The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ of ) =-1275.0
enthalpy of combustion of oxygen(Δ of ) = zero
enthalpy of combustion of carbon dioxide(Δ of ) = -393.5
enthalpy of combustion of water(Δ of ) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Answer:
1.3×10⁻³ M
Explanation:
Hello,
In this case, given the dissociation reaction of acetic acid:
We can write the law of mass action for it:
Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change due to the dissociation extent, we are able to rewrite it as shown below:
Thus, via the quadratic equation or solve, we obtain the following solutions:
Obviously, the solution is 0.00133M which match with the hydronium concentration, thus, answer is: 1.3×10⁻³ M in scientific notation.
Regards.
Hello! I can help you with this. First, convert them into it’s written out standard form. 10^4 is 10,000. 10,00 * 1.26 is 12,600. 10,000 * 2.5 is 25,000. 12,600 + 25,000 = 37,600 or 3.76 * 10^4 in scientific notation. The answer in scientific notation is 3.76 * 10^4.
The best and most correct answer among the choices provided by your question is the fourth choice.
Copper <span>has an average atomic mass of about 63.5 </span>amu<span>.</span>
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