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3 years ago

When elements form compounds, the elements

Chemistry

1 answer:

kati45 [8]3 years ago

6 0

Answer:

Explanation:Elements form compound to form chemical bonds, chemical bonding occurs because atoms of an element become more stable by losing, gaining and sharing elections....

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When 0.25 moles of NH3 are produced according to the reaction below, how many moles of H2 must have reacted?

marissa [1.9K]

Answer:

Explanation:

by stoichiometry

nNH3/2=nH2/3

0.25/2=nH2/3

nH2=0.25×3/2

nH2=0.375 mol

so its 0.38 mol

6 0

2 years ago

Read 2 more answers

A 51.24-g sample of Ba(OH)2 is dissolved in enough water to make 1.20 liters of solution. How many mL of this solution must be d

g100num [7]

Answer:

0.40 L

Explanation:

Calculation of the moles of $Ba(OH)_2$ as:-

Mass = 51.24 g

Molar mass of $Ba(OH)_2$ = 171.34 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{51.24\ g}{171.34\ g/mol}$

$Moles= 0.2991\ mol$

Volume = 1.20 L

The expression for the molarity is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.2991\ mol}{1.20\ L}=0.24925\ M$

Thus,

Considering

$Molarity_{working\ solution}\times Volume_{working\ solution}=Molarity_{stock\ solution}\times Volume_{stock\ solution}$

Given that:

$Molarity_{working\ solution}=0.100\ M$

$Volume_{working\ solution}=1\ L$

$Volume_{stock\ solution}=?$

$Molarity_{stock\ solution}=0.24925\ M$

So,

$0.100\ M\times 1\ L=0.24925\ M\times Volume_{stock\ solution}$

$Volume_{stock\ solution}=\frac{0.100\times 1}{0.24925}\ L=0.40\ L$

<u>The volume of 0.24925M stock solution added = 0.40 L </u>

6 0

2 years ago

Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002

marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The $pK_a=4.756$

The value of the dissociation constant = $K_a$

$pK_a=-\log[K_a]$

$K_a=10^{-4.756}=1.754\times 10^{-5}$

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

$HAc\rightleftharpoons H^++Ac^-$

Initially

At equilibrium ;

(c-cα) cα cα

The expression of dissociation constant is given as:

$K_a=\frac{[H^+][Ac^-]}{[HAc]}$

$1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}$

$1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}$

$1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}$

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

$[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M$

The pH of the solution ;

$pH=-\log[H^+]$

$=-\log[0.0001901 M]=3.72$

3 0

3 years ago

The reaction that produces cellular energy in plants is called?

Setler79 [48]

Atp or the mitchodria

6 0

3 years ago

Enter your answer in the provided box. Calcium hydroxide may be used to neutralize (completely react with) aqueous hydrochloric

NNADVOKAT [17]

Answer:

49.95 g of HCl

Explanation:

Let's formulate the chemical equation involved in the process:

Ca(OH)2 + 2 HCl → CaCl2 + 2 H2O

This means that we need 1 mole of Calcium hydroxide to neutralize 2 moles of hydrochloric acid. From this, we calculate the quantity of HCl moles that would be neutralized by 0.685 moles of Ca(OH)2

1 mole Ca(OH)2 ---- 2 moles HCl

0.685 moles Ca(OH)2 ---- x = 1.37 moles HCl

Now that we know the quantity of HCl moles that would react, let's calculate the quantity of grams this moles represent:

1 mole of HCl ---- 36.46094 g

1.37 moles ------ x = 49.95 g of HCl

3 0

2 years ago