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2 years ago

Areas in Mexico and Central America are known as

Chemistry

2 answers:

bixtya [17]2 years ago

5 0

It is known as Mesoamerica.

Tanzania [10]2 years ago

3 0

They are known as “American Central”

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Propose the structure of the following HNMR data.

Varvara68 [4.7K]

Answer:

1-Chloropropane is likely the answer (attached a picture)

Explanation:

First off there are 3 peaks and 3 carbons which indicates to me that this will be a chain without any symmetry and that each carbon has hydrogens on it.

Second the triplet at 1.0 that integrates to 3 likely correlates to a CH3 (methyl) group. Peaks are very upfield triplets that integrate to three are almost always methyl peaks.

Third the triplet at 3.7 is indicative of being next to the halogen. Hydrocarbons by themselves do not have peaks that far downfield meaning that its shift could only be explained by the chlorine being involved. Also we know that this can't be next to the methyl group since its multiplicity is to low to be next to it.

That leaves the multiplet at 1.75 being the hydrogens on the middle carbon which also makes sense since it is more downfield then the methyl group (due to being closer to the chlorine) but is not far enough downfield to say the chlorine is there. It also makes sense that it is a multiple since it would be a hextet due to the adjacent 5 hydrogens which can't always be resolved.

I hope this helps and let me know if anything is unclear or needs further explanation.

7 0

2 years ago

The titration of 25.00 ml a 0.125 m hclo4 solution requires 27.07 ml of koh to reach the endpoint. what is the concentration of

agasfer [191]

Answer : The concentration of the $KOH$ is, 0.115 M

Explanation :

Using dilution law,

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of an acid = 1

$n_2$ = acidity of a base = 1

$M_1$ = concentration of $HClO_4$ = 0.125 M

$M_2$ = concentration of KOH = ?

$V_1$ = volume of $HClO_4$ = 25 ml

$V_2$ = volume of NaOH = 27.07 ml

Now put all the given values in the above law, we get the concentration of the $KOH$ .

$1\times 0.125M\times 25ml=1\times M_2\times 27.07ml$

$M_2=0.115M$

Therefore, the concentration of the $KOH$ is, 0.115 M

5 0

3 years ago

Zebra mussels have no???

olchik [2.2K]

Zebras do ave muscles

4 0

3 years ago

The atomic number is the number of protons in the atom of the element

ipn [44]

Answer:

True

Explanation:

If you look closley at the nucleus, you don't count the neutrons just the prtons which then effect the electrons.

Good luck :)

5 0

3 years ago

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$

= (-3399.6) + (1560)

= -1839.6 KJ

5 0

3 years ago