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2 years ago

How many moles of NaOH are required to prepare 2.90 L of 1.8 M NaOH?

2 answers:

5 0

Here is a couple of useful formula to use when you get problems like this.

$Molarity= \frac{moles}{Liters}$ -----> $moles= (Molarity) x (Liters)$

since they are asking for moles, we are using the one on the right.

Molarity= 1.8 M
Liters= 2.90 L

moles= (1.8 M) x (2.90 L)= 5.2 moles of NaOH

marusya05 [52]2 years ago

3 0

V=2.90 L
c=1.8 mol/L

n(NaOH)=vc

n(NaOH)=2.90L*1.8mol/L=5.22 mol

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9 months ago

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2 years ago

Baking soda (NaHCO3) and vinegar (HC2H3O2) react to form sodium acetate, water, and carbon dioxide. If 42.00 g of baking soda re

Setler [38]

Answer:

0.5 mole of CO₂.

Explanation:

We'll begin by calculating the number of mole in 42 g of baking soda (NaHCO₃). This can be obtained as follow:

Mass of NaHCO₃ = 42 g

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= 23 + 1 + 12 + 48

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Mole of NaHCO₃ = 42/84

Mole of NaHCO₃ = 0.5 mole

Next, balanced equation for the reaction. This is given below:

NaHCO₃ + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O + CO₂

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of CO₂

Finally, we shall determine the number of mole of CO₂ produced by the reaction of 42 g (i.e 0.5 mole) of NaHCO₃. This can be obtained as follow:

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of CO₂.

Therefore, 0.5 mole of NaHCO₃ will also react to produce 0.5 mole of CO₂.

Thus, 0.5 mole of CO₂ was obtained from the reaction.

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2 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

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2 years ago