to neutralize a 40.0 ml sample of 0.650 m hcl using titration, what volume (ml) of the 0.800 m naoh would you need at the equivalence point of the titratio

Answers

Answer 1
Answer:

The volume of the NaOH that we would now need for the neutralization reaction is32.5 mL.

What is neutralization?

Neutralization is a chemical reaction in which an acid reacts with a base to form a salt and water. In this reaction, the hydrogen ions from the acid combine with the hydroxide ions from the base to form water

Number of moles of HCl = 40/1000 L * 0.65

= 0.026 moles

Since the reaction is 1:1, there would be 0.026 moles

of the NaOH used as such we would now have that;

Volume of the NaOH = 0.026 moles/0.8 M

= 32.5 mL

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If two gases are present in a container, the total pressure in the container is equal toa.the sum of the pressures that are exerted by each of the two gases.
b.twice the sum of the pressures that are exerted by the individual gases.
c.the sum of the pressures that each gas would exert if they occupied twice the volume.
d.the sum of the pressures that each gas would exert if they occupied half the volume.

Answers

Answer: The correct option is (a).

Explanation:

If two gases are present in a container, the total pressure in the container is equal to the sum of the all the individual pressures exerted by the individual gases in a container.

Individual pressure exerted by the individual gas in a container is called partial pressure of that individual gas.

So, total pressure of gas in container is:

p^o=p_a+p_b

Where p^o = Total pressure exerted by the gases in the container.

p_a = Partial pressure of gas 'a'

p_b = partial pressure of gas 'b'

The answer is a, because the question is referring to partial pressures. Partial pressures are a mere sum, no halving or doubling of anything is necessary.


So, A

Hope this helped :)

Which sample of water contains particles having the highest average kinetic energy ?

Answers

25mL if water as the highest average of the kinetic energy

Why do the elements at the bottom of the periodic table have lower ionization energies than their group/family partners at the top of the periodic table?

Answers

Ionization energy is the energy required to remove the losely bounded electron from an isolated gaseous atom of an element, so if an electron is more attracted towards nucleus it will require higher energy. On increasing size of an atom the electrons fall distant from the nucleus and will observe less effective nuclear energy hence less amount of energy will be required to remove them.

On moving down the group, the size of elements increases hence effective nuclear charge will decrease thus ionization energy will decrease.

Elements at the bottom of the periodic table have lower ionization energies compared to their group or family partners at the top of the periodic table because, they have more energy levels.  

Ionization energy decreases down the group as less energy is required to remove outer most electrons as energy levels increases.

Further Explanation

Ionization energy

  • Ionization energy is the energy required to remove outermost electrons from the outermost energy level. Energy is required to remove an electron from an atom.
  • The closer an electron is to the nucleus the more energy is required, since the electron is more tightly bound to the atom thus making it more difficult to remove, hence higher ionization energy.
  • Ionization energy increases across the periods and decreases down the group from top to bottom.  
  • Additionally, the ionization energy increases with subsequent removal of a second or a third electron.

First ionization energy  

  • This is the energy required to remove the first electron from the outermost energy level of an atom.
  • Energy needed to remove the second electron to form a divalent cation is called the second ionization energy.

Trend in ionization energy  

1. Down the group (top to bottom)

  • Ionization energy decreases down the groups in the periodic table from top to bottom.
  • It is because as you move down the group the number of energy levels increases making the outermost electrons get further from the nucleus reducing the strength of attraction to the nucleus.
  • This means less energy will be required compared to an atoms of elements at the top of the groups.

2. Across the period  (left to right)

  • Ionization energy increases across the period from left to right.
  • This can be explained by an increase in nuclear energy as extra protons are added to the nucleus across the period increasing the strength of attraction of electrons to the nucleus.
  • Consequently, more energy is needed to remove electrons from the nucleus.

