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elena55 [62]
3 years ago
5

Me podrían decir los nombres comunes de los siguientes hidrocarburos!! URGENTE AYUDA

Chemistry
2 answers:
blsea [12.9K]3 years ago
8 0

Answer:

No Se Espanol

Explanation:

Ilya [14]3 years ago
4 0

Answer:

fdhhdhdhdhhshdgdvdvebehbehdhehdhhddhdf

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You are given sodium acetate, 1m hcl, nahco3 and na2co3. Determine which of these four you would need and then show calculations
valentina_108 [34]

solution:

the given compounds are sodium acetate, 1M HCL,NaHCO₃ and Na2CO₃

pH of the buffer solution is 4.7

the value of pKa of sodium bicarbonate is 6.37

the value of pKa of acetic acid is 4.7

calculate concentration of acetic acid by using the following forumula

pH=pKa+lag[salt]/[acid]

substitute the pH and Pka values in the formula.

4.7=4.7+log[salt]/[acid]

log[salt]/[acid]=0

thus, the concentration ratio of the salt and acid should be equal to each other.

Thus, concentration of sodium acetate is 0.05M

Concentration of sodium acetate= concentration of acid

= 0.05M

Volume of the buffer solution is 100mL

The buffer solution can be prepared as 0.05M of 50mL sodium acetate will react with 0.05M of 50mL of 0.05M of HCL.

The chemical equation for neutralization of the weak base with strong can be represented as show as

CH₃COONa+HCL-->CH₃COOH+NaCL


5 0
3 years ago
Volume increases, pressure stays the same, temperature
Mila [183]
Temperature decreases (?)
4 0
3 years ago
What is the entropy change of the surroundings
KiRa [710]

Answer: The entropy change of the surroundings will be -17.7 J/K mol.

Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol

Amount of Acetone given = 10.8 g

Number of moles is calculated by using the formula:

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

Molar mass of acetone = 58 g/mol

Number of moles = \frac{10.8}{58}=0.1862moles

If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then

0.1862 moles will have = \frac{32.3}{1}\times 0.1862=5.828kJ/mol

To calculate the entropy change for the system, we use the formula:

\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}

Temperature = 56.2°C = (273 + 56.2)K = 329.2K

Putting values in above equation, we get

\Delta S_{sys}=\frac{5.828}{329.2}=0.0177kJ/Kmol=17.7J/Kmol   (Conversion Factor: 1 kJ = 1000J)

At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,

\Delta S_{system}+\Delta S_{surrounding}=0

\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol

5 0
3 years ago
What volume, in mL, of carbon dioxide gas is produced at STP by the decomposition of 0.242 g calcium carbonate (the products are
damaskus [11]

Answer:

54.21 mL.

Explanation:

We'll begin by calculating the number of mole in 0.242 g calcium carbonate, CaCO3.

This is illustrated below:

Mass of CaCO3 = 0.242 g

Molar mass of CaCO3 = 40 + 12 +(16x3) = 40+ 12 + 48 = 100 g/mol

Mole of CaCO3 =?

Mole = mass /Molar mass

Mole of CaCO3 = 0.242/100

Mole of CaCO3 = 2.42×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

CaCO3 —> CaO + CO2

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole CaO and 1 mole of CO2.

Next, we shall determine the number of mole of CO2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole of CO2.

Therefore,

2.42×10¯³ mole of CaCO3 will also decompose to produce 2.42×10¯³ mole of CO2.

Therefore, 2.42×10¯³ mole of CO2 were obtained from the reaction.

Finally, we shall determine volume occupied by 2.42×10¯³ mole of CO2.

This can be obtained as follow:

1 mole of CO2 occupies 22400 mL at STP.

Therefore, 2.42×10¯³ mole of CO2 will occupy = 2.42×10¯³ x 22400 = 54.21 mL

Therefore, 54.21 mL of CO2 were obtained from the reaction.

7 0
3 years ago
What is a liquid mixture whose parts are evenly blended?.
Eva8 [605]

Answer:

A homogeneous mixture is a type of mixture in which the individual substances are evenly mixed.

Explanation:

3 0
2 years ago
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