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1 year ago

an atmosphere is considered hazardous if it contains a hazardous gas in excess of 10 percent of the hazardous material's:

Chemistry

1 answer:

svlad2 [7]1 year ago

6 0

Lower flammable limit means the lowest concentration of a material that will propagate a flame.

What is hazardous atmosphere?

It is an atmosphere that may expose employees to risk of death, incapacitation, impairment of ability to self-rescue, injury, or acute illness from one or more of following causes

Flammable gas, vapor, or mist in excess of 10 percent of lower flammable limit (LFL)
Airborne combustible dust at concentration that meets or exceeds its LFL

What is lower flammable limit?

It means the lowest concentration of a material that will propagate a flame.
The LFL is usually expressed as percent by volume of material in air (or other oxidant)
Atmospheres with concentration of flammable vapors at or above 10 percent of lower explosive limit (LEL) are considered hazardous when located in confined spaces.
However, atmospheres with flammable vapors below 10 percent of LEL are not necessarily safe. Such atmospheres are too lean to burn

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Greeley [361]

Answer:

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3 years ago

Which is the best description of the chemical reaction shown here in aqueous solution? ch3ch2cooh (aq) ↔ ch3ch2coo- (aq) + h+ (a

AnnZ [28]

The given reaction:
ch3ch2cooh (aq) ↔ ch3ch2coo- (aq) + h+ (aq)
is called a reversible reaction.

This means that, the reaction does not reach an end point.
In this type of reactions, reactants react together to form products, while products combine together to form reactants.
So, the reaction proceeds in both direction forming both reactants and products.

6 0

2 years ago

Read 2 more answers

For the diprotic weak acid h2a, ka1 = 3.2 × 10-6 and ka2 = 6.1 × 10-9. what is the ph of a 0.0650 m solution of h2a? what are th

Stolb23 [73]

Given:

Diprotic weak acid H2A:

Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9.
Concentration = 0.0650 m

Balanced chemical equation:

H2A ===> 2H+ + A2-
0.0650 0 0
-x 2x x
------------------------------
0.065 - x 2x x

ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)

solve for x and determine the concentration at equilibrium.

5 0

3 years ago

quizlet nitrogen can have variable number of nuetrons(isotopes) it can also have varible number of electrons ehile becomon an io

mafiozo [28]

Nitrogen has 7 protons, 7 neutrons, and 7 electrons.

Seven protons, seven neutrons, and seven electrons make up nitrogen-14.

Utilize the atomic number and mass number of an atom to determine the number of subatomic particles it contains: Atomic number Equals proton count. Electron count equals atomic number. Atomic number - mass number equals the number of neutrons.

Seven protons, seven neutrons, and seven electrons make up the atom of nitrogen. The nucleus is the collection of protons and neutrons that make up the center of an atom. The 7 electrons, which are much smaller than the nucleus, orbit it in what is known as orbits. Since nitrogen-14 is a neutral atom, the number of protons in its nucleus must match the number of electrons around it.

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6 0

1 year ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

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 $CO_2=13.00g$

Mass of $H_2O=2.662g$

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

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

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 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $CHO_2$

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 = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

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

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

3 years ago