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Vitek1552 [10]
3 years ago
6

How many atoms are in this formula? 4 O2 A 2

Chemistry
1 answer:
Effectus [21]3 years ago
4 0

Answer:

8 atoms as 4*2=8

Explanation:

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Using only the periodic table, rank the elements in each set in order of increasing size: (a) Se, Br, Cl; (b) I, Xe, Ba.
mezya [45]

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A. Cl, Se, Br

B. I, Xe, Ba

Explanation:

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3 years ago
Hace cuanto el hombre estudia el universo
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Answer:

El Big Bang fue el momento, hace 13.800 millones de años, cuando el universo comenzó como una pequeña y densa bola de fuego que explotó. La mayoría de los astrónomos utilizan la teoría del Big Bang para explicar cómo comenzó el universo.

Explanation:

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3 years ago
True/False Indicate whether the sentence or statement is true or false. The number 1 234 000 in scientific notation is equal to
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False bc 1.234 x 10^5 is 123400
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3 years ago
2NO + 3MnO2 + 4H â 2NO3- + 3Mn2 + 2H2O For the above redox reaction, assign oxidation numbers and use them to identify the eleme
mixer [17]

Answer:

Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).

Explanation:

Hello there!

In this case, according to the given redox reaction, we rewrite it as a convenient first step:

2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O

Next, we assign the oxidation numbers as follows:

2N^{2+}O^{2-} + 3Mn^{4+}O^{-2}_2 + 4H^+ \rightarrow 2(N^{5+}O^{2-}_3)^- + 3Mn^{2+} + 2H^+_2O^{2-}

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).

Regards!

4 0
2 years ago
A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has dec
Delicious77 [7]

Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

3 0
3 years ago
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