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2 years ago

When combined in the correct ratio hydrogen and oxygen atoms can form water as shown below

Chemistry

1 answer:

Dimas [21]2 years ago

8 0

Answer:molecule of a compound

Explanation:

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Draw a line-bond structure for CBrN. Explicitly draw all H atoms. You do not have to include lone pairs in your answer. In cases

DochEvi [55]

Answer:

Br - C ≡ N

Explanation:

To draw the Lewis line-bond structure we need to bear in mind the octet rule, which states that in order to gain stability each atom tends to share electrons until it has 8 electrons in its valence shell.

C has 4 e⁻ in its valence shell so it will form 4 covalent bonds.

Br has 7 e⁻ in its valence shell so it will form 1 covalent bond.

N has 5 e⁻ in its valence shell so it will form 3 covalent bonds.

The most stable structure that respects these premises is:

Br - C ≡ N

It does not have any H atom.

8 0

2 years ago

2HCl+Ca(OH)2⟶CaCl2+2H2O

RUDIKE [14]

Isn’t it just 3 mols?

5 0

2 years ago

Please help, any and all is very much appreciated!

iogann1982 [59]

1. True
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3 0

2 years ago

A certain protein was found to contain 0.292% manganese by mass. Determine the minimum molecular mass of the protein.

Lisa [10]

$55u \ \ \ \ \ \ \Rightarrow \ \ \ \ \ 0,292\%\\
x \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ 100\%\\\\
x = \frac{55u*100\%}{0,292\%}\approx18835,62u$

4 0

3 years ago

Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if

mrs_skeptik [129]

The final volume of the methane gas in the container is 6.67 L.

The given parameters;

initial volume of gas in the container, V₁ = 2.65 L
initial number of moles of gas, n₁ = 0.12 mol
additional concentration, n = 0.182 mol

The total number of moles of gas in the container is calculated as follows;

$n_t = 0.12 + 0.182 = 0.302 \ mol$

The final volume of gas in the container is calculated as follows;

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L$

Thus, the final volume of the methane gas in the container is 6.67 L.

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5 0

2 years ago