The equation shows the formation of carbon through an induced fusion reaction between a helium nucleus, also called an alpha particle, and a beryllium atom. Explain how this equation shows that protons and neutrons are conserved during a fusion reaction. Cite specific mass numbers.

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.

The answer is b electromagnetic i believe, i hope i ya out!

    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