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geniusboy [140]
3 years ago
13

For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ

e anion is HPO4 3- c) is basic d) Is acidic
Chemistry
1 answer:
Komok [63]3 years ago
6 0

Answer:

Check the explanation

Explanation:

Answer – Given, H_3PO_4 acid and there are three Ka values

K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}

The transformation of H_2PO_4- (aq) to HPO_4^2-(aq)is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

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Use this equation for the following problems: 2NaN3 --> 2Na+3N2
olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

<em>2NaN₃ → 2Na + 3N₂,</em>

  • It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

<em>Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?​ </em>

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

  • Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

  • We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

  • We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

<em><u>using cross multiplication:</u></em>

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

  • Finally, we can get the grams of NaN₃ needed:

<em>mass = no. of moles x molar mass</em> = (0.517 mol)(65.0 g/mol) =<em> 33.6 g.</em>

<em />

<em>Q2: How many mL of N₂ result if 8.3 g Na are also produced?</em>

  • We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

  • We can get the no. of moles of N₂ produced with 0.36 mol of Na:

<em><u>using cross multiplication:</u></em>

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

  • We can get the mass of 0.54 mol of N₂:

mass = no. of moles  x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

  • Now, we can get the mL of 15.12 g of N₂:

<em><u>using cross multiplication:</u></em>

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

<em>∴ The volume of N₂ result </em>= (1.0 L)(15.12 g)/(0.92 g) = <em>16.434 L = 16434 mL.</em>

4 0
2 years ago
What is the formula, when rubidium reacts with tellurium?
natita [175]

Answer:

Like other alkali metals, rubidium metal reacts violently with water. As with potassium (which is slightly less reactive) and caesium (which is slightly more reactive), this reaction is usually vigorous enough to ignite the hydrogen gas it produces.

Explanation:

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5 0
2 years ago
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Study the following reaction carefully. What classification should this reaction have? 4Al + 3O 2 2Al 2 O 3
nikitadnepr [17]

The reaction is a synthesis…

4Al + 3O_2 \longrightarrow 2Al_2O_3\\

… because <em>two substances are combining to make one other substance</em>.

4\stackrel{\hbox{0}}{\hbox{Al}} + 3\stackrel{\hbox{0}}{\hbox{O}}_2 \longrightarrow 2\stackrel{\hbox{+3}}{\hbox{Al}}_2\stackrel{\hbox{-2}}{\hbox{O}}_3\\

It is also a <em>reduction-oxidation reaction</em> because the <em>oxidation number of Al increases</em> from 0 to +3 and the <em>oxidation number of O</em> decreases from 0 to -2.

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3 years ago
Convert 7.89 x 10^8 molecules of water to liters.
NNADVOKAT [17]

Answer:

2.94 x 10⁻¹⁴L

Explanation:

To solve this problem, we have to assume that the condition of this water is at standard temperature and pressure, STP.

At STP;

       1 mole of gas have a volume of 22.4L

So, let us find the number of moles of this water first;

         6.02 x 10²³ molecules can be found in 1 mole of a substance

          7.89 x 10⁸ molecules will contain \frac{7.89 x 10^{8} }{6.02 x 10^{23} }   = 1.31 x 10⁻¹⁵mole of water

So;

        Volume of water  = 22.4 x 1.31 x 10⁻¹⁵  = 2.94 x 10⁻¹⁴L

5 0
3 years ago
The overall change in enthalpy of a reaction depends only on
serious [3.7K]
<h3><u>Answer;</u></h3>

A. the reactants and the products.

<h3><u>Explanation</u>;</h3>
  • The enthalpy of a reaction is the heat released or absorbed by a reaction at constant pressure.
  • Enthalpy is important as it tells us the amount of heat energy present in a system. In chemical reactions it tells us the amount of heat energy lost or gained.
  • <em><u>Changes in enthalpy are determined as the reactants are converted into products. Therefore, enthalpy depends on the reactants and the products. In other words it depends on the heat of reactants and that of products.</u></em>
4 0
3 years ago
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