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nlexa [21]
3 years ago
11

If you decreased the volume of a sample of gas by a factor of three while maintaining a constant

Chemistry
2 answers:
anyanavicka [17]3 years ago
4 0

If you decreased the volume of a sample of gas by a factor of three while maintaining a constant pressure, how would the absolute temperature of the gas be affected?

✔ it would decrease threefold

A gas at 300 K and 4.0 atm is moved to a new location with a temperature of 250 K. The volume changes from 5.5 L to 2.0 L. What is the pressure of the gas at the new location?

✔ 9.2 atm

What is the partial pressure of 0.50 mol Ne gas combined with 1.20 mol Kr gas at a final pressure of 730 torr?

✔ 215 torr

Yuki888 [10]3 years ago
4 0

Answer:

I did the assignment

Explanation:

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Where does oxidation occur in an electrochemical cell?
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Option No. 1 at the cathode in both an electrolytic cell and a voltaic cell

Negative ion arrives at the anode, attracted by electrostatic forces, where they lose electrons. Loss of electrons is defined as oxidation.

The reverse process occur at the negative electrode, where positive ions pick up electrons. They are reduced.
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4 years ago
Read 2 more answers
For the following balanced equation: 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) a) How many moles of HNO3 will r
s344n2d4d5 [400]

Answer:

a) <u>26.67 moles HNO3 </u>

b) <u>0.33 moles NO</u>

c) <u>0.40 moles NO is produced</u>

d)<u>.157 moles Cu</u>

e) <u>0.105 moles NO</u>

f) <u>26.4 grams HNO3</u>

g) <u>Cu is in excess</u>

h) <u>2.41 grams Cu remain</u>

i) <u>2.37 grams NO</u>

Explanation:

Step 1: Data given

Molar mass of Cu = 63.55 g/mol

Molar mass of HNO3 = 63.01 g/mol

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)

a) How many moles of HNO3 will react with 10 moles of Cu?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 10 moles Cu we need 8/3 *10 = <u>26.67 moles HNO3 </u>

b) How many moles of NO will form if 0.50 moles of Cu reacts?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.50 moles Cu we'll have 2/3 *0.50 = <u>0.33 moles NO</u>

c) If 0.80 moles of H2O forms, how much NO must also form?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

If 0.80 moles H2O is produced, 0.80/2 = <u>0.40 moles NO is produced</u>

d) How many moles of Cu are in 10.0 grams of Cu?

Moles Cu = 10.0 grams / 63.55 g/mol = 0.157 moles

In 10.0 grams Cu we have <u>0.157 moles Cu</u>

e) If 10.0 g of Cu reacts, how many moles of NO will form?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll have 2/3 * 0.157 = <u>0.105 moles NO</u>

f) If 10.0 g of Cu reacts, how many grams of HNO3 are required?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll need 0.419 moles HNO3

This is 0.419 moles * 63.01 g/mol = <u>26.4 grams HNO3</u>

g) If 10.0 g of Cu and 20.0 g of HNO3 are put together in a reaction vessel, which one will be in excess?

Moles Cu = 0.157 moles

Moles HNO3 = 20.0 grams / 63.01 g/mo = 0.317 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

The limiting reactant is HNO3. It will completely be consumed (0.317 moles). <u>Cu is in excess.</u> There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

h) How many grams of the excess substance will be left over?

There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

This is 0.038 moles * 63.55 g/mol = 2.41 grams

i) How many grams of NO will form in the reaction described in part g?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.317 moles HNO3 we'll have 0.317/4 = 0.0793 moles NO

This is 0.079 mol * 30.01 g/mol =<u> 2.37 grams NO</u>

3 0
3 years ago
Drag each label to the correct location on the chart. Classify each reaction as endothermic or exothermic.
fomenos

A reaction is exothermic if Δ<em>H</em> (or \Delta H _\text{rxn} in some textbooks) is negative:

  • H₂ + Br → 2 HBr, ΔH < 0.
  • CH₄ + 2 O₂ → CO₂ + 2 H₂O, ΔH < 0.

A reaction is endothermic if Δ<em>H</em> is positive:

  • 2 NH₃ → N₂ + 3 H₂, ΔH > 0.
  • 2 HCl → H₂ + Cl₂ ΔH > 0.
<h3>Explanation</h3>

The enthalpy of a system is the sum of its internal energy. ΔH < 0 indicates that the reactants lose internal energy in the reaction. Energy conserves, and those internal energies must have converted to some other form of energy. They typically end up as thermal energy. The reaction will release heat since it is exothermic.

Similarly, ΔH > 0 indicates that the reactants gains internal energy in the reaction. Energy conserves. As a result, the reaction must have gained energy from its surroundings. The reaction will be endothermic since it absorbs heat.

5 0
3 years ago
Help me with my work please thanks?!
masya89 [10]

Answer:

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