All categories

3 years ago

3. Define isobar 75 points

Chemistry

1 answer:

steposvetlana [31]3 years ago

5 0

Answer:

meteorology A line drawn on a map or chart connecting places of equal or constant pressure.

nuclear physics Either of two nuclides of different elements having the same mass number.

thermodynamics A set of points or conditions at constant pressure.

You might be interested in

Calculate the molarity of a solution containing 15.2 grams of NaCl dissolved in 2.5 L of solution.

Advocard [28]

Answer:

en español por favor

Explanation:

no seeeed perdona

8 0

3 years ago

The danger from radon gas would most likely be greatest in:.

Semmy [17]

The danger from radon gas would most likely be greatest in well-insulated homes. It is a toxic gas.

<h3>What is radon gas?</h3>

Radon is a naturally radioactive gas generated from different elements such as uranium or radium.

Radon is generated when these radioactive materials naturally decompose in the soils and/or rocks.

Radon is highly toxic and therefore this gas must be avoided to maintain healthy conditions in a home.

Learn more about the radon gas here:

brainly.com/question/1121893

5 0

2 years ago

The most concentrated solution is

aev [14]

The most concentrated solution is b

6 0

3 years ago

Read 2 more answers

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 $C_6H_4O_2$

6 0

3 years ago

Simplest mole ratio for glucose

EastWind [94]

Answer:1. In Glucose: C : H : O = 1 : 2 : 1

2. In Sulfuric acid: H : S : O = 2 : 1 : 4

3. In Butene: C : H = 1 : 2

4 0

2 years ago