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ladessa [460]
4 years ago
12

You wish to make a 0.339 m h2so4 solution from a stock solution of 12.0 m h2so4. how much concentrated acid must you add to obta

in a total volume of 100.00 ml of the dilute solution?
Chemistry
1 answer:
Alex_Xolod [135]4 years ago
8 0
When preparing diluted solutions from concentrated solutions , we can use the following equation;
c1v1 =c2v2
Where c1 and v1 are the concentration and volume of the concentrated solution
c2 is the concentration of the diluted solution to be prepared
v2 is the volume of the diluted solution
Substituting the values;
12.0 M x v1 = 0.339 M x 100 mL
v1 = 2.825 mL needs to be taken from the stock solution
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Calculate the change in temperature that occurs when 3.50 x 10^-2kg of copper heated by 635 J of energy. Ccu = 0.385 j/gxC
vovikov84 [41]

Answer:

The change in T° is 47.1 °C

Explanation:

Calorimetry formula to solve this:

Q = m . C . ΔT

We replace the data gien:

635 J = 3.50×10⁻² kg . 0.385 J/g°C . ΔT

In units of C, we have g and the mass (m) is in kg. So let's convert it from kg to g → 3.50×10⁻² kg . 1000 g / 1kg = 35 g. Now we can determine the ΔT:

635 J = 35 g . 0.385 J/g°C . ΔT

635 J / 35 g . 0.385 J/g°C = ΔT

47.1°C = ΔT

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The 1H-NMR of a compound with molecular formula C6H12O2 consists of four signals: 1.1 (triplet, integrating to 3 Hydrogens), 1.2
Tamiku [17]

Answer:

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Explanation:

1. Information from formula

The formula is C₆H₁₂O₂. A six-carbon alkane would have the formula C₆H₁₄. The deficiency of two H atoms indicates the presence of either a ring or a double bond.

2. Information from the spectrum

(a) Triplet-quartet

A 3H triplet and a 2H quartet is the classic pattern for a CH₃CH₂- (ethyl) group

(b) Septet-doublet

A 1H septet and a 6H doublet is the classic pattern for a -CH(CH₃)₂ (isopropyl) group

(c) The rest of the molecule

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The rest of the molecule must have the formula CO₂ and one unit of unsaturation. That must be a C=O group.

The compound is either

CH₃CH₂-COO-CH(CH₃)₂ or (CH₃)₂CH-COO-CH₂CH₃.

(d) Well, which is it?

The O atom of the ester function should have the greatest effect on the H atom on the adjacent carbon atom.

The CH of an isopropyl is normally at 1.7. The adjacent O atom should pull it down perhaps 3.2 units to 4.9.  

The CH₂ of an ethyl group is normally at 1.2. The adjacent O atom should pull it down to about 4.4.

We see a signal at 5.0 but none near 4.4. The compound is isopropyl propionate.

3. Summary

My peak assignments are shown in the diagram below.

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5. Fe0 (Fe: 56, O: 16) ---> 56+16=72
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