Answer:
Positron emission
Explanation:
In positron emission, a proton transforms into a neutron. This alteration results in a daughter nucleus with its mass number increased by 1, while the atomic number remains unchanged. The formation of a new neutron boosts the neutron count in the daughter nucleus, thus enhancing the N/P ratio.
Concurrently, a positron is emitted along with an anti-neutrino to ensure spin conservation.
Answer:
Oxygen's mass percent in Fe(OH)3 is 44.92%
Explanation: The mass percentage is a means of indicating the concentration of a specific element within a compound. It is determined through the ratio of the element's mass to the compound's total mass, multiplied by 100.
•First calculate the overall mass of the compound
•Fe's molar mass = 55.85 g/mol
•O's molar mass = 16 g/mol
•H's molar mass = 1 g/mol
Using these values, we can compute the molecular mass of Fe(OH)3 = 55.85 g/mol + (16 g/mol)3 + (1 g/mol)3
=55.85 g/mol + 48 g/mol + 3 g/mol
=106.85 g/mol
Mass percent of an element = mass of element/total mass of compound × 100
In the case of 3 oxygen atoms present within the compound, the mass of oxygen totals 48 g/mol
Mass percent of oxygen= 48 g/mol/106.85 g/mol × 100
= 0.4492×100= 44.92%
[[TAG_31]]Thus, the mass percent of oxygen in Fe(OH)3 amounts to 44.92%[[TAG_32]]
In the case of a 100m Race, displacement equals the distance traveled. If we divide this equation by the time t (assuming t represents the time taken to finish the 100m race), we derive that velocity equals speed. Conversely, in a 400m race where a full lap is completed, the racer’s starting and ending positions overlap, leading to displacement equaling 0, while the distance is not zero (400m). Therefore, it follows that displacement does not equal distance, leading to the conclusion that velocity does not equal speed.
Answer:
Refer to the explanation.
Explanation:
Formation reactions involve the creation of one mole of a compound from its elements in their standard states.
NaBr (s)
The equation for the standard formation is
Na (s) + (1/2)Br₂ (g) → NaBr (s)
As per appendix C, the standard heat of formation for NaBr(s) is
ΔH∘f = -359.8 kJ/mol.
SO₃ (g)
The equation for the standard formation is
S (s) + (3/2) O₂ (g) → SO₃ (g)
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ΔH∘f = -395.2 kJ/mol.
Pb(NO₃)₂ (s)
The equation for the standard formation is
Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)
According to appendix C, the standard heat of formation for Pb(NO₃)₂(s) is
ΔH∘f = -451.9 kJ/mol.
I hope this is helpful!
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