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3 years ago

Half life in chemistry I wasn't here when we did this

Chemistry

1 answer:

tatuchka [14]3 years ago

6 0

A half-life is the amount of time it takes for half of a substance to decay.

1.) After one half-life - 13.5mg.
After two - 6.75mg
Three - 3.375mg

2.) 250mg of Iron-59 will remain.

3.) 14 days is the half-life of Phospherous-32.

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A sulfuric acid solution containing 571.3 g of h2so4 per liter of aqueous solution has a density of 1.329 g/cm3. Part a calculat

loris [4]

Mass percentage of a solution is the amount of solute present in 100 g of the solution.

Given data:

Mass of solute H2SO4 = 571.3 g

Volume of the solution = 1 lit = 1000 ml

Density of solution = 1.329 g/cm3 = 1.329 g/ml

Calculations:

Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g

Therefore we have:

571.3 g of H2SO4 in 1329 g of the solution

Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987

Mass percentage of H2SO4 (%w/w) is 42.99 %

3 0

3 years ago

Calculate the number of atoms of sodium in a 4.5-gram sample

allochka39001 [22]

Answer:

1.18×10²³ atoms.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms.

From the above concept, 1 mole of sodium also contains 6.02×10²³ atoms.

1 mole of sodium = 23 g.

Thus,

23 g of sodium contains 6.02×10²³ atoms.

Therefore, 4.5 g of sodium will contain = (4.5 × 6.02×10²³)/23 = 1.18×10²³ atoms.

From the above calculation,

4.5 g of sodium contains 1.18×10²³ atoms.

7 0

3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

Write a general word equation for a neutralization reaction.

xenn [34]

Hydrochloric acid + Sodium hydroxide =Sodium chloride +water

8 0

3 years ago

What does this do to the electrons outside the nucleus in the gaseous atoms

AleksandrR [38]

Answer:

Explanation:

As you know, ionization energy is the energy needed to remove one mole of electrons from one mole of atoms in the gaseous state

energy

→

−

Right from the start, you can tell that the harder it is to remove an electron from an atom, the higher the ionization energy will be.

Now, the periodic trends for ionization energy can be describe as follows

ionization energy increases as you move from left to right across a period

ionization energy decreases as you go down a group

As you mentioned, if you compare the first ionization energies for oxygen and chlorine using these two trends, you will get conflicting results.

If you follow the way ionization energy increases across period, chlorine would have a higher ionization energy, since it's closer to the noble gases.

On the other hand, if you go by how ionziation energy decreases from top to bottom in a group, oxygen would have higher ionization energy, since it's located in period 2, as compared with period 3 for chlorine.

As it turns out, the trend for groups overpowers the trend for periods. As aresult, oxygen will have a higher ionization energy than chlorine.

This happens because the smaller oxygen atom has its outermost electrons held tighter by the nucleus. By comparison, chlorine's outermost atoms are located further away from the nucleus.

Not only that, but they are screened from the charge of the nucleus better, since they're located on the third energy level.

Oxygen's outermost electrons are screened by

electrons, while chlorine's are screened by

electrons.

All these factors will make chlorine's outermost electrons a little easier to remove, which implies a smaller ionization energy than that of oxygen.v

6 0

3 years ago