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vampirchik [111]
3 years ago
13

Two unknown compounds are tested. Compound 1 contains 15.0g of hydrogen and 120.0g of oxygen. Compound 2 contains 2.0g of hydrog

en and 32.0g of oxygen. Are the compounds the same? Explain your answer.
Chemistry
1 answer:
ivann1987 [24]3 years ago
7 0

Answer:

The both compounds are different.

Explanation:

In order to confirm weather both compounds are same we will check the mole ration. If it is same the compounds will be same.

Given data:

For compound 1.

Mass of hydrogen = 15 g

Mass of oxygen = 120 g

Moles of hydrogen and oxygen = ?

Number of moles of hydrogen = 15 g/ 1g/mol = 15 mol

Number of moles of oxygen = 120 g/ 16 g/mol = 7.5 mol

Total number of moles = 22.5 mol

% of hydrogen = 15 /22.5 × 100 = 66.7%

% of oxygen = 7.5 / 22.5× 100 = 33.3%

For compound 2:

Mass of hydrogen = 2 g

Mass of oxygen = 32 g

Moles of hydrogen and oxygen = ?

Number of moles of hydrogen = 2 g/ 1g/mol = 2 mol

Number of moles of oxygen = 32 g/ 16 g/mol = 2 mol

Total number of moles = 4 mol

% of hydrogen = 2 /4 × 100 = 50%

% of oxygen = 2 / 4× 100 = 50%

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Yuliya22 [10]

Answer:

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Explanation:

However, an alpha hydrogen is the hydrogen that is found on the alpha, α-carbon in an organic molecule. Alpha carbon is referred to the first carbon that is attached to a functional group. Generally, compounds that do not have alpha carbon do not have alpha hydrogen. For example, first member of all functional groups do not usually have alpha carbon and hence do not have alpha hydrogen.

Also, Alkanes, alkenes and alkynes do not have also

6 0
3 years ago
Given the following information: benzoic acid = C6H5COOH hydrocyanic acid = HCN C6H5COOH is a stronger acid than HCN (1) Write t
Maurinko [17]

Answer:

The net ionic equation is

C6H5COOH+ CN-= C6H5COO- + HCN

Explanation:

From the ionic equation

C6H5COOH + Na+ + CN- = C6H5COO- + Na+ + HCN

Only sodium is the spectator ion, so it cancels out, since C6H5COOH and HCN do not ionize completely they are left undissociated

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How is the periodic law demonstrated in halogens
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Halogens is defined as the group of 7 periodic table. As, every periodic table contains 7 valence electrons and they only need 1 more to complete an outer shell, that is why they are extremely reactive. And according to the law that recurring patterns of the properties of elements arise when they are arranged in order of increasing atomic number. As the halogen all act very similarly with each other in chemical reaction, it is true.
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For the reaction represented by the equation Fe + H2O ® Fe2O3 + H2, how many moles of iron(III) oxide are produced from 285 g of
schepotkina [342]

Balanced equation: 2Fe + 3H2O → Fe2O3 +3H2

Convert g to mols:

285/55.845 = 5.1034 mols

Mole ratio of Iron and Iron (III) Oxide: 2:1

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8 0
3 years ago
How many atoms are there in 8.88 g Si?
Mariana [72]

Answer:

\boxed {\boxed {\sf 1.90 \times 10^{23} \ atoms \ Si}}

Explanation:

We are asked to find how many atoms are in 8.88 grams of silicon.

<h3>1. Grams to Moles </h3>

First, we convert grams to moles. We use the molar mass or the mass of 1 mole of a substance. These values are found on the Periodic Table as they are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up silicon's molar mass.

  • Si:  28.085 g/mol

We will convert using dimensional analysis. Set up a conversion factor with the molar mass.

\frac { 28.085 \ g \  Si}{1 \ mol \ Si}

We are converting 8.88 grams of silicon to moles, so we multiply by this value.

8.88 \ g \ Si *\frac { 28.085 \ g \  Si}{1 \ mol \ Si}

Flip the fraction so the units of grams of silicon cancel.

8.88 \ g \ Si *\frac{1 \ mol \ Si} { 28.085 \ g \  Si}

8.88  *\frac{1 \ mol \ Si} { 28.085 }

\frac {8.88} { 28.085 } \ mol \ Si

0.316183015845 \ mol \ Si

<h3>2. Moles to Atoms </h3>

Next, we convert moles to atoms. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of silicon.

Set up another conversion factor.

\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}

Multiply by the number of moles we calculated.

0.316183015845\ mol \ Si *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}

The units of moles of silicon cancel.

0.316183015845 * \frac {6.022 \times 10^{23} \ atoms \ Si}{1}

0.316183015845 * {{6.022 \times 10^{23} \ atoms \ Si}

1.90405412 \times 10^{23} \ atoms \ Si

<h3>3. Significant Figures</h3>

The original measurement of 8.88 grams has 3 significant figures, so our answer must have the same.

For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 0 in the hundredth place.

1.90 \times 10^{23} \ atoms \ Si

<u>8.88 grams of silicon contains 1.90 ×10²³ atoms of silicon.</u>

6 0
3 years ago
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