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Allisa [31]
3 years ago
13

Element X had a mass of 300g and a 1/2 life of 10 years. how many grams will remain after 40 years?

Chemistry
1 answer:
kiruha [24]3 years ago
3 0
<h3>Answer:</h3>

18.75 grams

<h3>Explanation:</h3>
  • Half-life refers to the time taken by a radioactive material to decay by half of the original mass.
  • In this case, the half-life of element X is 10 years, which means it takes 10 years for a given mass of the element to decay by half of its original mass.
  • To calculate the amount that remained after decay we use;

Remaining mass = Original mass × (1/2)^n, where n is the number of half-lives

Number of half-lives = Time for the decay ÷ Half-life

                                 = 40 years ÷ 10 years

                                  = 4

Therefore;

Remaining mass = 300 g × (1/2)⁴

                            = 300 g × 1/16

                             = 18.75 g

Hence, a mass of 300 g of an element X decays to 18.75 g after 40 years.

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How would a system in which both matter and energy are exchanged freely between the system and the surroundings be classified? (
Nata [24]

Answer:

An open system

Explanation:

An open system is a system in which both matter and energy are exchanged freely between the system and the surroundings.

An example is a pot of water boiling on the stove. The surroundings (the stove) can supply heat energy to the water and the water can escape into the atmosphere.

A <em>closed system</em> is a system in which energy but not matter is exchanged freely between the system and the surroundings.

An example is a pressure cooker on the stove. The surroundings (the stove) can supply heat energy to the food inside, but no matter can escape through the closed lid.

An <em>isolated system</em> is a system in which neither energy nor matter can be exchanged between the system and the surroundings.

An example is a thermos of hot soup. The cap prevents matter from escaping and the shiny interior reflects heat back into the soup.

8 0
3 years ago
One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate
Ugo [173]

Answer:

\delta H_{rxn} = -66.0  \ kJ/mole

Explanation:

Given that:

3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \  \ \delta H = -47.0 \ kJ/mole  -- equation (1)  \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)}  \ \ \delta H = -25.0 \ kJ/mole  -- equation (2)  \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole  -- equation (3)

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

3FeO_{(s)} + CO_{2(g)}    \to    Fe_3O_4_{(s)} + CO_{(g)}   \ \delta H = -19.0 \ kJ/mole  -- equation (4)

Multiplying  (2) with equation (4) ; we have:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

From equation (1) ; multiplying (-1) with equation (1); we have:

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

From equation (2); multiplying (3) with equation (2); we have:

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

Now; Adding up equation (5), (6) & (7) ; we get:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

<u>                                                                                                                      </u>

FeO  \ \ \ +  \ \ \ CO   \ \  \to   \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \  \delta H = - 66.0 \ kJ/mole

<u>                                                                                                                     </u>

<u />

\delta H_{rxn} = \delta H_1 +  \delta H_2 +  \delta H_3    (According to Hess Law)

\delta H_{rxn} = (-38.0 +  47.0 + (-75.0)) \ kJ/mole

\delta H_{rxn} = -66.0  \ kJ/mole

8 0
3 years ago
Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr
VikaD [51]

Answer : The value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

Explanation :

The balanced cell reaction will be,

Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)

The half-cell reactions are:

Oxidation reaction (anode) : Pb(s)\rightarrow Pb^{2+}(aq)+2e^-

Reduction reaction (cathode) : 2Ag^+(aq)+2e^-\rightarrow 2Ag(g)

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

\Delta G^o=-2\times 96500\times 0.93

\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol

Now we have to calculate the value of 'K'.

\Delta G^o=-RT\ln K

where,

\Delta G_^o =  standard Gibbs free energy  = -180 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K

K=3.6\times 10^{31}

Therefore, the value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

5 0
3 years ago
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uysha [10]

When comparing single bonds between atoms of comparable types, the stronger the bond is, the bigger the atom, the weaker it is.

The length of the X-H bond lengthens while the strength of the bond shortens with increasing halogen size (F-H strongest, I-H weakest). When comparing single bonds between atoms of similar sorts, the larger the atom, the weaker the bond. It can be explained by the fact that less energy is required to break the bond the bigger the atom's atomic size. The force of attraction from the nucleus to the outermost orbit will be less for iodine since it has a larger atom than the other elements in the group.

Learn more about single bonds here-

brainly.com/question/16626126

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Evidence that best supports the theory of biological evolution was obtained from the
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The homologous structures and the analogous structures of different species.
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