The titration of HCl with NaOH is an example of:
A. a weak acid-weak base titration,
B. a weak acid-strong base titration.
c. a strong acid-strong base titration.
D. a strong acid-weak base titration.

Answers

Answer 1

Answer:

I’m pretty sure it’s C

Explanation:

AP Ex


Related Questions

The substance ammonia has the following properties: normal melting point: 195.4 K normal boiling point: 239.8 K triple point: 5.9×10-2 atm, 195.3 K critical point: 111.5 atm, 405.5 K At temperatures above 405.5 K and pressures above 111.5 atm, NH3 is a _________ . NH3 does not exist as a liquid at pressures below atm. NH3 is a _________ at 5.90×10-2 atm and 249.5 K. NH3 is a _________ at 1.00 atm and 236.0 K. NH3 is a _________ at 24.6 atm and 185.6 K.

Answers

Answer:

a) Superficial fluid

b)  5.9*10^-2 atm

c) Gas

d) Liquid

e) Solid

Explanation:

a)  At temperatures above 405.5 K and pressures above 111.5 atm, NH3 is a superficial fluid because liquid and gases does not exit at temperature and pressure greater than  405.5 K and  111.5 atm

b) NH3 does not exist as a liquid at pressures below  5.9*10^-2 atm , That is below the triple point there is existence of liquid

c)  NH3 is a Gas at 5.90×10^-2 atm and 249.5 K.

d) NH3 is a  Liquid  at 1.00 atm and 236.0 K. because pressure and temperature ( standard ) is between the given normal melting and boiling point

e) NH3 is a solid at 24.6 atm and 185.6 K . because the pressure here is more than that of triple point while the temperature is lesser

Se prepara una concentración molar 0.5M de hidróxido de sodio (NaCH) con la finalidad de determinar el grado de acidez de una bebida embotellada. Calcula la cantidad de gramos de hidróxido de sodio (NaCH) que se requiere para preparar 5 litros de dicha solución.​

Answers

Answer:

100 g

Explanation:

Primero hay que calcular el número requerido de moles de NaOH, usando la definición de molaridad:

Molaridad = moles / litros0.5 M = moles / 5 Lmoles = 2.5 moles

Ahora hay que convertir 2.5 moles de NaOH en gramos, usando su masa molecular:

2.5 mol * 40 g/mol = 100 g

Se requieren 100 gramos de NaOH.

how much corn syrup should be added to water to make 200 mL of a 10% by volume solution​

Answers

....20 ml corn syrup

To make 200 mL of a 10% by volume solution, add 20 mL corn syrup to water.

What is volume solution?Volume percent of a solution is defined as the ratio of the volume of solute present in a solution to the volume of the solution as a whole. It means that the volume of a closed figure determines how much three-dimensional space it can occupy. In terms of numerical value, volume is the amount of three-dimensional space enclosed by a closed surface. For example, a substance's space can be solid, liquid, gas, or plasma, or any other space-occupying shape. The volume percentage of a solution can be calculated by dividing the volume of solute by the volume of solution and multiplying the result by 100. The basic formula for volume is length width height, whereas the basic formula for area of a rectangular shape is length width height. The calculation is unaffected by how you refer to the various dimensions: you can, for example, use 'depth' instead of 'height.'

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2+4 help please i will fail 6th grade

Answers

Answer:

6

Explanation:

lol

Which best describes the total mass of substances before and after a
combustion reaction?
A. The mass is the same because atoms are not created or
destroyed.
B. The mass is greater after the reaction because the number of
atoms increases.
C. The mass is the same because the molecules do not change.
D. The mass is less after the reaction because gases are produced.

Answers

Answer:

B. The mass is greater after the reaction because the number of

atoms increases.

A 500.0 g sample of aluminium, I initially at 25.0 degrees, absorbs heat from its surroundings and reaches a final temperature of 90.7 degrees. How much heat (in KJ) has been absorbed by the sample? To one decimal place
Specific heat= 0.9930j g-1 K-1 for aluminium

Answers

Q= (500.0)(0.9930)(90.7 - 25.0)
Q= 32620.05

A 500.0 g sample of aluminum, initially at 25.0 degrees, absorbs heat from its surroundings and reaches a final temperature of 90.7 degrees. 32.62245 kJ heat has been absorbed by the sample.

What is specific heat?

The term specific heat is defined as the amount of heat required to increase the temperature of 1 gram of a substance 1 degree Celsius (°C).

To calculate the amount of heat absorbed by the sample, use the formula:

Q = mcΔT

where Q is the amount of heat absorbed by the sample, m is the mass of the sample, c is the specific heat of aluminum, and ΔT is the change in temperature of the sample.

Substituting the given values into the formula, we get:

Q = 500.0 g × 0.9930 J/g·K × (90.7°C - 25.0°C)

Q = 500.0 g × 0.9930 J/g·K × 65.7 K

Q = 32,622.45 J

To convert the result to kilojoules (kJ), we divide by 1000:

Q = 32.62245 kJ

Thus, the amount of heat absorbed by the sample is 32.6 kJ.

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A balloon is inflated with 1.25 L of helium at a temperature of 298K. What is
the volume of the balloon when the temperature is 310K?

Answers

Answer:

What is

the volume of the balloon when the temperature is 310K?

determine the maximum amount of NaN03 that was produced during the experiment. Explain how you determined the amount

Answers

Answer:

9 moles of NaNO3 is obtained

Explanation:

The balanced chemical reaction equation for the reaction is;

Al(NO3)3 + 3NaCl-------> 3NaNO3 + AlCl3

Now, we have to determine the limiting reactant. The limiting reactant yields the least amount of NaNO3.

1 mole of Al(NO3)3 yields 3 moles of NaNO3

4 moles of Al(NO3)3 yields 4 * 3/1 = 12 moles of NaNO3

Also,

3 moles of NaCl yields 3 moles of NaNO3

9 moles of NaCl yields 9 * 3/3 = 9 moles of NaNO3

Hence, NaCl  is the limiting reactant and 9 moles of NaNO3 is obtained.

9 moles of NaNO₃ is obtained.

Balanced chemical reaction:

[tex]Al(NO_3)_3 + 3NaCl-----> 3NaNO_3 + AlCl_3[/tex]

From the reaction, it is seen that the limiting reactant yields the least amount of NaNO₃.

1 mole of Al(NO₃)₃ yields 3 moles of NaNO₃

4 moles of Al(NO₃)₃ yields 4 * 3/1 = 12 moles of NaNO₃

Also,

3 moles of NaCl yields 3 moles of NaNO₃

9 moles of NaCl yields 9 * 3/3 = 9 moles of NaNO₃

Hence, NaCl  is the limiting reactant and 9 moles of NaNO₃ is obtained.

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