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iris [78.8K]
3 years ago
9

Fire extinguishers that spray carbon dioxide on the fire, work very effectively because it forms a blanket around the burning ma

terial. But these carbon dioxide extinguishers cannot be used for wood fires. Why?
a. The carbon dioxide blanket itself acts as a fuel and starts burning along with the wood.

b. The extinguished wood can retain enough heat inside the carbon dioxide blanket to re-ignite when the gas disappears.

c. In the presence of carbon dioxide from the extinguisher, the burning wood reacts with oxygen to release heat and make the fire worse.

d. Carbon dioxide reacts with wood to give carbon and oxygen; the oxygen makes the fire worse and the carbon burns along with the wood.
Chemistry
1 answer:
Arte-miy333 [17]3 years ago
8 0

Answer:

I think it will option B it will retain enough heat

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Answer:

2C _{6}H _{14} +  19O _{2}→12CO _{2} +14H _{2} O

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Suppose that you add 24.3 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the adde
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Solution :

We know that :

$\Delta T_f = k_f.m$  and   $m=\frac{w_2}{m_2 \times w_1}$

Then, $\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$   ..................(1)

Where,

w_1 = amount of solvent (in kg)

w_2 = amount of solute (in kg)

m_2 = molar mass of solute (g/mole)

m = molality of solution (mole/kg)

Given :

\Delta T_f = 3.14\ ^\circ C,   k_f= 5.12\ ^\circ C/m

                              =5.12 \ ^\circ C/mole/kg

                              =5.12 \ ^\circ C \ kg/mole

w_1 = 0.250 kg,  w_2 = 24.3 g

Then putting this values in the equation is (1),

$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$

$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$

m_2= 158.49  g/mole

So, the molar mass of the unknown compound is 158.49 g/mole.

7 0
3 years ago
Explain how did you find the qostion
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How do protons in the nucleus behave?
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Measurements show that the enthalpy of a mixture of gaseous reactants increases by 130. kJ during a certain chemical reaction, w
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Answer:

167 kJ

Explanation:

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Change in energy, \delta = ?  

From the first law of thermodynamics:

ΔH = ΔE + PΔV,

Where, ΔH= change in enthalpy

ΔE = change in Internal energy  

Substituting values into the above formula —  

130 kJ = ΔE + -137  kJ

130 kJ = ΔE - 137 kJ

therefore,

ΔE = (130 + 137 ) kJ

ΔE = 167 kJ

Therefore, the change in energy of the gas mixture during the reaction is 167 kJ.  

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