Answer:
<em>o</em>-bromotoluene, <em>m</em>-bromotoluene and <em>p</em>-bromotoluene.
Explanation:
Hello,
In this case, on the attached picture you will find the reaction which yields <em>o</em>-bromotoluene as the first product, <em>m</em>-bromotoluene as the second product and <em>p</em>-bromotoluene as the last one since the substitution could be done at the second (ortho), third (meta) or fourth (para) carbons on the toluene.
Regards.
It looks like we are solving for a pressure. All that is required is some algebraic manipulation to find our pressure in mmHg.
Given:
(5.0 m³)(7.5 mmHg) = (P)(4.0m³)
Multiply:
37.5 = 4.0P
Divide:
9.375 = P
P = 9.4 mmHg (remember sig figs)
<h3>
Answer:</h3>
9.4 mmHg
HNO2 is the formula for Nitrous Acid
Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)
Answer:
-3670.33 J/K
Explanation:
Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.
Mathematically, change of Entropy can be expressed as,
ΔS = ΔH/T ....................................... Equation 1
Where ΔS = Change of entropy, ΔH = heat change, T = temperature.
ΔH = -(Lf×m).................................... Equation 2
Note: ΔH is negative because heat is lost.
Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg
Substitute into equation
ΔH = -(3.34×10⁵×3.0)
ΔH = - 1002000 J.
But T = 0 °C = (0+273) K = 273 K.
Substitute into equation 1
ΔS = -1002000/273
ΔS = -3670.33 J/K
Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C
