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kap26 [50]
3 years ago
6

Which of the following is an example of velocity

Chemistry
1 answer:
skelet666 [1.2K]3 years ago
3 0

Answer:

20 km/h North

Explanation:

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Using the spectra data below, which structure best matches this data? c4h8o 1h nmr triplet at 1.05 ppm (3h) singlet at 2.13 ppm
Delicious77 [7]

Answer:

Butan-2-one

Explanation:

1. 1700 cm⁻¹

A strong peak near 1700 cm⁻¹ is almost certainly a carbonyl (C=O) group.

2. Triplet-quartet

A triplet-quartet pattern indicates an ethyl group.

The 2H quartet is a CH₂ adjacent to a CH₃. The peak normally occurs at δ 1.3, but it is shifted 1.2 ppm downfield to δ 2.47 by an adjacent C=O group.

The 3H triplet at δ 1.05 is the methyl group. It, too, is shifted downfield from its normal position at δ 0.9. The effect is smaller, because the methyl group is further from the carbonyl.

3. 3H(s) at δ 2.13

This indicates a CH₃ group with no adjacent hydrogen atoms.

It is shifted 0.8 ppm downfield to δ 2.13 by the adjacent C=O group.

4. Identification

The identified pieces are CH₃CH₂-, -(CO)-, and -CH₃. There is only one way to put them together: CH₃CH₂-(C=O)-CH₃.

The compound is butan-2-one.

5 0
2 years ago
URGENT!!! What is the concentration in moles per litre of a solution that contains 7.68 g of H2SO4 in 150.0 mL of water?
hichkok12 [17]

Answer: 0.5225 mcl/l

Explanation:

5 0
2 years ago
Dissolution of KOH, ΔHsoln:
swat32

Using Hess's law we found:

1) By <em>adding </em>reaction 10.2 with the <em>reverse </em>of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)   ΔH  (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the <em>total </em>change in enthalpy (ΔH).    

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq)   ΔHsoln    (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq)     ΔHneut     (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from <u>differents changes</u> in various <em>reactions </em>can be calculated as the <u>sum</u> of all the <em>enthalpies</em> of all those <em>reactions</em>.      

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq)    (10.3)

We need to <em>add </em>reaction 10.2 to the <u>reverse</u> of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

<u>Canceling</u> the KOH(s) from both sides, we get <em>reaction 10.3</em>:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)    (10.3)

2) The change in enthalpy for <em>reaction 10.3</em> can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

\Delta H = \Delta H_{soln} + \Delta H_{neut}

The enthalpy of <em>reaction 10.1 </em>(ΔHsoln) changed its sign when we reversed reaction 10.1, so:

\Delta H = \Delta H_{neut} - \Delta H_{soln}

Therefore, the ΔHsoln must be <u>subtracted</u> from ΔHneut to get the total change in enthalpy ΔH.

Learn more here:

  • brainly.com/question/2082986?referrer=searchResults
  • brainly.com/question/1657608?referrer=searchResults  

I hope it helps you!

6 0
2 years ago
What are the rows of the periodic table of the elements also called?
lesya [120]

The rows are called Periods.

4 0
3 years ago
Read 2 more answers
Fluorite is..
denpristay [2]

Fluorite is harder than gypsum but softer than apatite. Thus, the correct option is B.

<h3>What is the hardness of any element?</h3>

The hardness of any element may be defined as the capability of a material to oppose the process of deformation and remains in actual shape precisely.  

According to the table of hardness scales by Mohs, the increasing order of given hardness of given elements is as follows:

Gypsum < Fluorite < Apatite.

Therefore, Fluorite is harder than gypsum but softer than apatite. Thus, the correct option is B.

To learn more about the Hardness of elements, refer to the link:

brainly.com/question/23721736

#SPJ1

5 0
1 year ago
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