The nuclear emission that is negatively charged is .
Further Explanation:
Radioactive decay:
Also known as nuclear decay, radioactivity, nuclear or radioactive disintegration. It is the process due to which an unstable atomic nucleus releases its energy in the form of various particles such as alpha particles, beta particles, and gamma particles.
Following are the types of radioactive decays:
1. Alpha decay
In this decay, alpha particles are emitted by unstable nuclei. An alpha particle is a helium nucleus with two protons and two neutrons. It has a charge of +2. The general equation for an alpha decay is as follows:
2. Beta decay
In this decay, beta particles are produced. This occurs when the nucleus decays continuously and emits an electron or a positron. The general equation for beta decay is as follows:
Beta decay is further classified as follows:
(a) decay: In this decay, an electron and an electron antineutrino are emitted. The atomic number of the element is increased by one. Its general equation is as follows:
(b) decay or positron emission: In this decay, a positron and an electron neutrino are emitted. The atomic number of the element is decreased by one. Its general equation is as follows:
3. Gamma decay
In this decay, gamma rays are emitted. Energy is emitted during gamma decay but the number of protons remains unaltered. The general equation for gamma decay is as follows:
4. Neutron emission
It is the type of radioactive decay that occurs when one or more neutrons are ejected from a nucleus. Neutrons are the subatomic particles with no charge. Since only neutrons are removed from the atom, the number of protons as well electrons remain unaltered. This results in the formation of an isotope of the same element.
Therefore the negatively charged nuclear emission is a beta particle because an electron that is a negatively charged species is lost in decay, a type of beta decay.
Learn more:
1. What nuclide will be produced in the given reaction? brainly.com/question/3433940
2. Calculate the nuclear binding energy: brainly.com/question/5822604
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Nuclear chemistry
Keywords: alpha particle, beta particle, gamma decay, positron, negatively charged, neutron emission, radioactivity, radioactive decay, unstable atomic nucleus.
How many molecules are in 2.47 moles of CC14? *
M1m1 = M2m2
where M1 is the concentration of the stock solution, m1 is the mass of the stock solution, M2 is the concentration of the new solution and m2 is its new mass.
M1m1 = M2m2
.925(m1) = .35(250)
m1 = 94.59 g
Answer: 94.59 grams of 92.5 % by mass solution will be needed.
Explanation:
Mass of sulfuric acid is 250 grams of 35 % by mass solution:
Mass of in 250 g of 35 % solution = 87.5 g
Mass of 92.5 % needed to make 35 % by mass solution.
Mass of the solution required = 94.59 g
94.59 grams of 92.5 % by mass solution will be needed.
The combustion reaction of the unknown compound can be written as follows:
Other part of the question is discussed below:
We know that 3.50 g of the compound produced 5.13 g of carbon dioxide and 2.10 g of water. This means that 1.63 g of oxygen were also produced in the reaction.
The mass of oxygen in the original sample is equal to the mass of oxygen produced in the reaction, so the original sample contained 1.63 g of oxygen.
The total mass of carbon and hydrogen in the original sample is equal to the mass of the sample minus the mass of oxygen, which is 3.50 g - 1.63 g = 1.87 g.
We can find the number of moles of hydrogen in the original sample by dividing the mass of hydrogen by the molar mass of hydrogen (1.008 g/mol). This gives us 1.87 g / 1.008 g/mol = 1.86 moles of hydrogen.
Therefore, the answer is 1.86
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By dividing the mass of water produced in combustion by the molar mass of Hydrogen in water, we find that the original compound contained 1.05 mol of Hydrogen.
To determine the mole amount of Hydrogen in the compound, you have to consider the reaction applied, and how it applies to the law of conservation of mass. Combustion of the compound produces carbon dioxide (CO2) and water (H2O). For water, one mol equals to the molar mass of H which is 1 g/mol, times the amount of H in water which is 2 to get 2 g/mol. Now knowing that 2.10 g of water were produced from combustion, we divide this by the molar mass of H-in-water to get the number of hydrogen moles in the original sample. So, the calculation will be 2.10 g/ 2 (g/mol) = 1.05 mol. So, the original compound contained 1.05 mol of hydrogen.
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