Muscles bones no on organs and no on skin
Answer:
The heat of reaction when hydrogen and oxygen combine to form water is :
<u>C. 571.6 kJ</u>
Explanation:
Enthalpy Change = The enthalpy change for the formation of 1 mole of the substance from their standard state is called the enthalpy of formation.
This is intensive quantity as it is fixed for 1 mole .
Intensive properties = Those properties which are independent on the amount of the substance are intensive properties.
The value of these quantities does not get halve if you divide the substance into two equal parts. example , density, refractive index.
However , the enthalpy of reaction is extensive. Because on increasing the amount the value of the enthalpy also get doubles
Hence for this reaction :

Its value is -285.8 kJ for 1 mole
And here two moles are present . so the value of molar enthalpy is:
-285.8 x 2 = -571.6 kJ
The true statements are B, C, and D
-first ( A ) the concentrations of reactants and products are equal is false, As the concentrations of reactants and products may be different from each other.
- But (B) the concentrations of reactants and product remains constant is true, as the equilibrium remains when there is no change in the concentration of the reactants and products.
-(c) reactants are being converted to products (and vise verse) is true also, as there are reactions still happened at a constant rate so it looks like nothing is happening.
Answer:
See below.
Step-by-step explanation:
Ethers react with HI at high temperature to produce an alky halide and an alcohol.
R-OR' + HI ⟶ R-I + H-OR'
<em>Benzylic ethers</em> react by an Sₙ1 mechanism by forming the stable benzyl cation.
- PhCH₂-OR + HI ⟶ PhCH₂-O⁺(H)R + I⁻ Protonation of the ether
- PhCH₂-O⁺(H)R ⟶ PhCH₂⁺ + HOR Sₙ1 ionization of oxonium ion
- PhCH₂⁺ + I⁻ ⟶ PhCH₂-I Nucleophilic attack by I⁻
If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:
ROH +HI ⟶ R-I + H-OH
Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).
The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).