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Lostsunrise [7]

3 years ago

13

(3) consider a the titration of 1.0 m sulfurous acid (h2so3, ka1 = 1.5e-2, ka2 = 1.0e-7) with 2.0 m naoh. what is the ph at the

equivalence point of the titration?

1 answer:

kupik [55]3 years ago

4 0

bonjours ma question est sin x=1/7 comment trouver les valeurs de x plz help my

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2. Under a pressure of 95kPa and a temperature of 25 C, a gas occupies 4.0 liters. What would the new

Alekssandra [29.7K]

Answer:

35

Explanation:

multiply and subtract

3 0

3 years ago

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]

<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 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. 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

Help help help help! Kingdom

Brrunno [24]

Answer:

all 4 of the middle ones are part of the nucleus

3 0

2 years ago

Read 2 more answers

A sample of argon gas has a volume of 795 mL at a pres-sure of 1.20 atm and a temperature of 116 ∘C. What is the final volume of

jek_recluse [69]

<u>Answer:</u> The volume when the pressure and temperature has changed is $1.6\times 10^2mL$

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

Let us assume:

$P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K$

Putting values in above equation, we get:

$\frac{1.20atm\times 795mL}{389K}=\frac{0.55atm\times V_2}{348K}\\\\V_2=\frac{1.20\times 795\times 348}{0.55\times 389}=1.6\times 10^3mL$

Hence, the volume when the pressure and temperature has changed is $1.6\times 10^2mL$

5 0

3 years ago

The electron configuration of Ne is:

Slav-nsk [51]

I think it is correct......

3 0

3 years ago