In the atom, each 3p orbital has 4 lobes, due to the existence of the planar nodes present in that orbital (they conform 2 lobes), and also the spherical node, that allows the conformation of the other 2 lobes.
Answer:
Zr. You should be able to see it on a periodic table.
Explanation:
The nuclear disintegration series of 238-U is the source of radon-222 in soil.
<h3>What is nuclear disintegration ?</h3>
- The process of nuclear disintegration is how an unstable atomic nucleus loses energy through radiation.
- A substance that has unstable nuclei is regarded as radioactive.
- Alpha, beta, and gamma decay are three of the most prevalent types of decay, and they all entail the emission of one or more particles.
- Ionizing radiation offers a health concern by destroying tissue and the DNA in genes because it can damage the atoms in living things.
- Alpha particles may be present in the ionising radiation that is released.
- As one atom transforms into another, radioactive decay involves the emission of a particle and/or energy.
- Helium ions are released from an atom's nucleus during alpha decay. A neutron in the nucleus is converted to a proton and electron during beta decay.
Learn more about radioactive decay here:
brainly.com/question/1770619
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<span>Answer: 0.070 m/s
Explanation:
1) balanced chemical equation:
given: 2HBr(g) → H2 (g)+Br2(g)
2) Mole ratios:
2 mol HBr : 1 mol H2
3) That means that every time 2 moles of HBr disappear 1 mol of H2 appears.
That is, the H2 appears at half rate than the HBr disappears.
∴ rate of appearance of H2 = rate of disappearance of HBr / 2 = 0.140 m/s / 2 = 0.070 m/s, which is the answer.</span>
Answer:
An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of −17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.
The answer to the above question is
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
Explanation:
To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows
ΔH = m×c×ΔT
= 0.205×4190×(79.9 -31.0) = 42002.655 J
Therefore fore the ice, we have
Total heat = mi×L + mi×ci×ΔTi = mi×334000 + mi × 2100 × (0 -−17.5) = 42002.655 J
370750×mi = 42002.655 J
or mi = 0.1133 kg
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
