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3 years ago

An unknown substance is heated from 10 °C to 25 °C. The table below shows the state of the substance at different temperatures.

Chemistry

2 answers:

bezimeni [28]3 years ago

7 0

Answer:

Its melting point is 17 °C.

Explanation:

The melting point of a substance is the temperature at which it changes state from solid to liquid. At the melting point the solid and liquid phase exist in equilibrium.

So, the melting point is 17 °C.

And this is shown in the figure attached.

astraxan [27]3 years ago

7 0

Answer:

someone already said it but the melting point is 17 degrees c

Explanation:

hope it helps :))))

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A low-pressure weather system comes into the city of Denver. The atmospheric pressure is 693 mmHg.693 mmHg. If 78.0%78.0% of dry

scoundrel [369]

Answer:

540.54 mmHg

Explanation:

We know that the partial pressure of a substance is defined as; Mole fraction * total pressure.

If the total amount of gases in the atmosphere is 100%, the mole fraction of nitrogen gas is now

78/100 = 0.78

Thus, partial pressure of nitrogen gas = 0.78 * 693 = 540.54 mmHg

6 0

2 years ago

The burning of methane gas, given below, is a redox reaction. which part of the reaction illustrates oxidation?

Effectus [21]

The burning of methane gas, given below, is a redox reaction. which part of the reaction illustrates oxidation, Ch4+O2---CO2+H2OCH4---CO2

8 0

3 years ago

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n

mylen [45]

The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide $(CO_2)$ , nitrogen dioxide $(NO_2)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876 g $CO_2$ , and 3.921 g $NO_2$ Determine the empirical formula of the compound.

Answer: The empirical formula for the given compound is $CaCN_2$

Explanation:

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

$Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2$

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of $CO_2=1.876g$

Mass of $NO_2=3.921g$

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

For calculating the mass of carbon:

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

So, in 1.876 g of carbon dioxide, $\frac{12}{44}\times 1.876=0.5116g$ of carbon will be contained.

For calculating the mass of nitrogen:

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, $\frac{14}{46}\times 3.921=1.193g$ of nitrogen will be contained.

For calculating the mass of calcium:

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, $\frac{40}{56}\times 2.389=1.706g$ of calcium will be contained.

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

Step 1: Converting the given masses into moles.

Moles of Calcium = $\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles$

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles$

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

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 0.0426 moles.

For Calcium = $\frac{0.0426}{0.0426}=1$

For Carbon = $\frac{0.0426}{0.0426}=1$

For Nitrogen = $\frac{0.0852}{0.0426}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CaCN_2$

3 0

3 years ago

True or false The atomic number of an element is always more than the mass number of that element.

Veseljchak [2.6K]

The sentence is true

8 0

3 years ago

Formaldehyde is a carcinogenic volatile organic compound with a permissible exposure level of 0.75 ppm. At this level, how many

Flauer [41]

Answer : The amount of formaldehyde permissible are, $5.4\times 10^{-6}g$

Explanation : Given,

Density of air = $1.2kg/m^3=1.2g/L$ $(1kg/m^3=1g/L)$

First we have to calculate the mass of air.

$\text{Mass of air}=\text{Density of air}\times \text{Volume of air}$

$\text{Mass of air}=1.2g/L\times 6.0L$

$\text{Mass of air}=7.2g$

Now we have to calculate the amount of formaldehyde.

Permissible exposure level of formaldehyde = 0.75 ppm = $\frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}$

Amount of formaldehyde in 7.2 g of formaldehyde = $7.2g\times \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}$

Amount of formaldehyde in 7.2 g of formaldehyde = $5.4\times 10^{-6}g$

Thus, the amount of formaldehyde permissible are, $5.4\times 10^{-6}g$

8 0

3 years ago