Keywords: Ionization energy, periodic table, energy levels, electrons

Learn more about

Level: High school  

Subject: Chemistry  

Topic: Periodic table and chemical families  

Sub-topic: Ionization energy

What is the molar mass of (NH₄)₃PO₄ in g/mol? A) 149.10 g/mol B) 113.01 g/mol C) 141.03 g/mol D) 61.99 g/mol

Answers

Answer: A) 149.10 g/mol

Explanation:

First, I list what I know, which are the molar masses of N, H, P, and O:

N = 14.01 g/mol
H = 1.01 g/mol
P = 30.97 g/mol
O = 15.99 g/mol

Now, I determine how many molecules of N, H, P, and O there are, starting with (NH₄)₃:

With just NH₄, there would be 1 N and 4 H, but since there are () around it with a subscript of 3, we multiply 3 by both N and H₄, so:

1*3 = 3 N
1*12 = 12 H

Now, we look at PO₄:

The subscript of 3 is before PO₄, so we do not apply that to PO₄:

1 P
4 O

Now, we know the molar masses and number of each of the elements, so we can determine the molar mass of the entire compound.

First, I calculate the total molar masses of each of the elements by multiplying the number of molecules of the element by the element's molar mass:

3 N = 3*14.01 g/mol = 42.03 g/mol
12 H = 12*1.01 g/mol = 12.12 g/mol
1 P = 1*30.97 g/mol = 30.97 g/mol
4 O = 4*15.99 g/mol = 63.96 g/mol

Then, I calculate the total molar mass of the compound by adding the total molar masses of each of the elements together:

42.03 + 12.12 + 30.97 + 63.96 = 149.08 g/mol

I probably used slightly different approximations of the molar masses of each element, but my answer is closest to choice A) 149.10 g/mol.

So, if there are no () around a compound, like PO₄, the subscript only applies to the element it's attached to, so there are 4 molecules of O and only 1 molecule of P.

If there are () around a compound with a subscript, like (NH₄)₃, the subscript after the () gets applied to each element inside the ().

I hope this helps! :)

Final answer:

The molar mass of the compound (NH₄)₃PO₄ is calculated by adding up the molar masses of all its constituent elements. The calculated molar mass is approximately 149.09 g/mol, which corresponds to answer choice (A) 149.10 g/mol.

Explanation:

To calculate the molar mass of a compound, you multiply the quantity of each element by its atomic mass and then sum them all up. The formula (NH₄)₃PO₄ contains Nitrogen (N), Hydrogen (H), Phosphorous (P), and Oxygen (O). The atomic masses of these elements are approximately 14.01 g/mol, 1.01 g/mol, 30.97 g/mol, and 16.00 g/mol respectively. So for the compound (NH₄)ᵣPO₄, the molar mass would be as follows:

(3*N) + (12*H) + (1*P) + (4*O) = (3*14.01) + (12*1.01) + 30.97 + (4*16.00) = 149.09 g/mol approximately

Therefore, the closest answer choice to the correct molar mass is (A) 149.10 g/mol.

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Given the reaction at equilibrium:N2(g) + 3H2(g) -->2NH3(g) + 91.8 kJ
What occurs when the concentration of H2(g) is increased?
(1) The rate of the forward reaction increases and the concentration of N2(g) decreases.
(2) The rate of the forward reaction decreases and the concentration of N2(g) increases.
(3) The rate of the forward reaction and the concentration of N2(g) both increase.
(4) The rate of the forward reaction and the concentration of N2(g) both decrease.

Answers

If the concentration of H_2 (g) is increased, the rate of the forward reaction increases and the concentration of N_2 (g)decreases.

Le Chartelier principle

The principle states that reactions in equilibrium being disturbed by factors such as temperature, the concentration of species in the reaction, etc, will experience a shift in the equilibrium so as to annul the effects of the disturbance.

Thus, adding more of the reactants ( H_2 (g) to the reaction will see more  products being synthesized. This means more  N_2 (g) will be consumed and its concentration will decrease accordingly.

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(1) The rate of the forward reaction increases and the concentration of N2(g) decreases.
N2 is constant 

A gas absorbs 3.4 kJ of heat and does 1.9 kJ of work. Calculate ΔE

Answers

Answer:

Answer is given below:

Explanation:

Given Data:

heat = 3.4kJ

work done is = 1.9 kJ

To Find:

ΔE=?

Formula:

ΔE = q + w

Solution:

ΔE = q + w

ΔE = 3.4 kJ + 1.9kJ

ΔE =  5.3 kJ