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3 years ago

Is flammability an example of physical property

Chemistry

2 answers:

Alina [70]3 years ago

6 0

Fammability is a chemical change because it occurs when one or more substances change into entirely new substances.

kipiarov [429]3 years ago

5 0

It's a chemical property

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Which type of front occurs when cold air and warm air are next to each other, but are at a standstill?

noname [10]

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3 years ago

Which statement is true of diffusion?

raketka [301]

Answer:

Molecules move from areas of high concentration to areas of low concentration.

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2 years ago

A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128

alisha [4.7K]

Answer: The final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

Explanation:

The given data is as follows.

mass = 7.62 g, $T_{2} = 10.8^{o}C$

Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.

q = $mC \times \Delta T$

= $7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$

Now, heat gained by lead will be calculated as follows.

q = $mC \times \text{Temperature change of lead}$

= $2.04 \times 0.128 \times (T - 11.0)$

According to the given situation,

Heat lost = Heat gained

$7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$ = $2.04 \times 0.128 \times (T - 11.0)$

T = $50.26^{o}C$

Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

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3 years ago

Select the options that are properties of electromagnetic waves:

kaheart [24]

Answer:

They propagate in materialistic media and non-materialistic media ( space ) .

They propagate in the space at constant velocity , which is 3 × 108 m/s .

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Explanation:

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3 years ago

Imagine that you have a 5.00 L gas tank and a 3.50 L gas tank. You need to fill one tank with oxygen and the other with acetylen

Lorico [155]

Answer:

77.14 atm of pressure should be of an acetylene in the tank.

Explanation:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

According to reaction, 2 moles of acetylene reacts with 5 moles of oxygen.

Moles of oxygen= $n_1$

Moles of acetylene = $n_2$

$\frac{n_1}{n_2}=\frac{5 mol}{2 mol}=\frac{5}{2}$

Volume of large tank with oxygen gas, $V_1 = 5.00 L$

Pressure of the oxygen gas inside the tank = $P_1=135 atm$

$RT=\frac{P_1V_1}{n_1}$ ..[1]

Volume of small tank with acetylene gas , $V_2= 3.50 L$

Pressure of the acetylene gas inside the tank = $P_2=?$

$RT=\frac{P_2V_1}{n_2}$ ..[2]

Considering both the gases having same temperature T, [1]=[2]

$\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}$

$P_2=\frac{P_1V_1\times n_2}{V_2\times n_1}$

$=\frac{135 atm\times 5.00 L\times 2}{3.50 L\times 5}=77.14 atm$

77.14 atm of pressure should be of an acetylene in the tank.

8 0

3 years ago