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3 years ago

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Calculate the approximate enthalpy of the reaction in joules. Estimate that 1.0 mL of vinegar has the same thermal mass as 1.0 m

L of water. It takes 4.2 joules to change 1.0 grm (1.0 mL) of water 1.0 Celcius. Is it endotherrmic or exothermic? Volume of vinegar 25 mL, Initial temp of vinegar 17 Celcius and final temp is 14 Celcius.

1 answer:

Nesterboy [21]3 years ago

5 0

Explanation:

The given data is as follows.

Density of vinegar = 1.0 g/ml

Specific heat capacity = 4.25 $J/g ^{o}C$

$T_{1}$ = $17 ^{o}C$ , and $T_{2}$ = $14 ^{o}C$

Relation between enthalpy and specific heat is as follows.

$\Delta H = mC \Delta T$

Hence, putting the values into the above formula as follows.

$\Delta H = mC \Delta T$

= $25 \times 1.0 \times 4.25 J/g ^{o}C \times -3^{o}C$ (as density = $\frac{mass}{volume}$ )

= - 315 J

Thus, we can conclude that the enthalpy of reaction is -315 J.

As the value is negative so, it means that heat is releasing. Hence, the reaction is exothermic in nature.

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Chemistry an atom of strontium has at least four different isotopes. what is different between an isotope of 86sr and an isotope

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Isotopes are atoms of the same element that differ in the number of neutrons.

Remember that all the atoms of an element have the same number of protons. So the only difference between isotopes of an element is the number of neutrons.

86 Sr means that the mass number of this isotope is 86. Also, remember that the mass number is the number of protons plus the number of neutrons.

87 Sr means that the mass number of this isotope is 87.

So, 86 Sr and 87 Sr differ 1 neutron.

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Squaring both sides, we get

$(\frac{r_{2}}{r_{1}})^{2} = \frac{M_{1}}{M_{2}}$

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2 years ago

Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbol

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Answer:

The Lewis structures are in image attached.

Explanation:

Lewis symbol is a representation of an element symbol along with its valence electrons around it in the form of dot(s).

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Sodium's atomic number is 11 in which only 1 electrons are present in its valence shell .So in order to gain noble gas stability it will loose 1 electron to completes its octet. In the Lewis symbol no dot shown as sodium has lost its 1 electron.

$Na=1s^22s^23p^63s^1$

$Na^+=1s^22s^23p^63s^0$

(c) Mg

Magnesium's atomic number is 12 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electrons to completes its octet.

In the Lewis symbol no dot shown as magnesium has lost its 2 electrons.

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(d)Ca

Calcium's atomic number is 20 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

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(e) K

Potassium's atomic number is 19 in which only 1 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 1 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 1 electron.

$K= 1s^22s^22p^63s^23p^64s^1$

$K^{+}=1s^22s^23p^6^23p^64s^0$

(f) Br

Bromine's atomic number is 35 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

$Br=1s^22s^22p^63s^23p^63d^{10}4s^24p^5$

$Br^-= 1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

(g) Sr

Strontium's atomic number is 38 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

$Sr=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2$

$Sr^{2+}=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^0$

(h) F

Florine's atomic number is 7 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet.

$F=1s^22s^22p^5$

$F^-=1s^22s^22p^6$

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3 years ago