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Ivahew [28]
2 years ago
12

What is the difference b/w combustion and synthesis reaction ?explain with example​

Chemistry
1 answer:
raketka [301]2 years ago
6 0

Answer:

<em>A </em><em>synthesis</em><em> </em><em>reaction </em><em>occurs </em><em>when </em><em>two </em><em>or </em><em>more </em><em>reactants</em><em> </em><em>combine </em><em>to </em><em>form </em><em>a </em><em>single </em><em>product </em><em>in </em><em>a </em><em>double </em><em>replacement </em><em>reaction </em><em>two </em><em>compounds</em><em> </em><em>exchange </em><em>elements. </em><em>Combustion</em><em> </em><em>reaction</em><em> </em><em>occurs </em><em>when </em><em>a </em><em>substance</em><em> </em><em>react</em><em>s</em><em> </em><em>quickly</em><em> </em><em>with </em><em>oxygen </em><em>combustion</em><em> </em><em>is </em><em>commonly</em><em> </em><em>called </em><em>burning. </em><em>Example</em><em>'</em><em>-</em><em> </em><em>for </em><em>synthesis</em><em> </em><em>reaction</em><em> </em><em>is </em><em>iron+</em><em> </em><em>oxygen </em><em>—</em><em>≥</em><em> </em><em>water </em><em>.</em><em> </em><em>Example:</em><em>-</em><em> </em><em>for </em><em>combustion</em><em> </em><em>reaction </em><em>is </em><em>often </em><em>gaseous </em><em>products</em><em>.</em>

<u><em>may </em><em>be </em><em>this </em><em>was </em><em>ur </em><em>answer</em><em> </em><em>I </em><em>hope </em><em>it </em><em>might </em><em>help</em><em> </em><em>u</em></u>

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Explanation:

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3 years ago
All physical changes can be identified at the macroscopic level. true or false
Aliun [14]

Answer:

True

Explanation:

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3 years ago
1. How is melting point of a solid determined?
Oxana [17]

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4 0
3 years ago
The formation constant for the reaction ag (aq) 2nh3(aq) ag(nh3)2 (aq) is kf = 1.7 × 107 at 25°c. what is δg° at this temperatur
anzhelika [568]

The value of ΔG° at this temperature is -18034.18 J/mol

Calculation,

Given information

formation constant (Kf)= 1.7 × 10^{7}

Universal gas constant (R) = 8.314 J/K• mol

Temperature = 25° C = 25 °C + 273 = 300 K

Formula used:

ΔG° = -RT㏑Kf

By putting the valur of R,T, Kf we get the value of ΔG°

ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 × 10^{7}

ΔG° = -2494.2㏑ 1.7 × 10^{7} = -18034.18 J/mol

So, change in standard Gibbs's free energy is -18034.18 J/mol

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1 year ago
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A. True

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3 years ago
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