Classify each of these as an element, compound, or mixture:

A 10.8ml sample of sulfuric acid titrated with 80.0 ml of 0.200 m mg solution. What is the concentration of the sample given the

IRISSAK [1]

Answer:

1.48 M

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

Mg + H2SO4 —> MgSO4 + H2

Step 2:

Determination of the number of mole of Mg in 80.0 mL of 0.200 M Mg solution. This is illustrated below:

Molarity of Mg = 0.200 M

Volume of solution = 80 mL = 80/1000 = 0.08L

Mole of Mg =?

Molarity = mole /Volume

0.2 = mole /0.08

Mole = 0.2 x 0.08

Mole of Mg = 0.016 mole.

Step 3:

Determination of the number of mole of H2SO4 that reacted. This is illustrated below:

Mg + H2SO4 —> MgSO4 + H2

From the balanced equation above,

1 mole of Mg reacted with 1 mole of H2SO4.

Therefore, 0.016 mole of Mg will also react with 0.016 mole of H2SO4.

Step 4:

Determination of the concentration of the acid.

Mole of H2SO4 = 0.016 mole.

Volume of acid solution = 10.8 mL = 10.8/1000 = 0.0108 L

Molarity =?

Molarity = mole /Volume

Molarity = 0.016/0.0108

Molarity of the acid = 1.48 M

Therefore, the concentration of acid is 1.48 M

8 0

4 years ago

Because it is a cycle,the water cycle has no ___________ or ________

natima [27]

No beginning or end I believe

7 0

3 years ago

What do you always wear when working with chemicals or flying objects

irga5000 [103]

Safety goggles. a lab coat if available. and gloves.

8 0

3 years ago

Read 2 more answers

How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56

neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.0156

Thus,

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.0156=e^{-k\times t}$

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}$

$t = 6\times t_{1/2}$

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

6 0

4 years ago

In the reaction below, if a total of 10.0 g of zinc and hydrochloric acid react completely, what is the total mass of zinc chlor

Sloan [31]

Answer:

Not enough information to tell.

Explanation:

What is given?

Mass of Zn = 10.0 g,

Mass of HCl = 10.0 g,

Molar mass of Zn = 65.4 g/mol,

Molar mass of HCl = 36.4 g/mol,

Molar mass of ZnCl2 = 136.2 g/mol,

Molar mass of H2 = 2 g/mol.

Step-by-step solution:

First, let's convert 10.0 g of each reactant to moles using their respective molar mass:

$\begin{gathered} 10.0\text{ g Zn}\cdot\frac{1\text{ mol Zn}}{65.4\text{ g Zn}}=0.153\text{ moles Zn,} \\ \\ 10.0\text{ g HCl}\cdot\frac{1\text{ mol HCl}}{36.4\text{ g HCl}}=0.275\text{ moles HCl.} \end{gathered}$

Now, let's identify what is the limiting reactant. Let's see how many moles of ZnCl2 can be produced by 0.153 moles of Zn if 1 mol of Zn reacted produces 1 mol of ZnCl2, and how many moles of ZnCl2 can be produced by 0.275 moles of HCl if 2 moles of HCl reacted produces 1 mol of ZnCl2:

$\begin{gathered} 0.153\text{ moles Zn}\cdot\frac{1\text{ mol ZnCl}_2}{1\text{ mol Zn}}=0.153\text{ moles ZnCl}_2, \\ \\ 0.275\text{ moles HCl}\cdot\frac{1\text{ mol ZnCl}_2}{2\text{ moles HCl}}=0.138\text{ moles ZnCl}_2. \end{gathered}$

You can realize that the limiting reactant, in this case, is HCl because is the first reactant consumed first and this reactant 'impose' the limit to produce the products.

So now, let's find how many moles of H2 are being produced by 0.275 moles of HCl if 2 moles of HCl reacted produces 1 mol of H2:

$0.275\text{ moles HCl}\cdot\frac{1\text{ mol H}_2}{2\text{ moles HCl}}=0.138\text{ moles H}_2.$

The final step is to convert each number of moles of each product to grams using their respective molar mass, as follows:

$\begin{gathered} 0.138\text{ moles ZnCl}_2\cdot\frac{136.2\text{ g ZnCl}_2}{1\text{ mol ZnCl}_2}=18.8\text{ g ZnCl}_2, \\ \\ 0.138\text{ moles H}_2\cdot\frac{2\text{ g H}_2}{1\text{ mol H}_2}=0.276\text{ g H}_2. \end{gathered}$

We're producing 18.8 g of zinc chloride (ZnCl2), and 0.276 g of hydrogen (H2), so based on this logic the answer would be not enough information to tell.

5 0

1 year ago