CHEMISTRY HIGH SCHOOL

After doing multiple titrations, your NaOH solution is determined to have a mean concentration value of 0.100 M. Given you are to assume your unknown acid is 75.0% KHP, how many grams of your unknown will you need to use 15.00 mL of your 0.100 M standardized NaOH

Answers

Answer 1

Answer:

Answer: The mass of unknown acid needed is 0.230 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaOH solution = 0.100 M

Volume of solution = 15.00 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.100M=\frac{\text{Moles of NaOH}}{0.015L}\n\n\text{Moles of NaOH}=(0.100mol/L* 0.015L)=0.0015mol$

The chemical reaction for the reaction of KHP and NaOH follows

$KHC_8H_4O_4(aq.)+NaOH\rightarrow KNaC_8H_4O_4(aq.)+H_2O(l)$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of KHP.

So, 0.0015 moles of NaOH will react with = $(1)/(1)* 0.0015=0.0015mol$ of KHP

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of KHP = 0.0015 moles

Molar mass of KHP = 204.22 g/mol

Putting values in above equation, we get:

$0.0015mol=\frac{\text{Mass of KHP}}{204.22g/mol}\n\n\text{Mass of KHP}=(0.0015mol* 204.22g/mol)=0.306g$

We are given:

Mass of unknown acid = 75 % of Mass of KHP

So, mass of unknown acid = $(75)/(100)* 0.306=0.230g$

Hence, the mass of unknown acid needed is 0.230 grams

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Enter the molecular formula for butane, C4H10.
express in a chemical formula

Answers

Answer: the molecular formula is C₄H₁₀

You can also express it as: CH₃ - CH₂ - CH₂ - CH₃.

Explanation:

1) The molecular formula is the chemical formula for a covalent compound. It shows the elements present in one molecule of the compound: type of atoms and the number of each.

The number of each type of atoms is indicated by the subscript to the right of the symbol.

2) Butane is the alkane with 4 carbon atoms.

3) The general formula for the alkanes is CₓH₂ₓ₊₂, meaning thar for x atoms of C there are 2x+2 atoms of H. Therefore, for 4 C atoms there are 2(4)+2 = 8 + 2 = 10 H atoms.

That is why the molecular formula of butane is C₄H₁₀, meaning that each molecule of this alkane has 4 atoms of carbon and 10 atoms of hydrogen.

The other expression: CH₃ - CH₂ - CH₂ - CH₃, shows that this compounds is a chain, in which each C has 4 bonds, one with other C atom to the right, one with other C atom to the right and the difference (2 or 3) with H atoms.

The molecular formula of butane is $\boxed{{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}$ .

Further explanation:

The molecular formula is a chemical formula that depicts the total number and kinds of atoms present in a molecule. For example, molecular formula of carbon dioxide is ${\text{C}}{{\text{O}}_2}$ .

Hydrocarbon is a term for the organic compounds that consist of hydrogen and carbon only.

Types of hydrocarbons:

1. Saturated hydrocarbons

The simplest hydrocarbons that are composed of only single bonds are called saturated hydrocarbons. These hydrocarbons have the general formula of ${{\text{C}}_n}{{\text{H}}_(2n + 2)}$ , where n is the number of carbon atoms. These hydrocarbons have suffix “ane” in their names. Examples of such hydrocarbons are methane, hexane, and propane.

2. Unsaturated hydrocarbons

These have one or more multiple bonds in them. These hydrocarbons have suffix “ene” or “yne”, depending on whether there is a double or triple bond between them. Hydrocarbons comprising of double bonds are called alkenes and those having triple bonds are called alkynes.

The name of butane includes the suffix “ane”. This implies it is a saturated hydrocarbon and contains only single bonds in it. The prefix “but” indicates the presence of four carbon atoms in this molecule.

Substitute 4 for n in the general formula of alkane $\left( {{{\text{C}}_n}{{\text{H}}_(2n + 2)}} \right)$ to find out the formula of butane.

$\begin{aligned}{\text{Molecular formula of butane}} &= {{\text{C}}_{\text{4}}}{{\text{H}}_{\left( {{\text{2}}\left( {\text{4}} \right){\text{ + 2}}} \right)}}\n&= {{\text{C}}_4}{{\text{H}}_(10)}\n\end{aligned}$

Learn more:

Calculate the moles of ions in the solution: brainly.com/question/5950133
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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Stoichiometry of formulas and equations

Keywords: molecular formula, butane, C4H10, 4, ane, ene, yne, alkane, alkyne, alkene, saturated hydrocarbon, unsaturated hydrocarbon.

