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3 years ago

Name physical properties

Chemistry

1 answer:

Paraphin [41]3 years ago

8 0

Answer:

Some examples of physical properties are:

color (intensive)

density (intensive)

volume (extensive)

mass (extensive)

boiling point (intensive): the temperature at which a substance boils.

melting point (intensive): the temperature at which a substance melts.

Explanation:

Hope this helped! <3

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What is the frequency of 3.98 x 10^-77

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Answer:

What is the frequency of a 6.9 x 10-13 m wave? 3.00 x 108 = 6.9x10-13 mly). GAMMA. V = 4.35 x 10 20 5-11. 3. What is the wavelength of a 2.99 Hz wave?

Missing: 3.98 ‎77

Explanation:

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3 years ago

3. What noble gas would be part of the electron configuration notation for Mn?

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Answer:

Argon {Ar}

Explanation:

The noble gas used for a condensed electron configuration is the one before the element which you are configuring.

In this case, the element (Mn) is manganese

The noble gas that is before this element is Argon which is the row above it

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This is the equation for the combustion of propane.

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Answer:

Explanation:

Reactants are the compounds which reaction produce the products. In general terms this can be expressed symbolically as follows:

reactants -> products

Other phenomena like heat are omitted because are not always present, that is, only compounds are included. Therefore, in this reaction the reactants are C3H8 (propane) and O2 ( oxygen) and the products are CO2 (carbon dioxide) and H2O (water)

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2 years ago

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When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

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3 years ago

How long would it take for 1.50 mol of water at 100.0 âc to be converted completely into steam if heat were added at a constant

defon

To convert boiling water to steam, that would involve heat of vaporization. The heat of vaporization for water at atmospheric conditions is: ΔHvap = 2260 J/g.

Molar mass of water = 18 g/mol

Q = mΔHvap = (1.50 mol water)(18 g/mol)(2260 J/g) = 61,020 J

Time = Q/Rate = (61,020 J)(1 s/20 J) = 3051 seconds

In order to express the answer in three significant units, let's convert that to minutes.
Time = 3051 s * 1min/30 s = 102 min

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2 years ago