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NeX [460]
3 years ago
9

21. Solve the simultaneous equations graphically taking the values of

Chemistry
1 answer:
tigry1 [53]3 years ago
3 0

Answer:

take the l my gang tfdfhngtyhggggggfggg

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Identify whether longhand notation or noble-gas notation was used in each case below.
n200080 [17]

Answer:

The given electronic configuration is long hand notation.

Explanation:

Long-hand notation of representing electronic configuration is defined as the arrangement of total number of electrons that are present in an element.

Noble-gas notation of representing electronic configuration is defined as the arrangement of valence electrons in the element. The core electrons are represented as the previous noble gas of the element that is considered.

The given electronic configuration of potassium (K):  

The above configuration has all the electrons that are contained in the nucleus of an element. Thus, this configuration is a long-hand notation.

6 0
3 years ago
Pls help fast, I will give brainliest!
beks73 [17]

Answer:

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Explanation:

4 0
3 years ago
Cyclobutane, C4H8, consisting of molecules in which four carbon atoms form a ring, decomposes, when heated, to give ethylene. Th
frosja888 [35]

Answer:

The concentration of cyclobutane after 875 seconds is approximately 0.000961 M

Explanation:

The initial concentration of cyclobutane, C₄H₈, [A₀] = 0.00150 M

The final concentration of cyclobutane, [A_t] = 0.00119 M

The time for the reaction, t = 455 seconds

Therefore, the Rate Law for the first order reaction is presented as follows;

\text{ ln} \dfrac {[A_t]}{[A_0]} = \text {-k} \cdot t }

Therefore, we get;

k = \dfrac{\text{ ln} \dfrac {[A_t]}{[A_0]}}  {-t }

Which gives;

k = \dfrac{\text{ ln} \dfrac {0.00119}{0.00150}}  {-455} \approx 5.088 \times 10^{-4}

k ≈ 5.088 × 10⁻⁴ s⁻¹

The concentration after 875 seconds is given as follows;

[A_t] = [A₀]·e^{-k \cdot t}

Therefore;

[A_t] = 0.00150 × e^{5.088 \times 10^{-4} \times 875}  = 0.000961

The concentration of cyclobutane after 875 seconds, [A_t] ≈ 0.000961 M

6 0
3 years ago
The object has a mass of 5 grams and a volume of 15 mL what is density
CaHeK987 [17]

Answer:

1/3 g/mL

Explanation:

density= mass/ volume

5 0
3 years ago
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