Why can sand stop fire?


Answer 1
Answer: It will cover up all oxygen from the fire. And sand has no natural flammability because it is simply a crushed up mineral, so it can not make the fire any bigger.
Answer 2
Answer: Because the fire has no oxygen to expand. it's really suffocating it.

Related Questions


What do oxidation reactions produce during the electrolysis of water?


Answer : Oxidation reaction produces oxygen and water molecule during electrolysis of water.


Electrolysis : It is a type of process in which the chemical decomposition produced by passing an electric current through the solution that containing the ions.

Electrolysis of water : It is basically the decomposition of water into oxygen and hydrogen gas through a process of electric current.

Oxidation reaction is defined as the reaction in which a substance looses its electron to attain stability.

The chemical reactions during electrolysis of water are:

At cathode:  

At anode:  

Overall reaction:

Oxidation reaction occurs at anode.

Hence, oxidation reaction produces oxygen and water molecule during electrolysis of water.

Hydrogen gas, oxygen gas and water

4H2O (l) ---> 2H2O (l) + O2 (g) + 2H2 (g)

Which compounds have the empirical formula CH2O


Chemical Compounds, hope this helps.


A chemistry student is given 600. mL of a clear aqueous solution at 37.° C. He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 21.° C. At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitate, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 0.084 kg.Using only the information from above, can you calculate the solubility of X at 21.° C?If yes, calculate it. Be sure your answer has a unit symbol and the right number of significant figures.



  • Yes, it is 14. g of compound X in 100 ml of solution.


The relevant fact here is:

  • the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.

That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.

Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.

With that, the solubility can be calculated from the followiing proportion:

  • 84. g solute / 600 ml solution = y / 100 ml solution

      ⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.

The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.

The answer is 14. g of solute per 100 ml of solution.


How do minerals affect igneous rock formations?


Mineral composition affects the classification of igneous rock. in simplified classification, igneous to is are classified by the type of feldspar present, by the I type a of feldspar present, or the absence of quartz. in case of neither present, then by the type of iron or magnesium present.

What is the most susceptible to damage from ionizing radiation. sorft tissue, paper, wood, lead


Soft tissue would be the correct answer
Tissues do the most dammage to susceptible radiation

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500. mL flask with 2.3 atm of ammonia gas at 32. °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 0.69 atm. Calculate the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.



Chemical reaction equation for the give decomposition of is as follows:.


And, initially only is present.

The given data is as follows.

  = 2.3 atm at equilibrium

    = 0.69 atm



                        = 0.23 aatm

So, = 2.3 - 2(0.23)

                       = 1.84 atm

Now, expression for will be as follows.




                     = 0.0224


Thus, we can conclude that  the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture is .


The movement of one or more from one reactant to another is called


Answer: Redox Reaction


Redox reaction is the key chemical events in an oxidation-reduction also called Redox. It is the net movement of electrons from one reactants to another


Elements that have properties in common are sometimes classified as a "chemical family". Select the element(s) that are in the same chemical family as gold and silver.


Answer is: copper (Cu).

Gold (Au) and silver (Ag) are noble metals and fall into the transitional metal family on the periodic table in group number 11.

Group 11 is also known as the coinage metals, due to their former usage.

Copper, silver and gold all occur naturally in elemental form.

Copper has atomic number 29, it has 29 protons and 29 electrons.

Electron configuration of copper atom: ₂₉Cu [Ar] 3d¹⁰ 4s¹.

Copper (Cu) is higher in activity series than silver and gold, so copper lose electron and silver gain electrons.


If 3.0×105 j of heat are added to the ice, what is the final temperature of the system?


A wet-chemistry biochemical analyzer was assessed for in-practice veterinary use. Its small size may mean a cost-effective method for low-throughput in-house biochemical analyses for first-opinion practice. The objectives of our study were to determine imprecision, total observed error, and acceptability of the analyzer for measurement of common canine and feline serum analytes, and to compare clinical sample results to those from a commercial reference analyzer. Imprecision was determined by within- and between-run repeatability for canine and feline pooled samples, and manufacturer-supplied quality control material (QCM). Total observed error (TEobs) was determined for pooled samples and QCM. Performance was assessed for canine and feline pooled samples by sigma metric determination. Agreement and errors between the in-practice and reference analyzers were determined for canine and feline clinical samples by Bland-Altman and Deming regression analyses. Within- and between-run precision was high for most analytes, and TEobs(%) was mostly lower than total allowable error. Performance based on sigma metrics was good (σ > 4) for many analytes and marginal (σ > 3) for most of the remainder. Correlation between the analyzers was very high for most canine analytes and high for most feline analytes. Between-analyzer bias was generally attributed to high constant error. The in-practice analyzer showed good overall performance, with only calcium and phosphate analyses identified as significantly problematic. Agreement for most analytes was insufficient for transposition of reference intervals, and we recommend that in-practice-specific reference intervals be established in the laboratory.
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