All categories

3 years ago

WHAT ARE SOME FUN ALCHSELTER EXPERIMENTS?

Chemistry

1 answer:

Vesna [10]3 years ago

4 0

You can put an egg in Vinegar an egg in Olive oil

You might be interested in

50.0 mL of a solution of HCl is combined with 100.0 mL of 1.05M NaOH in a calorimeter. The reaction mixture is initially at 22.4

VikaD [51]

Answer:

2.1 M is the molarity of the HCl solution.

Explanation:

$HCl+NaOH\rightarrow H_2O+NaCl$

Molarity of HCl solution = $M_1=?$

Volume of HCl solution = $V_1=50.0mL$

Ionizable hydrogen ions in HCl = $n_1=1$

Molarity of NaOH solution = $M_2=1.05 M$

Volume of NaOH solution = $V_2=100.0 mL$

Ionizable hydroxide ions in NaOH = $n_2=1$

$n_1M_1V_1=n_2M_2V_2$ (neutralization )

$M_1=\frac{M_2V_2}{V_1}=\frac{1.05M\times 100.0 mL}{50.0 mL}$

$M_1=2.1 M$

2.1 M is the molarity of the HCl solution.

3 0

3 years ago

How many σ and π bonds are in this molecule? A chain of five carbon atoms. There is a double bond between the first and second c

kati45 [8]

Answer:

Sigma bonds: 10

Pi bonds: 4

Explanation:

The compound described must be CH2=CH-CO-CH≡CH. If we look at the compound closely, we will notice that there are 10 sigma bonds and 4 pi bonds.

There are three pi bonds between carbon atoms and one pi bond between a carbon and an oxygen atom (C=O). All these can easily be seen in the structure of the formula chosen in this answer.

5 0

4 years ago

Which students propose a conservation strategy that represents reducing the use of

andre [41]

Answer:

student A

Explanation:

5 0

3 years ago

Heyoo I need help pleaseeeeee

Rasek [7]

c) Current cannot flow

Reason: Because the switch is not closed.

6 0

3 years ago

Read 2 more answers

A) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains o

Mumz [18]

Answer:

For a: The empirical formula for the given compound is $CH$

For b: The empirical and molecular formula for the given organic compound are $C_{10}H_{20}O$

Explanation:

For a:

The chemical equation for the combustion of hydrocarbon follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x', and 'y' are the subscripts of Carbon and hydrogen respectively.

We are given:

Conversion factor used: 1 g = 1000 mg

Mass of $CO_2=5.86mg=5.86\times 10^{-3}g$

Mass of $H_2O=1.37mg=1.37\times 10^{-3}g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in $5.86\times 10^{-3}g$ of carbon dioxide, $\frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in $1.37\times 10^{-3}g$ of water, $\frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g$ of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.133\times 10^{-3}$ moles.

For Carbon = $\frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1$

For Hydrogen = $\frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

Hence, the empirical formula for the given compound is $CH$

For b:

The chemical equation for the combustion of menthol follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 0.2829 g

Mass of $H_2O$ = 0.1159 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.013g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.013) = 0.105 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.013g}{1g/mole}=0.013moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0105g}{16g/mole}=0.00065moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00065 moles.

For Carbon = $\frac{0.0064}{0.00065}=9.84\approx 10$

For Hydrogen = $\frac{0.013}{0.00065}=20$

For Oxygen = $\frac{0.00065}{0.00065}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

The empirical formula for the given compound is $C_{10}H_{20}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 156 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

$n=\frac{156g/mol}{156g/mol}=1$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O$

Hence, the empirical and molecular formula for the given organic compound are $C_{10}H_{20}O$

3 0

4 years ago