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mina [271]
3 years ago
5

Write the complete symbol, including mass number and atomic number, for each atom.

Chemistry
1 answer:
Reil [10]3 years ago
8 0

Answer:

a) 58Ni

b) 43Ti

c) 34P

d) 18O

e) 58Ni

f) 43Ti

g)34P

h) 18O

i) 54Cr

Explanation:

<em>a) contains 28 protons and 30 neutrons:</em>

An atom with 28 protons and 30 neutrons has an atomic number of 28 and a mass number of (30+28) = 58.

The atom with atomic number 28 is Ni.

<em>b) contains 22 protons and 21 neutrons:</em>

An atom containing 22 protons and 21 neutrons has an atomic number of 22 and an atomic mass of (22+21) = 43

The atom with atomic number 22 is Titanium (Ti).

This is the 43Ti isotope of titanium

<em>c)  contains 15 electrons and 19 neutrons</em>

An atom with 15 electrons would have the same amount of , 15 protons (considering it to be neutral) and has 19 neutrons.

Thus, the atomic number of the atom would be 15 and the mass number would be (15+19) = 34

The atom with atomic number 15 is phosphorus (P).

This is the 34P isotope of phosphorus

<em>d) an oxygen atom with 10 neutrons:</em>

Oxygen has an atomic number of 8, thus there are 8 protons in an atom of oxygen.

The mass number of the oxygen atom with 10 neutrons would be (8+10) = 18

This is 18O - isotope.

<em>e) contains 28 protons and 30 neutrons:</em>

An atom with 28 protons and 30 neutrons has an atomic number of 28 and a mass number of (30+28) = 58.

The atom with atomic number 28 is Ni.

<em>f) contains 22 protons and 21 neutrons:</em>

An atom containing 22 protons and 21 neutrons has an atomic number of 22 and an atomic mass of (22+21) = 43

The atom with atomic number 22 is Titanium (Ti).

This is the 43Ti isotope of titanium

<em>g) contains 15 electrons and 19 neutrons:</em>

An atom with 15 electrons would have the same amount of , 15 protons (considering it to be neutral) and has 19 neutrons.

Thus, the atomic number of the atom would be 15 and the mass number would be (15+19) = 34

The atom with atomic number 15 is phosphorus (P).

This is the 34P isotope of phosphorus

<em>h) an oxygen atom with 10 neutrons:</em>

Oxygen has an atomic number of 8, thus there are 8 protons in an atom of oxygen.

The mass number of the oxygen atom with 10 neutrons would be (8+10) = 18

This is 18O - isotope.

<em>i) The atomic number of Cr is 24 </em>

The mass number is given to be 54.

This is the 54Cr isotope and has 24 protons and 30 neutrons

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<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ

Hence, the standard heat for the given reaction is -138.82 kJ

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as given that rate is first order with respect to ICl and second order with respect to H2

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5 0
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Nicotine is 74.1% carbon, 8.6% hydrogen, and 17.3% nitrogen by mass. What is its molecular formula if its molar mass is 162.26 g
WINSTONCH [101]

first we need to find the empirical formula of nicotine

empirical formula is the simplest ratio of whole numbers of elements making up a compound

the percentage compositions for each element has been given. So we can calculate for 100 g of the compound.

masses of elements and the number of moles

C - 74.1 g - 74.1 g/12 g/mol = 6.17 mol

H - 8.6 g - 8.6 g / 1 g/mol = 8.6 mol

N - 17.3 g - 17.3 g / 14 g/mol = 1.23

divide all by the least number of moles

C - 6.17 / 1.23 = 5.01

H - 8.6 / 1.23 = 6.99

N - 1.23 / 1.23 = 1.00

when the atoms are rounded off to the nearest whole numbers

C - 5

H - 7

N - 1

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we have to find what the mass of 1 empirical unit is

mass - 5 x 12 g/mol + 7 x 1 g/mol + 14 g/mol = 81 g

molecular mass is 162.26 g/mol

we have to find how many empirical units make up 1 molecule

number of empirical units = molecular mass / mass of 1 empirical unit

= 162.26 g/mol / 81 g = 2.00

there are 2 empirical units

molecular formula is - 2 (C₅H₇N)

molecular formula - C₁₀H₁₄N₂

4 0
3 years ago
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