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nevsk [136]
3 years ago
5

Which of the following is the best name for a compound made from nitrogen and oxygen? (N2O3) nitrogen oxide nitrogen II oxide ni

trogen trioxide dinitrogen trioxide
Chemistry
2 answers:
xxMikexx [17]3 years ago
6 0
Dinitrogen Trioxide is the answer trust me
sergejj [24]3 years ago
3 0

The best name for the compound is DINITROGEN TRIOXIDE.

Each chemical compound is always represented by a chemical symbol, which shows the ratio at which each atom of the elements of the compound are combine together and this is often used in naming the compound. Looking at the compound given in the question, the compound is made up of two atoms of nitrogen and three atoms of oxygen and this fact was used in naming the compound. In naming chemical compounds, 'Di' stands for 2 while 'Tri' stands for 3. Since there are two nitrogen and three oxygen atoms in the compound, that was why it was named dinitrogen trioxide.

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What is the empirical formula of a compound composed of 30.5 g potassium (k) and 6.24 g oxygen (o)?
VLD [36.1K]
K5O2

convert grams to moles, divide both by the smallest mole mass, multiply that until hole.

30.5 g K ÷ 39.10 = .78 mol
6.24 g O ÷ 16 = .39 mol
.78 mol ÷ .39 mol = 2.5
.39 mol ÷ .39 mol = 1
2.5 x 2 = 5
1 x 2 = 2
K5O2

5 0
3 years ago
Im literally failing some classes can some one help me with them
galben [10]

Answer:in can

Explanation:

5 0
3 years ago
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I cant seem to figure out ANY of these... help when you can pls :.)
Nonamiya [84]
The first one is some reaction with water even I am studying the same
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One of many cells in the retina of the eye that responds to low light. VOCABULARY
Ugo [173]

Answer:

I hope this helps you!

Explanation:

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3 0
3 years ago
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A 100mL reaction vessel initially contains 2.60x10^-2 moles of NO and 1.30x10^-2 moles of H2. At equilibrium the concentration o
Tcecarenko [31]

Answer:

Kc=17.7

Explanation:

Hello,

In this case, for the given reaction:

2H_2(g) + 2NO(g)\rightleftharpoons 2 H_2O(g) + N_2(g)

In such a way, the initial concentrations are:

[H_2]_0=\frac{1.30x10^{-2}mol}{0.1L}=0.130M

[NO]_0=\frac{2.60x10^{-2}mol}{0.1L}=0.260M

Thus, at equilibrium, the change x, due to the chemical reaction extent, turns out:

x=\frac{[NO]_{0}-[NO]_{eq}}{2} =\frac{0.260M-0.161M}{2}=0.049M

Thus, the rest of the concentrations at equilibrium are:

[H_2]_{eq}=0.130M-2(0.049M)=0.032M

[H_2O]_{eq}=2(0.049M)=0.098M

[N_2]_{eq}=0.049M

In such a way, the equilibrium constant for the reaction, result as follows, even when on the statement the NO is excluded, because it participates in the equilibrium:

Kc=\frac{[H_2O]_{eq}^2[N_2]_{eq}}{[H_2]_{eq}^2[NO]_{eq}^2}=\frac{(0.098M)^2(0.049M)}{(0.032M)^2(0.161M)^2} \\\\Kc=17.7

Best regards.

4 0
3 years ago
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