All categories

2 years ago

Methane gas, which may be present in confined spaces such as a basement, can cause death by __________.

Chemistry

1 answer:

stiks02 [169]2 years ago

4 0

Methane gas, which may be present in confined spaces such as a basement, can cause death by asphyxiation.

<h3>What are the dangers of methane gas in confined space?</h3>

Low amounts are safe for inhalation. A significant amount can replace oxygen in the atmosphere. Reduced oxygen availability can cause symptoms like weariness, emotional instability, fast breathing, and rapid heart rate. Insufficient oxygen can cause nausea, vomiting, collapse, convulsions, coma. The onset of symptoms is accelerated by physical activity. Organs including the heart and brain can suffer long-term damage from a lack of oxygen.

Contact with Skin: Not irritating. The skin might become chilled or frozen when in direct touch with the liquid gas (frostbite). Itching and numbness are signs of mild frostbite. Burning and rigidity are signs of more serious frostbite.

No evidence of carcinogenicity.

learn more about methane gas refer:

brainly.com/question/25649765

#SPJ4

You might be interested in

Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as molarity, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

n = moles (unit mol)

V = volume (unit L)

Find the molar mass (M) of potassium hydroxide.

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

Calculate the moles of potassium hydroxide.

$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

Convert the volume of water to litres.

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

Calculate the concentration.

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0

3 years ago

Calculate the standard entropy of vaporization of ethanol at its boiling point, 352 K. The standard molar enthalpy of vaporizati

Korvikt [17]

Answer : The correct option is, (b) +115 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{vap}}{T_b}$

where,

$\Delta S$ = change in entropy

$\Delta H_{vap}$ = change in enthalpy of vaporization = 40.5 kJ/mol

$T_b$ = boiling point temperature = 352 K

Now put all the given values in the above formula, we get:

$\Delta S=\frac{\Delta H_{vap}}{T_b}$

$\Delta S=\frac{40.5kJ/mol}{352K}$

$\Delta S=115J/mol.K$

Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K

8 0

4 years ago

If you were asked to convert 4.2 × 1022 atoms of aluminum to moles, which of the following should you use for the conversion?

DanielleElmas [232]

4.2×1022NA N_{A} is the avagadro number

7 0

3 years ago

What is the total number of moles of NaCl(s) needed to make 3.0 liters of a 2.0 M NaCl solution? *

RideAnS [48]

Answer:

6 moles of NaCl are needed to make 3.0 liters of a 2.0 M NaCl solution.

Explanation:

A 2M solution means that there are 2 moles of solute (NaCl in this case) in 1 liter of solution:

1 L solution-----2 moles of NaCl

3 L solution----x= (3 L solutionx2 moles of NaCl)/1 L solution= 6 moles of NaCl

5 0

3 years ago

If you can continue to add more solute to a solution, the solution is said to be ____________________

Savatey [412]

Answer:

Unsaturated

Explanation:

In order to successfully answer this question, we need to think about the solubility of solutes in specific solvents, typically water.

A solution is considered to be unsaturated if at a given temperature and volume of water we may still add more solute and it will dissolve;
A solution is considered to be saturated if at a given temperature and volume of water we have a maximum amount of solute dissolved and trying to add more solute results in undissolved crystals that can be seen in the solution;
A solution is considered to be oversaturated (or supersaturated) i at a given temperature and volume of water we exceeded the maximum amount of a solute that could possibly dissolve.

In this case, if we can continue to add more solute to a solution and the solute dissolves, we may state that we are still at a point in which we have an unsaturated solution.

4 0

3 years ago

Read 2 more answers