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Rashid [163]
3 years ago
14

Neutralization is a process that occurs whenever an acid reacts with a base in the mole ratios specified by the balanced equatio

n. however, not all neutralization reactions produce neutral solutions. what factors can determine the outcome of a neutralization reaction?
Chemistry
1 answer:
labwork [276]3 years ago
7 0
The factors that can determine the outcome of a "neutralization reaction" include:
__________________________________________________________
1) temperature; 

2) concentration (of the solution); 

3) type of acid used;

4) type of alkali used;

5) type of reaction; 
     (e.g. whether the reaction is "endothermic" or "exothermic");
_____________________________________________________
 
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What type of central-atom orbital hybridization corresponds to each electron-group arrangement:
galina1969 [7]

Tetrahedral arrangement is resulted upon mixing one s and three p atomic orbitals, resulting in 4 hybridized sp^3 orbitals → sp^3 hybridization.

<h3>What is orbital hybridization?</h3>

In the context of valence bond theory, orbital hybridization (or hybridisation) refers to the idea of combining atomic orbitals to create new hybrid orbitals (with energies, forms, etc., distinct from the component atomic orbitals) suited for the pairing of electrons to form chemical bonds.

For instance, the valence-shell s orbital joins with three valence-shell p orbitals to generate four equivalent sp3 mixes that are arranged in a tetrahedral configuration around the carbon atom to connect to four distinct atoms.

Hybrid orbitals are symmetrically arranged in space and are helpful in the explanation of molecular geometry and atomic bonding characteristics. Usually, atomic orbitals with similar energies are combined to form hybrid orbitals.

Learn more about Hybridization

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3 0
1 year ago
10. Which of the following best describes the reaction 2VO3– (aq) + Zn (s) + 8H+ (aq) → 2VO2 (aq) + Zn2+ (aq) + 4H2O (l)? ______
Oduvanchick [21]
Correct answer is option E. <span>It is a redox reaction in which Zn is oxidized at the anode, and V is reduced at the cathode.

Reason:
In above reaction, the oxidation state of VO3- is +5, while that of VO2 is +4. Thus there is reduction of V from +5 to +4
In case of Zn, oxidation state of Zn is increased from 0 to +2, Thus process is referred as oxidation. </span>
3 0
2 years ago
What did timbuktu contributed to mali's important as a kingdom
Alex73 [517]
Timbuktu was considered a MASSIVE trade center in Mali. That is how all of the foreign goods came into the kingdom. It also had education centers which was very important in Mali.
5 0
3 years ago
Read 2 more answers
Design a test to determine whether thorium-234 also emits particles. First, explain how Rutherford’s experiment measured positiv
liubo4ka [24]

The characteristics of the α and β particles allow to find  the design of an experiment to measure the ²³⁴Th particles is:

  • On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.
  • The neutrons cannot be detected in this experiment because they have no electrical charge.

In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.

The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.

The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.

Thorium has several isotopes, with different rates and types of emission:

  • ²³²Th emits α particles, it is the most abundant 99.9%
  • ²³⁴Th emits β particles, exists in small traces.

In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.

Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.

In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:

  • On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.
  • The neutrons cannot be detected in this experiment because they have no electrical charge.

Learn more about radioactive emission here: brainly.com/question/15176980

7 0
2 years ago
A cylindrical piece of metal is 4.5 dm in height with radius of 5.50 x 10^-5 km.
adell [148]

Answer:

a) V=4.3x10^3mL

b) V=4.3x10^6mm^3

c) \rho=1.5x10^5g/L

Explanation:

Hello,

a) In this case, the given height in cm is:

h=4.5dm*\frac{1m}{10dm}* \frac{100cm}{1m}=45cm

And the radius in cm is:

r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{100cm}{1m}=5.5cm

Thus, the volume in cubic centimeters which is also equal in mL (1cm³=mL) is:

V=\pi (5.5cm)^2*45cm\\\\V=4.3x10^3cm^3=4.3x10^3mL

b) In this case, the given height in mm is:

h=4.5dm*\frac{1m}{10dm}* \frac{1000mm}{1m}=450mm

And the radius in mm is:

r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{1000mm}{1m}=55mm

Thus, the volume in cubic millimeters is:

V=\pi (55mm)^2*450mm\\\\V=4.3x10^6mm^3

c) Finally, since 1000 mL equal 1 L, the required density in g/L turns out:

\rho=\frac{m}{V}=\frac{6.54x10^5g}{4.3x10^3mL}*\frac{1000mL}{1L}\\   \\\rho=1.5x10^5g/L

Best regards.

8 0
3 years ago
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