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bazaltina [42]
3 years ago
11

Which subatomic particles had negligible mass and travels around outside the nucleus? A protons B neutron C electron D orbital ?

??
Chemistry
1 answer:
kap26 [50]3 years ago
8 0
C. Electron. It’s a negatively charged particle that orbits around the nucleus. Usually, there are multiple electrons going around the nucleus in different orbitals (the circle around the nucleus in which the electron travels). The mass is much smaller compared to that of a proton or neutron and can be ignored when calculating the mass of an atom.
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In the experiment, you added water to the reaction vessel after the reaction was complete, but we did not discuss why. Based on
chubhunter [2.5K]

Answer:

A.

Explanation:

Water was added to the reaction after the completion of the reaction so as to lower the solubility if the product in the solution therefore, the product can be precipitated out. On adding water the reaction moves in forward direction and more product is formed. (By Le Chatelier's principle). Thus, the precipitation occurs. Hence, option A is correct.

5 0
4 years ago
When the equation Ca(OH)2 + HBr → (products) is completed and balanced, one term in the equation will be
NISA [10]
<span>Answer for the given question is CaBr2. Although, the given equation requires balancing of by adding one more HBr and one more H2o in resultant. The given equation will product at least one CaBr2. Hence the answer for the given equation is Calcium bromide i.e. CaBr2.</span>
4 0
3 years ago
Please answer fast, I will give brainliest if correct!
gizmo_the_mogwai [7]

Answer:

Explanation: 30 min

5 0
3 years ago
The question is in the picture below
Rus_ich [418]

Answer:

\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

\bf{\text{A} \rightarrow 2\text{B}}                  (Δ\text{H}_1)  -----(1)

Since we have 2B, multiply the whole of II. by 2:

\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}       (2Δ\text{H}_2) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is \text{A} \rightarrow 2\text{C} +2\text{D}.

Reversing III. gives us a negative enthalpy change as such:

\bf{2\text{D} \rightarrow \text{E}}                  (-Δ\text{H}_3) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of \text{A} \rightarrow 2\text{C} +\text{E}, which is also the equation of interest.

Adding all three together:

\text{A} \rightarrow 2\text{C}+\text{E}            (\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 })

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

4 0
1 year ago
The carbon cycle does not really involve matter. True False
stiks02 [169]
False everything involves matter

8 0
3 years ago
Read 2 more answers
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