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lisabon 2012 [21]
3 years ago
6

2Fe + 3Cl2 → 2FeCl3 which of the following options gives the correct reactant ratio?

Chemistry
1 answer:
Mademuasel [1]3 years ago
6 0

<u>Answer:</u> The reactant ratio in the given chemical equation will be: Fe:Cl_2=2:3

<u>Explanation:</u>

Mole ratio is defined as the ratio of the amount of moles of two substances that are participating in a chemical reaction.

In the given chemical equation:

2Fe+3Cl_2\rightarrow 2FeCl_3

The reactants are Fe and Cl_2 and the product is FeCl_3

The mole ratio is basically the stoichiometric ratio of the chemical compounds taking part in a chemical reaction.

The mole ratio of reactants is stoichiometric ratio of Fe and Cl_2, which is:

Fe:Cl_2=2:3

Hence, the reactant ratio in the given chemical equation will be: Fe:Cl_2=2:3

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Oxalic Acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and
blondinia [14]

Answer:

HCO₂

Explanation:

From the information given:

The mass of the elements are:

Carbon C = 26.7 g;     Hydrogen H = 2.24 g     Oxygen O = 71.1 g

To determine the empirical formula;

First thing is to find the numbers of moles of each atom.

For Carbon:

=26.7 \ g\times \dfrac{1 \ mol }{12.01 \ g} \\ \\ =2.22 \ mol \ of \ Carbon

For Hydrogen:

=2.24 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =2.22 \ mol \ of \ Hydrogen

For Oxygen:

=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen

Now; we use the smallest no of moles to divide the respective moles from above.

For carbon:

\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon

For Hydrogen:

\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen

For Oxygen:

\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen

Thus, the empirical formula is HCO₂

4 0
3 years ago
What is the relationship between melting/boiling point and intermolecular forces
sattari [20]
Boiling points are a measure of intermolecular forces. The intermolecular forces increase with increasing polarization of bonds. Boiling point increases with molecular weight, and with surface area.
8 0
2 years ago
An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an
Westkost [7]

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that pH = -log[H^{+}], and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

pH = pKa - log \frac{[HA]}{[A^{-}]}

Where [HA] is the concentration of the acid, and [A^{-}] is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same  stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of OH^- will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of OH^- reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of OH^-, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

6.72 = pKa - log \frac{0.01}{0.0025}

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

8 0
3 years ago
Expressed in scientific notation(6.0 x 104) (3.1 x 10-1) =
levacccp [35]
The answer would be 1.86 x 10^4
4 0
3 years ago
You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH sol
oee [108]

Answer:

0.2788 M

1.674 %(m/V)

Explanation:

Step 1: Write the balanced equation

NaOH + CH₃COOH → CH₃COONa + H₂O

Step 2: Calculate the reacting moles of NaOH

0.03575 L \times \frac{0.1950mol}{L} = 6.971 \times 10^{-3} mol

Step 3: Calculate the reacting moles of CH₃COOH

The molar ratio of NaOH to CH₃COOH is 1:1.

6.971 \times 10^{-3} molNaOH \times \frac{1molCH_3COOH}{1molNaOH} = 6.971 \times 10^{-3} molCH_3COOH

Step 4: Calculate the molarity of the acetic acid solution

M = \frac{6.971 \times 10^{-3} mol}{0.02500L} =0.2788 M

Step 5: Calculate the mass of acetic acid

The molar mass of acetic acid is 60.05 g/mol.

6.971 \times 10^{-3} mol \times \frac{60.05g}{mol} =0.4186 g

Step 6: Calculate the percentage of acetic acid in the solution

\frac{0.4186g}{25.00mL}  \times 100\% = 1.674 \%(m/V)

6 0
3 years ago
Read 2 more answers
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