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Vlad1618 [11]
3 years ago
6

Morphine has the formula c17h19no3. it is a base and accepts one proton per molecule. it is isolated from opium. a 0.685 −g samp

le of opium is found to require 8.91 ml of a 1.16×10−2 m solution of sulfuric acid for neutralization. part a assuming that morphine is the only acid or base present in opium, calculate the percent morphine in the sample of opium.
Chemistry
1 answer:
kotykmax [81]3 years ago
7 0
2 C₁₇H₁₉NO₃ + H₂SO₄ → Product
Moles of H₂SO₄ = M x V(liters) = 0.0116 x 8.91/1000 = 1.033 x 10⁻⁴ mole 
moles of morphine = 2 x moles of H₂SO₄ = 2.066 x 10⁻⁴
Mass of morphine = moles x molar mass of morphine = 2.066 x 10⁻⁴ x 285.34 
                             = 0.059 g
percent morphine = \frac{mass of morphine}{mass of opium} x 100
                             = \frac{0.059}{0.685} x 100 = 8.6 %   
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10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C
Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

Q=mc\Delte T

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c =  0.385 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c =  0.903 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c =  2.42 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c =  4.18J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c =  0.128 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0
3 years ago
During the formation of an ionic compound, the element with the higher ionization energy will gain electrons.
Lesechka [4]
False because you have to take it out and do it right
8 0
3 years ago
PLS HELP ME I WILL GIVE BRAINLIEST TO U I JUST NEED ANSWERS ASAP
elena55 [62]

Answer:

is this multiple choice just wondering

4 0
3 years ago
If 48 g of magnesium reacts with oxygen gas, how many grams of magnesium oxide will be formed according to the following equatio
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4 0
3 years ago
Read 2 more answers
The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates:
alexira [117]

Answer:

Kc = 6x10⁻⁶

Explanation:

For the reaction:

4NH₃(g) + 3O₂(g) ⇄ 2N₂(g) + 6H₂O(g)

Kc is defined as:

Kc =[N₂]² [H₂O]⁶ / [NH₃]⁴ [O₂]³

The equilibrium concentrations of the gases is -Because volume of the container is 1.00L-:

[N₂] = 2X = 1.96x10⁻³; <em>X = 9.8x10⁻⁴</em>

[H₂O] = 6X; 6ₓ9.8x10⁻⁴ = 5.88x10⁻³

[NH₃] = 0.0150M - 4X = 0.01108M

[O₂] = 0.0150M - 3X = 0.01206M

Replacing in Kc expression:

Kc =[1.96x10⁻³]² [5.88x10⁻³]⁶ / [0.01108M]⁴ [0.01206M]³

<h3>Kc = 6x10⁻⁶</h3>

8 0
3 years ago
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