Chemistry, 26.12.2019 18:31 deandrathomas34

When 5 g of barium chloride reacted with 3 g of sodium sulfate the products formed were 3.5 g of barium sulfate and x gram of sodium chloride what will be the amount of sodium chloride ph5 g of barium chloride reacted with 3 g of sodium sulfate the products formed were 3.5 g of barium sulfate and x gram of sodium chloride what will be the amount of sodium chloride formed

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Answer from: jjmage2206

The amount of sodium chloride formed is 0.03 moles which are contained in 1.75 g

Explanation:

The reaction is this:

BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl

5g 3g 3.5g x

First of all, let's calculate the moles of everything

Mol = Mass / Molar mass

5g / 208.23 g/m = 0.024 mol BaCl₂

3g / 142.06 g/m = 0.0211 mol Na₂SO₄

3.5g / 233.39 g/m = 0.0150 mol BaSO₄

We can now determinate, the limiting reactant to work

Ratio between reactants is 1:1, so 1 mol of chloride reacts with 1 mol of sulfate. Limiting agent is the Na₂SO₄ ( I need 0.024 mol to react the BaCl₂, and I only have 0.0211 mol, BaCl₂ will remain without reaction)

Ratio is 1:1 between the sulfates so, 0.0211 mol of sodium sulfate produce 0.0211 mol of barium sulfate but I only produced 0.0150 mole of it.

This reaction has a % of yield.

0.0211 moles ___ 100 %

0.0150 moles ___ (0.0150 . 100)/0.0211 = 71.09 %

Now we can know the production of NaCl

Ratio is 1:2, so If i have 0.0211 mol of sodium sulfate I'll produce the double of NaCl (0.0211 .2) = 0.0422 mol

As the reaction has a 71.09% yield, I'll produce

0.0422 mol . 0.7109 = 0.03mol

Molar mass NaCl = 58.45 g/m

0.03 m . 58.45 g/m = 1.75 g

Answer from: Quest

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.

Answer from: Quest

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.

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