Which of the following pairs share the same empirical formula? Select one:
a. CH4 and C2H4
b. PbCl2 and PbCl4
c. N2O5 and NO2
d. C2H6 and C4H12

Answers

Answer: The correct option is d.

Explanation: Empirical formula is a chemical formula which has the simplest ratio of elements present in a chemical compound.

From the given following pairs, the pair which shares the same empirical formula is $C_2H_6\text{ and }C_4H_(12)$

Empirical formula of $C_2H_6$ is $CH_3$ (by dividing the coefficients of $C_2H_6$ by 3)

Empirical formula of $C_4H_(12)$ is $CH_3$ (by dividing the coefficients of $C_4H_(12)$ by 3)

Both the chemical formulas have same empirical formula. Hence, the correct option is d.

C2H6 and C4H12 share the same emperical formula of CH3 as the both compounds can be reduced by the smallest subscript in each compound. So C2H6 will be reduced by 2 making CH3 and C4H12 can be reduced by 4 making CH3

The _____ agent is the substance in a redox reaction that accepts electrons.B. oxidizing
C. buffering
D. half-reaction

Answers

B. oxidizing

In the reaction the oxidizing agent is reduced.

the answer is oxidizing look it up

Also i took the test

2. The process of splitting an atom into two lighter atoms is calledOA. nuclear separation.
OB. nuclear fusion.
C. nuclear fission.
OD nuclear disintegration.

Answers

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For the reaction:6Li(s)+N2(g)→2Li3N(s)
Determine:
the mass of N2 needed to react with 0.536 moles of Li.
the number of moles of Li required to make 46.4 g of Li3N.
the mass in grams of Li3N produced from 3.65 g Li.
the number of moles of lithium needed to react with 7.00 grams of N2.

Answers

Explanation:

$6Li(s)+N_2(g)\rightarrow 2Li_3N(s)$

1. Mass of $N_2$ needed to react with 0.536 moles of Li.

According to reaction, 6 moles of Li reacts with 1 mol of $N_2$ .

Then 0.536 moles of Li will react with:

$(1)/(6)* 0.536$ moles of $N_2$ that is 0.0893 moles.

Mass of $N_2[tex] gas needed:</strong></p><p><strong>[tex]=28 g/mol* 0.0893 mol=2.5004 g$

2.The number of moles of Li required to make 46.4 g of $Li_3N$

Moles of $Li_3N=(46.4 g)/(35 g/mol)=1.3257 mol$

According to reaction the 2 moles of $Li_3N$ are produced from 6 moles of Li.

Then 1.3257 moles of $Li_3N$ will produced from:

$(6)/(2)* 1.3257=3.9771 moles$

3.9771 moles of lithium will needed.

3. The mass in grams of $Li_3N$ produced from 3.65 g Li.

Moles of Li $=(3.65 g)/(7 g/mol)=0.5214 moles$

According to reaction, 6 moles of Li gives 2 moles of $Li_3N$

Then 0.5214 moles of Li will give $(2)/(6)* 0.5214$ that is 0.1738 moles of $Li_3N$ .

Mass of $Li_3N=0.1738 mole* 35 g/mol=6.083 g$

6.083 grams of $Li_3N$ will be produced.

4. The number of moles of lithium needed to react with 7.00 grams of $N_2$ .

Moles of $N_2=(7.00 g)/(28 g/mol)=0.25 moles$

1 mol of $N_2$ reacts with 6 mol of Li

Then, 0.25 moles of $N_2$ will ftreact with :

$6* 0.25 moles=1.5 moles$ of lithium

1.5 moles of Li will be needed.

How many grams of Ca(OH)2 are needed to produce 600 ml of 1.22 M Ca(OH)2 solution?

Answers

Answer:

54.2 g of Ca(OH)₂

Explanation:

Let's determine the moles of solute, we should need

Molarity . volume (L) = moles

Let's convert 600 mL to L

600 mL/ 1000 = 0.6L

1.22 mol/L . 0.6L = 0.732 moles

Finally we must convert the moles to mass ( moles . molar mass)

0.732 mol . 74.08 g/mol = 54.2 g

Answer: 54.2 g Ca(OH)2

Explanation: Molarity is moles of solute / L solution

First convert mL to L

600 mL x 1L / 1000 mL = 0.6 L

Find moles of Ca(OH)2

n= M x L

= 1.22 M x 0.6 L

= 0.732 moles Ca(OH)2

Convert moles to mass using its molar mass of Ca( OH)2 = 74 g

0.732 moles Ca(OH)2 X 74 g Ca(OH)2 / 1 mole Ca(OH)2

= 54.2 g Ca(OH)2

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