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Veseljchak [2.6K]
3 years ago
9

An alloy is a mixture of what?

Chemistry
1 answer:
Nastasia [14]3 years ago
5 0

Answer:

An alloy is a mixture of chemical elements, which form an impure substance that retains the characteristics of a metal.

Explanation:

Have a nice day. Auf Wiedersehen.

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For resonance to occur, the object must experience a frequency equal to its what?
babunello [35]

Answer:

A. Natural Frequency

Explanation:

I took the test. (100%)

8 0
3 years ago
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pesticides and fertilizers can help farmers to produce more crops. However, overuse of these chemicals can result in?
nadezda [96]
Overuse of the same chemicals can result in the pest becoming immune to the pesticides. 
8 0
3 years ago
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°
miskamm [114]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1835 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)    \Delta H_1=157kJ   ( × 4)

(2) P_4(g)+6Cl_2(g)\rightarrow 4PCl_3(g)     \Delta H_2=-1207kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(4\times (-157))+(1\times (-1207))=-1835kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1835 kJ.

6 0
3 years ago
Balance the following redox equation and identify the element oxidized, the element reduced, the oxidizing agent, and the reduci
earnstyle [38]

The charge of Br changed from –1 to 0, therefore it is the element which is oxidized. Since it is oxidized then Br is also the reducing agent.

 

The charge of Mn changed from +4 to +2 therefore it is the element which is reduced. Since Mn is reduced, then MnO2 is the oxidizing agent.

 

The half –reactions are:

Br: 2Br --> Br2 + 2e-

Mn: MnO2 --> Mn2+

First balance oxygen by adding H2O:

MnO2 --> Mn2+ + 2H2O

Then balance hydrogen by adding H+ ions:

4H+ + MnO2 --> Mn2 + 2H2O

Then the appropriate electrons:

4e- + 4H+ + MnO2 --> Mn2 + 2H2O

 

Multiply the half-reaction of Br by 2 because the half-reaction of Mn has 4 electrons.

4Br --> 2Br2 + 4e-

 

Combine the two half reactions and cancel common factors:

4Br-  +  4H+  +  MnO2 --> 2Br2  +  Mn2  +  2H2O

6 0
3 years ago
1. Fill in the best answer for each of the following: a) ________________________ is the capacity to do work or to supply heat.
user100 [1]

Answer:

A. Energy

B. Chemical potential

C. Endothermic reaction

D.The law of conservation of mass

E. Transfers of energy

5 0
3 years ago
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