Convert 32 Fahrenheit to Celsius

On the table below, record your observations before and after the interactions and note if mass was conserved. Describe the chan

lilavasa [31]

Steel Wool + Oxygen (Fire) The steel wool is a grayish color and has a rough surface. Oxygen is transparent It looks like a powder, and like if it was rusted steel wool but then broken apart and turned into a powder Well Iron oxide is 7 grams and steel wool combined is 7 grams so you can say that the mass was conserved Yes, this is a chemical change because the steel wool rusted and rust is a chemical change, so iron oxide is cause because if a physical change.

Egg + Heat The egg has a yellow yolk in the middle while there is some type of liquid at the border of the yolk, but it is clear, the heat is hot but transparent The egg turned into a delicious food called an omelet what was yolk staid as a liquid but got a tad harder, but the transparent border around it turned white It was conserved because the eggshell was 4 grams and the fried egg is 41 grams It is a chemical change because it results in the formation of new particles, and the chemical bonds break up and new ones are formed.

Water + Heat The water is a clear liquid, while the heat is very hot but transparent The water turned into a type of oxygen -Water Vapor- If 5 g of water becomes a gas it becomes 5 g of water vapor. The mass of the liquid water is simply transferred into the mass of the newly formed water vapor. There was no chemical reaction because the water vapor can be turned back to water also it just changed from a liquid to a gas and did not change its composition

Zinc + Hydrogen Chloride Zinc: a white/silver metallic solid. Hydrogen chloride (dilute hydroelectric acid) a transparent, colorless liquid with a very low pH (acidic). Zinc "dissolved" in hydrogen chloride, while emitting a colorless gas. The liquid remains a colorless liquid, possibly still having a low pH from the unused acid. The colorless gas collected in a test tube gives a popping sound when ignited with a burning wooden splint, so it is not air embedded in the zinc, nor dissolved in hydroelectric acid. Well although the zinc chloride is 12 grams and not 15 the gas that was released was 3 grams and as we all know 3 + 12 is 15 so you could say that the mass was conserved The production of a new substance (most probably hydrogen) from the reaction of the two reactants. When a few drops of the liquid product are evaporated on a watch glass, a white residue is left. When a few drops of the liquid hydroelectric acid are evaporated on a watch glass, there is no residue. This proves that a new product (hydrogen gas), (white powder, zinc chloride) is produced instead of zinc being physically dissolved in hydroelectric acid.

Sodium Hydroxide + Copper Sulfate Sodium hydroxide is a turbid solution and copper sulfate is in form of bright blue crystals. When their solutions are mixed with each other, a pale blue precipitate of basic copper hydroxide & a solution of neutral salt sodium sulfate will be formed. The sodium hydroxide and the copper sulfate combined make a total of 67 grams and the product is split because the sodium sulfate is 47 grams and the copper hydroxide is 20 grams but all together it is still 67 grams so you could say that the mass was conserved The proof of the reaction is the appearance of pale blue precipitates of basic copper hydroxide & a solution of neutral salt sodium sulfate.

5 0

3 years ago

Balance the reaction for the combustion of pentane: ?C5H12+?O2→?CO2+?H2O Enter the four coefficients in order, separated by comm

Daniel [21]

Answer:

The four coefficients in order, separated by commas are 1, 8, 5, 6

Explanation:

We count the atoms in order to balance this combustion reaction. In combustion reactions, the products are always water and carbon dioxide.

C₅H₁₂ + ?O₂→ ?CO₂ + ?H₂O

We have 12 hydrogen in right side and we can balance with 6 in the left side. But the number of oxygen is odd. We add 2 in the right side, so we have 24 H, and in the product side we add a 12.

As we add 2 in the C₅H₁₂, we have 10 C, so we must add 10 to the CO₂ in the product side.

Let's count the oxygens: 20 from the CO₂ + 12 from the water = 32.

We add 16 in the reactant side. Balanced equation is:

2C₅H₁₂ + 16O₂→ 10CO₂ + 12H₂O

We also can divide by /2 in order to have the lowest stoichiometry

C₅H₁₂ + 8O₂→ 5CO₂ + 6H₂O

6 0

3 years ago

combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

masya89 [10]

Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

6 0

3 years ago

Balance the following skeleton reaction in acidic solution (if the coefficient is 1, put 1; if the coefficient is 0, put 0) and

Mashcka [7]

Answer:

2 CrO42- + 3N2O + 10 H+ -----> 2Cr3+ + 6NO + 5H2O

Explanation:

2 CrO42- + 3N2O + 10 H+ -----> 2Cr3+ + 6NO + 5H2O

Oxidizing agent: -----------------------------> CrO42-

Reducing agent: ----------------------------> N2O

explanation:

in CrO4-2 oxdiation state of Cr = +6

in Cr+3 oxidation state = +3

+6 oxidation state changed from +3 it is reduction .

so CrO4-2 is oxidizing agent

atomatically

N2O should be reducing agent

7 0

2 years ago

A 4.00g sample of helium has a volume of 24.4L at a temperature of 25.0oC and a pressure of 1.00 atm. The volume of the helium i

Masteriza [31]

Answer:

0.41 moles.

Explanation:

Given that:

Mass of helium = 4.00 g

Initial Volume = 24.4 L

initial Temperature = 25.0 °C =( 25 + 273) = 298 K

initial Pressure = 1.00 atm

The volume was reduced to :

i.e

final volume of the helium - 10.4 L

Change in ΔV = 24.4 - 10.4 = 10.0 L

Temperature and pressure remains constant.

The new quantity of gas can be calculated by using the ideal gas equation.

PV = nRT

n = $\frac{PV}{RT}$

n = $\frac{1.00*10.0}{0.082057*298}$

n = 0.4089 moles

n = 0.41 moles.

7 0

2 years ago