Given the unbalanced equation: _a1 + _cuso4 _aqx(so4)3 + _cuwhen the equation is balanced using the smallest whole -number coefficients, what is the coefficient of a1

Radon-222 or rn-222 has a half-life of 3.82 days. to determine the fraction of an original sample of rn-222 that remains unchanged after 7.646 days, we calculate the amount of rn-222 remaining after 7.646 days and divide it with the original amount. using 1 gram as a basis and using half-life calculation methods, we can determine that  0.249 or 0.25  grams of rn-222 will remain, and thus the fraction remaining is 0.25 or 1/4.

The correct answer is:   [c]:     "it is the minimum amount of material needed to sustain a fission  reaction."

    pb(so₄)₂this element is lead (iv) sulfate. lead is a multivalent ion. but we know that sulfate ions, so₄²⁻, have a charge of -2. in order to have the ratios match up, we need pb⁴⁺ (two times as many pb⁴⁺  than so₄²⁻).the molar mass can be calculated by finding the mass of one mole of this compound. consider the following calculation,  where amu represents the atomic mass unit. the amu for each atom involved in this compound can be found on a periodic table.  molar mass = amu pb + 2(amu s + 4 * amu o)from a periodic table (with masses rounded to the nearest tenth),      amu pb = 207.2 g    amu s =  32.1 g    amu o = 16.0 gtherefore    molar mass =  207.2 + 2(32.1 + 4*16.0) = 399.4 gthe molar mass is therefore 399.4 grams per mol

Molar mass ca = 40.078 g/mol  molar mass cacl2 = 110.984 g/mol  % ca = 40.078 x 100 / 110.984 = 36.1