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Drupady [299]
3 years ago
12

A scientist has a container with a substance inside. At first, the molecules in the substance move away from each other. Later,

the molecules move around each other. What change did the scientist observe in the substance
Chemistry
1 answer:
den301095 [7]3 years ago
3 0

Answer:

before the substance was a gas it was a liquid.

Explanation:

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What is the value of work done on an object when a 50–newton force moves it 15 meters in the same direction as the force?
victus00 [196]

Answer:

work done = 750 J

Explanation:

Given data:

Force on object = 50 N

Distance covered = 15 m

Work done = ?

Solution:

W = F. d

W = work done

F = force

d = distance

Now we will put the values in formula.

W = 50 N × 15 m

W = 750 N. m = 750 J

7 0
3 years ago
Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2S(g) 3O2(g)2H2O(l) 2SO2
Marizza181 [45]

Answer:

\Delta _rH=-1124.14kJ/mol

Explanation:

Hello!

In this case, since the standard enthalpy change for a chemical reaction is stood for the enthalpy of reaction, for the given reaction:

2H_2S(g) +3O_2(g)\rightarrow 2H_2O(l) +2SO_2(g)

We set up the enthalpy of reaction considering the enthalpy of formation of each species in the reaction at the specified phase and the stoichiometric coefficient:

\Delta _rH=2\Delta _fH_{H_2O,liq}+2\Delta _fH_{SO_2,gas}-2\Delta _fH_{H_2S,gas}-3\Delta _fH_{O_2,gas}

In such a way, by using the NIST database, we find that:

\Delta _fH_{H_2O, liq}=-285.83kJ/mol\\\\\Delta _fH_{SO_2, gas}=-296.84kJ/mol\\\\\Delta _fH_{O_2,gas}=0kJ/mol\\\\\Delta _fH_{H_2S,gas}=-20.50kJ/mol

Thus, we plug in the enthalpies of formation to obtain:

\Delta _rH=2(-285.73kJ/mol)+2(-296.84kJ/mol)-2(-20.50kJ/mol)-3(0kJ/mol)\\\\\Delta _rH=-1124.14kJ/mol

Best regards!

8 0
3 years ago
B. The following reaction takes place in a basic solution. (7 points)
Natali5045456 [20]

<u>Answer:</u> The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

<u>Explanation:</u>

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)

To balance the given redox reaction in basic medium, there are few steps to be followed:

  • Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electrons

Oxidation half-reaction: NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-

Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-

  • Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions

Oxidation half-reaction: NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-         ( × 3)

Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-             ( × 2)

The half-reactions now become:

Oxidation half-reaction: 3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-

Reduction half-reaction: 2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-

  • Add the equations and simplify to get a balanced equation

Overall redox reaction: 3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

3 0
3 years ago
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