Answer: Flammability is a material's ability to burn in the presence of oxygen.
Explanation: Chemical properties can be observed only when the substance changes into one or more different substances through chemical reactions or transformations. One of the chemical properties is flammability.
Flammability is a material's ability to burn in the presence of oxygen.
Remember, oxygen doesn't burn. Precisely flammable substances obtain substances that burn. Oxygen remains an oxidizing agent, which means it supports the combustion process. Oxygen causes other objects to catch fire at low temperatures and burns hotter and faster. But oxygen itself does not burn. Consequently, if you at present deliver fuel and fire, adding oxygen will provide the fire.
Carbon dioxide is the result of combustion. An example can be seen in firewood in a fireplace. One of the chemical properties of carbon-based wood is having the ability to burn. Chemically the wood turns into carbon dioxide when it burns and leaves a residue of ash. Furthermore, this ash residue cannot be turned back into the wood. Chemical changes result in new substances.
Consider an example of a combustion reaction to methane gas:
Our balanced equation for methane combustion implies that every one CH₄ molecule reacts with two O₂ molecules. The product of combustion is one carbon dioxide molecule and two steam or water vapor molecules.
Answer:
12.44 g
Explanation:
2C4H10 + 13O2 = 8CO2 + 10H2O
n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).
n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).
Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.
mass of CO2 produced =
M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol
M = 0.5656/2 * 44
M = 0.2828 * 44
M = 12.44 of CO2
Answer:
37.1°C.
Explanation:
- Firstly, we need to calculate the amount of heat (Q) released through this reaction:
<em>∵ ΔHsoln = Q/n</em>
no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.
<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>
∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.
Q = m.c.ΔT,
where, Q is the amount of heat released to water (Q = 2781.87 J).
m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).
c is the specific heat capacity of water (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).
∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)
∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.
<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>
Answer: 1 proton, 0 electron and 1 neutron are present in 1 H⁺ isotope.
Answer: 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.
Explanation:
According to the dilution law,
where,
= concentration of stock solution = 12.0 M
= volume of stock solution = ?
= concentration of diluted solution= 6.00 M
= volume of diluted acid solution = 500 ml
Putting in the values we get:
Thus 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.
