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3 years ago

What volume of 0.350 m koh is required to react completely with 24.0 ml of 0.650 m h3po4?

Chemistry

1 answer:

BartSMP [9]3 years ago

7 0

The complete balanced chemical equation for this is:

3KOH + H3PO4 --> K3PO4 + 3H2O

First we calculate the number of moles of H3PO4:

moles H3PO4 = 0.650 moles / L * 0.024 L = 0.0156 mol

From stoichiometry, 3 moles of KOH is required for every mole of H3PO4, therefore:

moles KOH = 0.0156 mol H3PO4 * (3 moles KOH / 1 mole H3PO4) = 0.0468 mol

Calculating for volume given molarity of 0.350 M KOH:

Volume = 0.0468 mol / (0.350 mol / L) = 0.1337 L = 133.7 mL

Answer:

133.7 mL KOH

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In at least 4 complete sentences, describe the similarities and differences between Avogadro's Law and Charles' Law.

viktelen [127]

answer

Avogadro's law states that, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles present. In other words, equal volumes of gases at the same pressure and temperature contain the same number of molecules - this is true regardless of their physical properties or chemical nature.

This number of molecules is

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and is known as Avogadro's number,

Matematically, Avogadro's law can be written like this

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Avogadro's law, as well as Boyle's law and Charles' law, are special cases of the ideal gas law,

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→

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The ideal gas law can also be written to incorporate

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7 0

3 years ago

Equal masses of substance A and substance B are at 25°C. Substance A has a lower specific heat capacity than substance B. If 50

liraira [26]

Dang bro that stuff is really hard I’m defiantly on a lower grade lol

6 0

3 years ago

In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

Answer: The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

Explanation:

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$ .......(1)

Calculating for first dilution:

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

Calculating for second dilution:

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

8 0

3 years ago

Does potassium nitrate (KN03) incorporate ionic bonding, covalent bonding, or both? Explain.

ladessa [460]

Answer: $KNO_3$ incorporates both ionic bonding and covalent bonding.

Explanation:

A covalent bond is formed when an element shares its valence electron with another element. This bond is formed between two non metals.

An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element and the element which accepts the electrons is known as electronegative element. This bond is formed between a metal and an non-metal.

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here potassium is having an oxidation state of +1 called as $K^{+}$ cation and nitrate $NO_3^{-}$ is an anion with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $KNO_3$ . $NO_3^-$ is formed by sharing of electrons between two non metals nitrogen and oxygen.

Thus $KNO_3$ incorporates both ionic bonding and covalent bonding.

4 0

4 years ago

What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according

ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

Step 1: The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

Step 2: Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

Step 3: Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

Step 4: Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

4 0

3 years ago