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3 years ago

how would you write equation to show how buffers H2CO3 and NaHCO3 behave when (a) HCl is added and (b) NaOH is added.

1 answer:

8 0

H2CO3 <---> H+ + HCO3-
NaHCO3 <---> Na+ + HCO3-

When acid is added in the buffer, the excess H+ of that acid reacts with HCO3- to form H2CO3, and due to this NaHCO3 dissociates into HCO3- to attain the equilibrium. and hence there is no net effect of H+ due to pH remain almost constant. when a base is added to the buffer, the OH- ion of base react eith H+ ion present in buffer, then to attain equilibrium of H+ ion, the H2CO3 dissociates to produce H+ ion, but now there is the excess of HCO3- due to which Na+ ion react with them to attain equilibrium of HCO3-. hence there is again no net change in H+ ion due to which pH remain constant.....

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2 years ago

Which statement best describes what happens to the stability of hydrogen and oxygen atoms when they form bonds in a water molecu

ioda

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2 years ago

On a warm summer day in Oslo, the temperature is 21 degree C and the relative humidity is 70%. What is the absolute humidity? Wh

lesya [120]

Explanation:

According to the psychrometric chart at dry bulb temperature of $21^{o}C$ and RH 70%, the absolute humidity = 0.011 kg/kg dry air

Formula to calculate humid volume is as follows.

$\nu_{H} m^{3}/kg$ dry air = $\frac{22.41}{273}TK (\frac{1}{28.97} + \frac{1}{18.02}H)$

= $(2.83 \times 10^{-3} + 4.56 \times 10^{-3} H) T K$

= $(2.83 \times 10^{-3} + 4.56 \times 10^{-3} \times 0.011) \times 294$

= 0.847 $m^{3}/kg$ dry air

Hence, humid volume of air is 0.847 $m^{3}/kg$ dry air.

Specific enthalpy of dry air = specific heat capacity of dry air × dry bulb temperature

= $1.006 kJ/kg ^{o}C \times 21^{o}C$

= 21.126 kJ/kg dry air

Hence, the specific enthalpy of the air is 21.126 kJ/kg dry air.

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2 years ago

g When 2.50 g of methane (CH4) burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole

Anna [14]

Answer:

-800 kJ/mol

Explanation:

To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).

First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:

Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol

moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄

Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:

ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol

Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).

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2 years ago

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just olya [345]

Your answer would be an Arrhenius base

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1 year ago