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stiv31 [10]
3 years ago
11

A sample of gas has a mass of 0.560 g . Its volume is 125 mL at a temperature of 85 ∘C and a pressure of 757 mmHg .

Chemistry
1 answer:
-BARSIC- [3]3 years ago
4 0

Answer:

132g/mole

Explanation:

using the formula PV=nRT should be used to solve for the number of moles (n).  R is a constant which is 62.3637 L mmHG/mole K.

Inorder for your units to match you will have to convert 125ml to .125L and the temperature of 85C to K . you do that by adding 273 to the 85C and get 358K.  Once you solve for n then you use that number and divide by the number of grams from the question (.560g) since molar mass is grams/moles.

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A sample of 10.6 g of KNO3 was dissolved in 251.0 g of water at 25 oC in a calorimeter. The final temperature of the solution wa
finlep [7]

Answer:

36.55kJ/mol

Explanation:

The heat of solution is the change in heat when the KNO3 dissolves in water:

KNO3(aq) → K+(aq) + NO3-(aq)

As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.

To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:

<em>Moles KNO3 -Molar mass: 101.1032g/mol-</em>

10.6g * (1mol/101.1032g) = 0.1048 moles KNO3

<em>Change in heat:</em>

q = m*S*ΔT

<em>Where q is heat in J,</em>

<em>m is the mass of the solution: 10.6g + 251.0g = 261.6g</em>

S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-

And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C

q = 261.6g*4.184J/g°C*3.5°C

q = 3830.87J

<em>Molar heat of solution:</em>

3830.87J/0.1048 moles KNO3 =

36554J/mol =

<h3>36.55kJ/mol</h3>

<em />

6 0
2 years ago
Balance the following half-reaction: (acidic) NO−3(aq) → NO2(aq) Express your answer as a half-reaction. Identify all of the pha
Marrrta [24]

Answer:

2H⁺  + NO₃⁻  + 1e⁻ →  NO₂  + H₂O

Explanation:

NO₃⁻  →  NO₂

In left side, Nitrogen acts with +5 by oxidation number

In right side, the oxidation number is +4

This is a reduction reaction, because the oxidation number has decreased. So the N has gained electrons.

NO₃⁻  + 1e⁻ →  NO₂

In acidic medium, we have to add water, where there are less oxygens to ballance the amount. We have 2 O in left side, and 3 O in right side, so we have to add 1 H₂O on left side.

NO₃⁻  + 1e⁻ →  NO₂  + H₂O

Now that oxygens are ballanced, we have to ballance the hydrogens by adding protons in the opposite side

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8 0
3 years ago
Imagine that you have a 5.00 L gas tank and a 3.50 L gas tank. You need to fill one tank with oxygen and the other with acetylen
Lorico [155]

Answer:

77.14 atm of pressure should be of an acetylene in the tank.

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Moles of acetylene =n_2

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Volume of small tank with acetylene gas ,V_2= 3.50 L

Pressure of the acetylene gas inside the tank = P_2=?

RT=\frac{P_2V_1}{n_2} ..[2]

Considering both the gases having same temperature T, [1]=[2]

\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}

P_2=\frac{P_1V_1\times n_2}{V_2\times n_1}

=\frac{135 atm\times 5.00 L\times 2}{3.50 L\times 5}=77.14 atm

77.14 atm of pressure should be of an acetylene in the tank.

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