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DedPeter [7]
3 years ago
5

C. Another radioisotope of americium exists which has an atomic mass of 242.

Chemistry
1 answer:
ollegr [7]3 years ago
7 0

Answer:

1 g  

Explanation:

The half-life of Am-242 (16 h) is the time it takes for half of it to disappear.

We can make a table of the mass left after each half-life.

\begin{array}{cccc}\textbf{No. of} & & \textbf{Percent} & \textbf{Mass}\\\textbf{Half-lives} & \textbf{Time/h} & \textbf{Remaining} & \textbf{Remaining/g}\\0& 0 & 100 & 8\\1 & 16 &50 & 4\\2 & 32 & 25 & 2\\3 & 48 & 12.5 & 1\\4 & 64 & 6.25 & 0.5\\\end{array}

The mass remaining after 48 h  is 1 g.

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Which of the atoms below are least likely to violate the octet rule?
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Answer:

F i think i pretty sure F!!!!!!!!

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3 years ago
Lewis dot structure for N3-
labwork [276]

Answer :

The steps involved in the electron dot structure of N^{3-} are :

First we have to determine the total number of valence electron in N^{3-}.

Number of valence electrons in N = 5

The charge on N is (-3). So, we add 3 electrons.

Total number of valence electrons = 5 + 3 = 8 electrons

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3 years ago
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For a particular reaction, Δ=−111.4 kJ and Δ=−25.0 J/K.
vichka [17]

Answer:

\Delta G =-103.95kJ

Explanation:

Hello there!

In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:

\Delta G =\Delta H-T\Delta S

It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:

\Delta G =-111.4kJ-(298K)(-25.0\frac{J}{K}*\frac{1kJ}{1000J} )\\\\\Delta G =-103.95kJ

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6 0
3 years ago
If the initial temperature of a movable cylinder was 50 degrees Celsius
slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

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3 years ago
SO NOT HALAL MODE :(
amid [387]

Answer:

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Explanation:

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