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>
I think <span>3H2+N2==>2NH3</span>
Answer:
No, there is no evidence that the manufacturer has a problem with underfilled or overfilled bottles, due that according our results we cannot reject the null hypothesis.
Explanation:
according to this exercise we have the following:
σ^2 =< 0.01 (null hypothesis)
σ^2 > 0.01 (alternative hypothesis)
To solve we can use the chi-square statistical test. To reject or not the hypothesis, we have that the rejection region X^2 > 30.14
Thus:
X^2 = ((n-1) * s^2)/σ^2 = ((20-1)*0.0153)/0.01 = 29.1
Since 29.1 < 30.14, we cannot reject the null hypothesis.
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
Answer:
The answer is given below.
Explanation:
We will consider the acid as HA and will set up an ICE table with the equilibrium dissociation of α.
AT pH 2.4 the initial H+ concentration will be 3.98^10-3 M
HA → H+ + A-
Initial concentration: 0.1 → 3.98 ^10-3 + 0
equilibrium concentration: 0.1(1-α) → 3.98 * 10-3 + 0.1α 0.1α
pKa of chloroacetic acid is 2.9
-log(Ka) = 2.9
Ka = 1.26 * 10-3
From the equation, Ka = [H+] * [A-] / [HA]
1.26 * 10-3 = (3.98 * 10-3 + 0.1α )* 0.1α / 0.1(1-α)
Since α<<1, we assume 1-α = 1
Solving the equation, we have: α = 0.094
Since this is the fraction of acid that has dissociated, we can say that % of base form = 100 * α= 9.4%
