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2 years ago

Give the postional isomers of 2 halopropane(CH3)2CH

Chemistry

1 answer:

s344n2d4d5 [400]2 years ago

6 0

Give the positional isomers of 2 halopropane

(CH3)2CH

answer:

is in file download it

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For each reaction, calculate how many moles of the barium product you will produce using stoichiometry and the balanced reaction

ankoles [38]

Answer:

Explanation:

given that

mass of Ba(NO3)2 = 1.40g

mass of NH2SO3H = 2.50 g

1)to determine the mole of Ba(NO3)2

2) to determine the mass of all three product formed in the reaction

reaction

Ba(NO3)2 + 2NH2SO3H → Ba(NH2SO3)2 + 2HNO3

<u>Solution</u>

we calculate the molar mass of each species by using their atomic masses

BA = 137.33g/mol

N = 14g/mol

O= 16g/mol

H = 1g/mol

S = 32g/mol

calculation

Ba(NO3)2 = Ba + 2N + 6O

= 137.33 + 2X 14 + 6 X 16

= 261.33g/mol

NH2SO3H = N + 3H + S+ 3O

=14 + 3X1 + 32 + 3X 16

= 97g/mol

Ba(NH2SO3)2 = Ba + 2N + 4H +2S +6O

= 137.33 + 2 X 14 + 4 X1 + 2X32 + 6 X 16

= 329.33g/mol

HNO3 = H + n + 3O

= 1 + 14 + 3 X 16

= 63g/mol

3 0

4 years ago

Which is the correct electron dot diagram for krypton (Kr)?

SCORPION-xisa [38]

Krypton is in group 18, and so it has 8 valence electrons. B is the answer

4 0

4 years ago

Cual es la formula estructural condensada del octano??

Arlecino [84]

$CH_{3} CH_{2} CH_{2} CH_{2} CH_{2} CH_{2} CH_{2} CH_{2} CH_{3}$ o $CH_{3}(CH_{2})_{6} CH_{3}$

6 0

3 years ago

Calculate Delta G for each reaction using Delta Gf values: answer kJ ...thank you

Leni [432]

Answer:

a) $\Delta G=2.6kJ$

b) $\Delta G=-979.57kJ$

c) $\Delta G=264.21kJ$

Explanation:

Hello,

In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:

$\Delta _fG_{H_2}=\Delta _fG_{I_2}=0kJ/mol\\\Delta _fG_{HI}=1.3kJ/mol\\\Delta _fG_{CO_2}=-394.4kJ/mol\\\Delta _fG_{CO}=-137.3 kJ/mol\\\Delta _fG_{NH_3}=16.7 kJ/mol\\\Delta _fG_{HCl}=-95.3kJ/mol\\\Delta _fG_{MnO_2}=465.37kJ/mol\\\Delta _fG_{Mn}=0kJ/mol\\\Delta _fG_{NH_4Cl}=-342.81kJ/mol$

So we proceed as follows:

$\Delta G=2\Delta _fG_{HI}-\Delta _fG_{H_2}-\Delta _fG_{I_2}\\\\\Delta G=2*1.3\\\\\Delta G=2.6kJ$

$\Delta G=\Delta _fG_{Mn}+2*\Delta _fG_{CO_2}-\Delta _fG_{MnO_2}-2*\Delta _fG_{CO}\\\\\Delta G=0+2*-394.4-465.37-2*-137.3\\\\\Delta G=-979.57kJ$

$\Delta G=\Delta _fG_{NH_3}+\Delta _fG_{HCl}-\Delta _fG_{NH_4Cl}\\\\\Delta G=16.7-95.3-(-342.81)\\\\\Delta G=264.21kJ$

Regards.

6 0

3 years ago

The hybridization for the B in Bcl3 is ?

Phoenix [80]

Answer:

sp²

Explanation:

You need to look at how many electron orbitals around the atom. Looking at the structure below, you can see that there are three electron orbitals. This gives you an sp² hybridization.

6 0

3 years ago

Give the postional isomers of 2 halopropane(CH3)2CH​

Give the postional isomers of 2 halopropane(CH3)2CH