All categories

3 years ago

Which one of these statements about carbon tetrachloride fire extinguishers is true?

Chemistry

1 answer:

Fofino [41]3 years ago

5 0

Answer;

-Prohibited from use by OSHA

Explanation;

-Carbon tetrachloride is a clear and colorless liquid with an odor similar to that of chloroform.It is employed as a chemical reagent for a number of purposes, as a raw material in chemical manufacture and is, used very widely as a reagent to extinguish fires.

-It volatizes very easily, is a non-conductor of electricity and, as noted above, freezes at very low temperatures.

-It is effective, due to the blanketing effect of its vaporized fumes, on many Class B or volatile, flammable liquid fires and because it will mix with certain volatile, flammable liquids and form a non-burning combination.

You might be interested in

Calculate the number of oxygen atoms in 3.65 mol Fe(ClO3)3

IRINA_888 [86]

Answer:

no thank you

Explanation:

5 0

3 years ago

A. Use the half-life of carbon 14 to find the value or the parameter b, and plot the function.

vesna_86 [32]

A use the half life of carbon 14 to find the value or the parameters b and plot the function

5 0

4 years ago

A 5.00 liter gas sample is collected at a temperature and pressure of 27.0 degree C and 1.20 atm. It is desired to transfer the

kipiarov [429]

Answer:

The temperature of the gas in the 3.00 liter container, must be 150K

Explanation:

Let's apply the Ideal Gas Law, to find out the moles

P . V = n . R . T

1,20 atm . 5L = n . 0,082 L.atm/mol.K . 300 K

(1,20 atm . 5L) / (0,082 mol.K/L.atm . 300 K) = n

6/24,6 mol = n = 0,244 moles

We have the moles now, so let's find the temperature in our new conditions.

P . V = n . R . T

1 atm . 3L = 0,244 moles . 0,082 L.atm/mol.K . T° in K

(1 atm . 3L / 0,244 moles . 0,082 mol.K/L.atm) = T° in K

3/20,008 K = T° in K = 150K

3 0

3 years ago

Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the following reaction equation. 2H202(aq)

inn [45]

Answer : The temperature for non-catalyzed reaction needed will be 456 K

Explanation :

Activation energy : The energy required to initiate the reaction is known as activation energy.

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

Since, the rate for both the reaction are equal.

$K_1=K_2$

$A\times e^{\frac{-Ea_1}{RT_1}}=A\times e^{\frac{-Ea_2}{RT_2}}$

$\frac{Ea_1}{T_1}=\frac{Ea_2}{T_2}$ ..........(1)

where,

$Ea_1$ = activation energy for non-catalyzed reaction = 75 kJ/mol

$Ea_2$ = activation energy for catalyzed reaction = 49 kJ/mol

$T_1$ = temperature for non-catalyzed reaction = ?

$T_2$ = temperature for catalyzed reaction = $25^oC=273+25=298K$

Now put all the given values in the above formula 1, we get:

$\frac{Ea_1}{T_1}=\frac{Ea_2}{T_2}$

$\frac{75kJ/mol}{T_1}=\frac{49kJ/mol}{298K}$

$T_1=456K$

Therefore, the temperature for non-catalyzed reaction needed will be 456 K

3 0

3 years ago

A) What volume of butane (C 4 H 10 ) can be produced at STP, from the reaction of 13.45 g of carbon with 17.65 L of hydrogen gas

UNO [17]

From the stoichiometry of the reaction, carbon is in excess and 5.856 g s left over.

<h3>What is the volume of butane produced?</h3>

The reaction can be written as; 4C(s) + 5H2(g) -----> C4H10(g)

Number of moles of C = 13.45 g/1 2g/mol = 1.12 moles

If 1 mole of hydrogen occupies 22.4 L

x moles of hydrogen occupies 17.65 L

x = 0.79 moles

Now;

4 moles of carbon reacts with 5 moles of hydrogen

1.12 moles of carbon reacts with 1.12 moles * 5 moles/4 moles

= 1.4 moles of hydrogen

Hence hydrogen is the limiting reactant here and carbon is in excess.

If 4 moles of carbon reacts with 5 moles of hydrogen

x moles of carbon reacts with 0.79 moles of hydrogen

x = 0.632 moles

Number of moles of carbon unreacted = 1.12 moles - 0.632 moles

= 0.488 moles

Mass of carbon unreacted = 0.488 moles * 12 g/mol

= 5.856 g

Volume of butane produced is obtained from;

5 moles of hydrogen produces 1 mole of butane

0.79 moles of hydrogen produces 0.79 moles * 1 mole/ 5 moles

= 0.158 moles

1 mole of butane occupies 22.4 L

0.158 moles of butane occupies 0.158 moles * 22.4 L/ 1 mole

= 3.53 L

Learn more about stoichiometry:brainly.com/question/9743981

#SPJ1

5 0

2 years ago