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horsena [70]
3 years ago
5

What did Lorenzo Romano Amedeo Carlo Avogadro theorize in 1811?

Chemistry
2 answers:
-BARSIC- [3]3 years ago
8 0

Answer:

He theorized that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

Explanation:

In 1811, Avogadro published an article in a scientific journal, where he distinguished between molecules and atoms. He argued (contrary to what was thought) that in the case of water, the hydrogen and oxygen "atoms" were actually "molecules." One molecule of oxygen would react with two molecules of hydrogen (H2O).

Thus he proclaimed his famous hypothesis: "Equal volumes of any gases contain the same number of molecules when measured under the same conditions of temperature and pressure."

wlad13 [49]3 years ago
3 0

Avogadro's law, which states that equal volumes of gases under the same conditions of temperature and pressure will contain equal numbers of molecules.

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For an object to sink
AfilCa [17]

Answer:

For and object to sink, it must have more density than the liquid in which it is placed. For example, if you have a glass of water and a metal spoon, the spoon will sink because it is both heavier than the water, therefore having more density.

5 0
3 years ago
What is the empirical formula for a substance that is composed of 40.66% carbon, 8.53% Hydrogen,23.72% Nitrogen, and 27.09% Oxyg
prohojiy [21]

Answer:

THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO

Explanation:

The steps involved in calculating the empirical formula of this substance in shown in the table below:

Element                            Carbon           Hydrogen          Nitrogen         Oxygen

1. % Composition              40.66              8.53                 23.72                27.09

2. Mole ratio =

%mass/ atomic mass       40.66/12         8.53/1          23.72/14            27.09/16

                                       =  3.3883            8.53              1,6943             1.6931

3. Divide by smallest

value (0.6931)          3.3883/1.6931    8.53/1.6931    1.6943/1.6931   1.6931/1.6931

                               =      2.001                  5.038           1.0007                      1

4. Whole number ratio        2                       5                   1                               1

The empirical formula = C2H5NO

7 0
3 years ago
What is the empirical formula for a compound which contains 67.1 zinc and the rest is oxygen
wolverine [178]

Answer:

The empirical formula is ZnO2

Explanation:

What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?

Step 1: Data given

Suppose the compound has a mass of 100.0 grams

A compound contains:

67.1 % Zinc  = 67.1 grams

100 - 67.1 = 32.9 % oxygen  = 32.9 grams

Molar mass of Zinc = 65.38 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate moles of Zinc

Suppose the compound is 100 grams

Moles Zn = 67. 10 grams / 65.38 g/mol

Moles Zn = 1.026 moles

Step 3: Calculate moles of O

Moles O = 32.90 grams / 16.00 g/mol

Moles O = 2.056 moles

Step 4: Calculate mol ratio

We divide by the smallest amount of moles

Zn: 1.026/1.026 = 1

O: 2.056/1.026 = 2

The empirical formula is ZnO2

To control this we can calculate the % Zinc for 1 mol

65.38 / (65.38+2*16) = 0.67.1 = 67.2 %

7 0
2 years ago
Which statement correctly describes extreme weather?
sveta [45]

Answer: A

Explanation:

Extreme weather events follow normal climate patterns.

5 0
2 years ago
Researchers used a combustion method to analyze a compound used as an antiknock additive in gasoline. A 9.394 mg sample of the c
LuckyWell [14K]

Answer:

The percent composition of the compound is 90.5 % C and 9.5 % H

Explanation:

Step 1: Data given

Mass of compound = 9.394 mg

Mass  of CO2 yielded = 31.154 mg

Mass of H2O yielded = 7.977 mg

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles CO2

moles of CO2 = (0.031154 g / 44.01 g/mol) = 7.08 * 10^-4 mol CO2

Step 3: Calculate moles C

moles of C = moles of CO2 * (1 mol C / 1 mol CO2)

moles of C = 7.08 * 10^-4 mol

Step 4: Calculate moles H2O

moles of H2O = (0.007977 g / 18.02 g/mol) = 4.43 * 10^-4 mol H2O

Step 5: Calculate moles of H

moles of H = moles of H2O * (2 mol H / 1 mol H2O)

moles of H =  4.43* 10^-4 *2 = 8.86 * 10^-4 mol H

Step 6: Calculate mass of C

mass C = moles C * molar mass C

mass C = 7.08 * 10^-4 mol*12.01 g/mol

mass C = 0.0085 grams

Step 7: Calculate mass of H

mass H = moles H * molar mass H

mass H = 8.86 * 10^-4 mol*1.01 g/mol

mass H = 0.000894 grams

Step 8: Calculate total mass of compound =

0.0085 grams + 0.000894 grams = 0.009394 grams = 9.394 mg

Step 9: Calculate the percent composition:  

% C = (8.50 mg / 9.394 mg) x 100 = 90.5%  

% H = (0.894 mg / 9.394 mg) x 100 = 9.5%

The percent composition of the compound is 90.5 % C and 9.5 % H

6 0
3 years ago
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