Explanation:

Step 1: Data given

The initial volume = 28.4 L

The initial pressure = 725 mm Hg ( = 725/760 atm) = 0.953947 atm

The initial temperature = 305 K

The new volume is 14.8 L

The new temperature = 375 K

Step 2: Calculate the new pressure

(P1*V1)/T1 = (P2*V2)/T2

⇒ with P1 = the initial pressure = 725 mmHg = 0.953947 atm

⇒ with V1 = the initial volume = 28.4 L

⇒ with T1  = The initial temperature = 305 K

⇒ with P2 = the new pressure = TO BE DETERMINED

⇒ with V2 = the new volume = 14.8 L

⇒ with T2 = the new temperature = 375 K

(0.953947 * 28.4)/305 = (P2 * 14.8)/375

P2 = 2.25 atm = 1710 mm Hg

The final pressure is 2.25 atm or 1710 mm Hg

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6 0
3 years ago
A gram of gasoline produces 45.0kj of energy when burned. gasoline has a density of 0.77/gml . how would you calculate the amoun
babymother [125]
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5 0
3 years ago
Consider the following balanced reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is exce
adoni [48]

Answer:

The answer is "10.84 g".

Explanation:

The formula for calculating the number for moles:

\text{Number of moles }= \frac{\ Mass}{ \ molar \ mass }

In the given acid nitric:

Owing to the nitric acid mass = 75.9 g

Nitric acid molar weight= 63\  \frac{g}{mol}

If they put values above the formula, they receive:

\text{moles in nitric acid} = \frac{75.9}{63}

                               =1.204 \ mol

In the given chemical equation:

3 NO_2 \ (g) + H_2O \ (l) \longroghtarrow 2 HNO_3 \ (aq) + NO\ (g)

In this reaction, 2 mols of nitric acid are produced by 1 mole of water.

So, 1.204 moles of nitric acid will be produced:

= \frac{1}{2} \times 1.204

=0.602\ \ \text{mol of water}.

We are now using Equation 1 in determining the quantity of water:

Water moles = 0.602\  mol

Water weight molar = 18.02 \ \frac{g}{mol}

\to 0.602 = \frac{\text{mass of mols}}{ 18.02}\\\\\to \text{mass of mols} = 0.602 \times 18.02\\\\

                         =10.84 \ g

8 0
4 years ago
What is the percent yield of NaCl if 31.0 g of CuCl2 reacts with excess NaNO3 to produce 21.2 g of NaCl?
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First you need to find the balanced chemical formula.
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Then you can find out how much NaCl 31.0g of CuCl₂ should produce using stoichiometry.  Divide 31.0g by the molar mass of CuCl₂ (134.446g/mol) to get 0.2306mol CuCl₂.  Than multiply 0.2306mol CuCl₂ by 2 to get 0.4612mol NaCl.  Than multiply 0.4612mol by the molar mass of NaCl (58.45g/mol) to get 26.95g of NaCl.
that means that 100% yield would give you 26.95g of NaCl so to find percent yield divide 21.2 by 26.95 to get 0.7867 which is 78.7% yield

Therefore the answer is 78.7% yield.
I hope this helps.  Let me know if anything is unclear
3 0
3 years ago
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