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2 years ago

Please help me!! Look at the picture to understand the question please!

Chemistry

1 answer:

Romashka-Z-Leto [24]2 years ago

5 0

Answer:

C. I

Explanation:

The elements are categorized in groups and periods. The periods are horizontal and the groups are vertical. The elements in the same group (vertical) have very similar characteristics.

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Gas burning in an engine is an example of a

pav-90 [236]

The answer is the first option. Gas burning in an engine is an example of a chemical change. Chemical change is a change where the substance changes in identity or form new substances after undergoing a process. In this case, the gas reacts with oxygen forming combustion products, commonly carbon dioxide and water.

8 0

3 years ago

A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of

zaharov [31]

Answer:

a) pH = 2.88

b) pH = 4.598

c) pH = 5.503

d) pH = 8.788

e) pH = 12.097

Explanation:

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.75 E-5 = [H3O+]*[CH3COO-] / [CH3COOH]

a) 0.0 mL KOH:

mass balance:

⇒ C CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M

charge balance:

⇒ [H3O+] = [CH3COO-]

⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])

⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0

⇒ [H3O+] = 1.314 E.3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.88

b) 5.0 mL KOH:

CH3COOH + KOH ↔ CH3COONa + H2O

∴ C CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)

⇒ C CH3COOH = 0.05 M

∴ C KOH = ((5 E-3)(0,200))/(0.025+5 E-3) = 0.033 M

mass balance:

⇒ C CH3COOH + C KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M

charge balance:

⇒ [H3O+] + [K+] = [CH3COO-]

⇒ [CH3COO-] = [H3O+] + 0.033

⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))

⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])

⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]

⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0

⇒ [H3O+] = 2.523 E-5 M

⇒ pH = 4.598

equivalent point:

(C*V)acid = (C*V)base

⇒ (0.100 M)*(0.025 L) = (0.200 M)( Vbase)

⇒ Vbase = 0.0125L = 12.5 mL

c) 10.0 mL KOH:

∴ C CH3COOH = 0.0143 M

∴ C KOH = 0.057 M

as in the previous point, starting from the mass and charge balances, we obtain:

⇒ [H3O+] = 3.1386 E-6 M

⇒ pH = 5.503

d) 12.5 mL KOH:

at the equivalence point, there is complete salt formation, then the pH is calculated through the salt:

CH3COO- + H2O ↔ CH3COOH - OH-

∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]

∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M

mass balance:

⇒ 0.066 = [CH3COOH] + [CH3COO-]..........(1)

charge balance:

⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M.........(2)

∴ [K+] = C CH3COO- = 0.066 M

(1) = (2):

⇒ [OH-] = [CH3COOH].......(3)

⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])

⇒ [OH-]² + 5.714 E-10[OH-] - 3.7712 E-11 = 0

⇒ [OH-] = 6.1408 e-6 m

⇒ pOH = 5.212

⇒ pH = 14 - pOH = 8.788

d) 15.0 mL KOH:

after the equivalence point there is salt and excess base (OH-); ph is calculated from excess base:

⇒ C KOH = ((0.015)(0.200) - (0.025)(0.100)) / (0.025 + 0.015) = 0.0125 M

⇒ [OH-] ≅ C KOH = 0.0125 M

⇒ pOH = 1.903

⇒ pH = 12.097

8 0

3 years ago

In order to predict the outcome of the reaction, write the molecular, full ionic, and net ionic equations for a mixture of aqueo

Elina [12.6K]

Answer:

See explanation

Explanation:

Full molecular equation;

2NH3(aq) + AgNO3(aq) -------> [Ag(NH3)2]NO3(aq)

Full ionic equation

2NH3(aq) + Ag^+(aq) + NO3^-(aq) --------> [Ag(NH3)2]^+(aq) + NO3^-(aq)

Net ionic equation;

2NH3(aq) + Ag^+(aq) --------> [Ag(NH3)2]^+(aq)

When Silver nitrate is mixed with a solution of aqueous ammonia, a white and cloudy solution was observed.

6 0

3 years ago

The following information is given for benzene, C6H6, at 1atm: boiling point = 80.1 °C Hvap(80.1 °C) = 30.7 kJ/mol specific heat

user100 [1]

Answer: The heat required for the process is 4.24 kJ

Explanation:

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

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

Given mass of benzene = 24.8 g

Molar mass of benzene = 78.11 g/mol

Putting values in above equation, we get:

$\text{Moles of benzene}=\frac{24.8g}{78.11g/mol}=0.318mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat absorbed = ?

n = number of moles = 0.318 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = 30.7 kJ/mol

Putting values in above equation, we get:

$30.7kJ/mol=\frac{q}{0.318mol}\\\\q=(30.7kJ/mol\times 0.318mol)=4.24kJ$

Hence, the heat required for the process is 4.24 kJ

6 0

2 years ago

What is the main energy conversion that occurs when you plug in a lamp

MrRa [10]

Elecrtical or Radiantal

4 0

3 years ago

Read 2 more answers