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3 years ago

If there are 20.0g of KOH available, how many grams Ca(OH)2 will form?

Chemistry

1 answer:

notka56 [123]3 years ago

6 0

Answer:

can u send the qsn 4rm the start so tt I can help

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Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

Answer: The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

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

$\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]$

For the given chemical reaction:

$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

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

$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0

3 years ago

Which arrow represents the activation energy for the forward reaction

ValentinkaMS [17]

Answer:

Explanation:

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4 0

2 years ago

Read 2 more answers

A scientific theory _______.

Butoxors [25]

describes what occurs every time in a given situation

3 0

3 years ago

Read 2 more answers

Convert 947400 s to cs

Gennadij [26K]

Answer:image

Explanation: image

3 0

3 years ago

A certain solid has a density of 8.0 g/cm3. if 4.0 g of this solid are poured into 4.00 ml of water, which drawing below most cl

Alexandra [31]

Answer is: the volume of water after the solid is added is 4.5 ml.
d(gold) = 8.0 g/cm³; density of gold.
m(gold) = 4 g; mass of gold.
V(gold) = m(gold) ÷ d(gold); volume of gold.
V(gold) = 4 g ÷ 8 g/cm³.
V(gold) = 0.5 cm³ = 0.5 ml.
V(water) = 4.00 ml = 4.00 cm³.
V(flask) = V(gold) + V(water).
V(flask) = 0.5 cm³ + 4 cm³.V = 4.5 cm³.

4 0

3 years ago