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Oduvanchick [21]

3 years ago

5

Give the IUPAC name for the following compound. Group of answer choices 6-Ethyl-6-methylhept-3-yne 6,6-Dimethyloct-3-yne 3,3-Dim

ethyloct-5-yne 3,3-Dimethyloct-4-yne

1 answer:

slamgirl [31]3 years ago

8 0

Answer:

3,3-Dimethyloct-4-yne · Contents · 1Structures · 2Names and Identifiers · 3Chemical and Physical Properties · 4Related Records · 5 Patents · 6 ..

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givi [52]

Answer: The enthalpy of formation of $SO_3$ is -396 kJ/mol

Explanation:

Calculating the enthalpy of formation of $SO_3$

The chemical equation for the combustion of propane follows:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

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

$\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-297kJ/mol\\\Delta H^o_{rxn}=-198kJ$

Putting values in above equation, we get:

$-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol$

The enthalpy of formation of $SO_3$ is -396 kJ/mol

4 0

3 years ago

Au(NO3)3 added with water? Aq or s?

Reptile [31]

Gold (III) nitrate in an aqueous solution is hydrolyzed with formation of gold (III) metahydroxide.

Au(NO₃)₃ → Au³⁺(aq) + 3NO₃⁻(aq)

Au³⁺ + H₂O ⇄ AuOH²⁺ + H⁺
AuOH²⁺ + H₂O ⇄ Au(OH)₂⁺ + H⁺
Au(OH)₂⁺ + H₂O → AuOOH·H₂O(s) + H⁺

Au(NO₃)₃(aq) + 2H₂O(l) = AuOOH(s) + 3HNO₃(aq)

4 0

3 years ago

10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C

Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

$Q=mc\Delte T$

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c = 0.385 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC$

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c = 0.903 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC$

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c = 2.42 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC$

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c = 4.18J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC$

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c = 0.128 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC$

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0

3 years ago

Four animals were fleeing from a wildfire. All of them were running at a speed of 30 miles per hour. Which one has the greatest

RUDIKE [14]

Answer: I don’t know lol

Explanation: I am so sorry I thought this was easy

6 0

3 years ago

Read 2 more answers

A scientist measures the standard enthalpy change for the following reaction to be 67.9 kJ:

bekas [8.4K]

Answer: The standard enthalpy of formation of $H_2O(g)$ is -252.1 kJ/mol.

Explanation:

The balanced chemical reaction is,

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactantss}]$

Putting the values we get :

$\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}]$

$67.9kJ=[(2\times 0)+(3\times H_f{H_2O})]-[(1\times -824.2kJ/mol)+3\times 0kJ/mol)]$

$H_f{H_2O}=-252.1kJ/mol$

Thus standard enthalpy of formation of $H_2O(g)$ is -252.1 kJ/mol.

3 0

2 years ago