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svp [43]
2 years ago
7

What happens when a fatty acid is reacted with naoh.

Chemistry
1 answer:
Dmitry [639]2 years ago
4 0
Saponification, salt of fatty acids are created, glycerol as well.
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A student is making a glucose solution. The student is using 12.55 g of glucose (C6H1206) and a 500 mL volumetric flask. Use thi
Ber [7]

Answer:

M = 0.138 M

Explanation:

Given data:

Mass of glucose = 12.55 g

Volume of solution = 500 mL

Molarity of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Number of moles of glucose:

Number of moles = mass/molar mass

Number of moles = 12.55 g/ 180.156 g/mol

Number of moles = 0.069 mol

Volume in L:

500 mL × 1 L /1000 mL

0.5 L

Molarity:

M = 0.069 mol / 0.5 L

M = 0.138 M

8 0
3 years ago
The tablet has a mass of 1.20 g and contains 700 mg of lithium carbonate.
kondaur [170]

Answer:

The right answer is "58.3%".

Explanation:

The given values are:

Mass of lithium carbonate,

= 700 mg

i.e.,

= 0.7 g

Mass of tablet,

= 1.20 g

Now,

The percentage of active ingredient will be:

=  \frac{Mass \ of lithium \ carbonate}{Mass \ of \ tablet}\times 100

On substituting the given values, we get

=  \frac{0.7}{1.20}\times 100

=  0.583\times 100

=  58.3%

4 0
3 years ago
For the reaction 2H2 + O2 --> 2H2O, how many grams of oxygen are needed to react 3 moles of hydrogen?
mr Goodwill [35]

Answer:

48 grams

Explanation:

The chemical equation for the reaction is the following:

2 H₂ + O₂ → 2 H₂O

That means that 2 moles of H₂ react with 1 mol of O₂ to produce 2 moles of H₂O. We convert the moles of oxygen (O₂) by using the molecular weight (MW) as follows:

MW(O₂) = 16 g/mol x 2 = 32 g/mol

mass of O₂ = 1 mol x 32 g/mol = 32 g

So, we have the following stoichiometric ratio: 32 g O₂/2 moles H₂. We have 3 moles of hydrogen (H₂), so we multiply the moles by the stoichiometric ratio to calculate how many grams are needed:

3 moles H₂ x 32 g O₂/2 moles H₂ = 48 g O₂

<em>Therefore, 48 grams of O₂ are needed to react with 3 moles of H₂.</em>

4 0
3 years ago
Are all science possibilities feasible?
nikitadnepr [17]

That a particular concept- the Alcubierre Warp Drive- might actually be feasible.... the major hurdles and opportunities presented by warp mechanics research

5 0
3 years ago
1. Identify all of the gas law equations that relate to the ideal gas law.
zheka24 [161]

Explanation:

1)  PV = nRT(ideal gas equation )

All of the gas laws equations that relate to the ideal gas law:

1. Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.  

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

2. Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

3. Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

4. Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

\frac{V_1}{n_1}=\frac{V_2}{n_2}

where,

V_1\text{ and }n_1 are the initial volume and number of moles

V_2\text{ and }n_2 are the final volume and number of moles

2)

Initial moles of gas = n_1=0.0400 mol

Initial volume of the gas = V_1=500 mL=0.5 L (1 mL = 0.001 L)

Final moles of gas = n_2=?

Final volume of the gas = V_2=1.00 L

Equation used to find the amount of gas added is Avogadro's law equation :

\frac{V_1}{n_1}=\frac{V_2}{n_2}

n_2=\frac{V_2\times n_1}{V_1}=\frac{1.00 L\times 0.0400 mol}{0.5 L}

n_2=0.08 mol

3 0
3 years ago
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