Increasing temperature increases solubility so that 480 grams (g) of sugar makes a saturated solution in 100 milliliters (mL) of
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This problem is providing us with the mass equivalent to one troy ounce. Thus, the troy ounces of gold in one short ton of average Nevada ore is required and found to be the 0.103 otz according to the following dimensional analysis.
<h3>Dimensional analysis</h3>In chemistry, a raft of problems do not always provide an equation in order to be solved yet dimensional analysis can be applied, so as to obtain the desired amount in the required units.
Thus, since this problem asks for try ounces in an average Nevada ore, which has 3.2 grams of gold per short ton of ore, one can solve the following setup in order to obtain the required answer in otz:
Where the short tons are cancelled out as well as the grams, in order to obtain:
Learn more about dimensional analysis: brainly.com/question/10874167
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:
8.5 mol H₂SO₄
Explanation:
It seems the balanced reaction the problem is referring to is absent, however the description matches the following balanced reaction:
- 2SO₂ + O₂ + 2H₂O → 2H₂SO₄
Now we <u>can convert 8.5 moles of SO₂ into moles of H₂SO₄</u>, using <em>the stoichiometric coefficients of the balanced reaction</em>:
- 8.5 mol SO₂ *
= 8.5 mol H₂SO₄