The mass of 1 curie of cesium-137 is approximately 3.7 grams.
To find the original mass of the cesium-137 sample, you can use the exponential decay formula: final amount = initial amount * (1/2)^(time/half-life). With the information provided, you would have: 12.5 = initial amount * (1/2)^(90.69/30.1). Solving for the initial amount gives you approximately 40 grams.
The element is cesium (Cs) and the cation is Cs^1+
To determine how much cesium would remain, it's essential to know the initial amount, the half-life of the specific cesium isotope in question (e.g., cesium-137 has a half-life of about 30 years), and the elapsed time. The remaining quantity can be calculated using the formula ( N = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}} ), where ( N_0 ) is the initial amount, ( t ) is the elapsed time, and ( T_{1/2} ) is the half-life. Without specific values, a precise answer cannot be given.
The number 87 when referred to francium is that element's atomic number. The most common isotope of francium is 223, which has a half-life of 22 minutes and decays by beta-negative emission into radium-223.
Cesium, or Cs, has 1 valence electron.
To find the original mass of the cesium-137 sample, you can use the exponential decay formula: final amount = initial amount * (1/2)^(time/half-life). With the information provided, you would have: 12.5 = initial amount * (1/2)^(90.69/30.1). Solving for the initial amount gives you approximately 40 grams.
Since the half-life of cesium-137 is about 30 years, 3 half-lives would have passed in 90 years. The first half-life would leave .5 mg of cesium-137. The second would leave .25 mg, and the third half-life would leave .175 mg of cesium-137.
One atom of cesium has a mass of 132.9054 amu, and one mole of cesium has a mass of 132.9054 grams, so five moles of cesium has a mass of 664.527 grams.Multiplying the mass of a particle (which can be an atom, molecule, etc.) by Avogadro's number (6.022x1023, the number of particles of a substance in 1 mole) will give you the mass of a mole of that particle, or molar mass, in grams. Avogadro's number is special because the molar mass of a substance will be the same number as its atomic mass, only in grams!
The element is cesium (Cs) and the cation is Cs^1+
The factors of 137 are 1 and 137, because it is prime.
It is a Cesium isotope, with the atomic mass of 112.It could also be an ion depending on how many total electrons it has.
To determine how much cesium would remain, it's essential to know the initial amount, the half-life of the specific cesium isotope in question (e.g., cesium-137 has a half-life of about 30 years), and the elapsed time. The remaining quantity can be calculated using the formula ( N = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}} ), where ( N_0 ) is the initial amount, ( t ) is the elapsed time, and ( T_{1/2} ) is the half-life. Without specific values, a precise answer cannot be given.
Cesium has 1 unpaired electron.
The number 87 when referred to francium is that element's atomic number. The most common isotope of francium is 223, which has a half-life of 22 minutes and decays by beta-negative emission into radium-223.
Cesium, or Cs, has 1 valence electron.
It's CsI, because cesium has a +1 charge and iodine has a -1 charge
The compound formed from cesium and bromine is cesium bromide, with the chemical formula CsBr. It is an ionic compound where cesium contributes a +1 charge and bromine contributes a -1 charge to form a balanced compound.