CHEMISTRY HIGH SCHOOL

Name 4 toxicants that are heavier than air

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Answer 1
Answer:

Answer:

CO2,H2S,HALON,FREON that's is the four toxicants that are heavier than air

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Assume that 13.5 g solid aluminum (Al) react with HCl to produce solid aluminum chloride (AlCl3) salt and gaseous hydrogen (H2) at standard temperature and pressure.How many moles of Al react? How many moles of H2 are produced? How many liters of H2 are produced?

C2H5OH is able to dissociate in water to conduct an electric current. This compound could be classified as

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The fact that when in solution C₂H₅OH conducts electricity means that it is an electrolyte so you can automatically cross out A and C (the definition of an electrolyte is a substance that will conduct electricity when in a solution).  I think the correct answer is B due to the -OH group (the - indicates a covalent bond) which allows the molecule to give up a proton (H⁺).
 I hope this helps.  Let me know if anything is unclear.

The following three solutions are mixed: 100.0mL of 0.100M Na₂SO₄, 50.0mL of 0.300M ZnCl₂, and 100.0mL of 0.200M Ba(CN)₂. Cyanide (CN-) solubilities are not in most tables, but Ba(CN)₂ is soluble, Zn(CN)₂ is not. a. What ionic compounds will precipitate out of solution? b. What is the molarity of each ion remaining in the solution, assuming complete precipitation of all insoluble compounds, and assuming that volumes are additive?

Answers

a. To determine the ionic compounds that will precipitate out of solution, we need to consider the solubility rules. According to the solubility rules:

1. All sodium (Na+) salts are soluble, so Na₂SO₄ will remain in solution.
2. Zinc (Zn2+) salts are generally soluble, except for zinc sulfide (ZnS) and zinc hydroxide (Zn(OH)₂). However, in this case, we are adding ZnCl₂ to the solution, which contains chloride (Cl-) ions. Chloride ions form soluble salts with most cations, including Zn2+. Therefore, ZnCl₂ will remain in solution.
3. Barium (Ba2+) salts are generally soluble, except for barium sulfate (BaSO₄) and barium carbonate (BaCO₃). However, in this case, we are adding Ba(CN)₂ to the solution, which contains cyanide (CN-) ions. Cyanide ions form insoluble salts with most cations, including Ba2+. Therefore, Ba(CN)₂ will precipitate out of solution as Ba(CN)₂ is not soluble.

b. Assuming complete precipitation of all insoluble compounds and that volumes are additive, we can calculate the molarity of each ion remaining in the solution.

For Na₂SO₄:
- Sodium (Na+) ion concentration: 2 * 0.100 M = 0.200 M
- Sulfate (SO₄2-) ion concentration: 0.100 M

For ZnCl₂:
- Zinc (Zn2+) ion concentration: 0.300 M
- Chloride (Cl-) ion concentration: 2 * 0.300 M = 0.600 M

For Ba(CN)₂:
- Barium (Ba2+) ion concentration: 0.200 M
- Cyanide (CN-) ion concentration: 2 * 0.200 M = 0.400 M

Therefore, the molarity of each ion remaining in the solution, assuming complete precipitation of all insoluble compounds and that volumes are additive, are as follows:
- Sodium (Na+) ion: 0.200 M
- Sulfate (SO₄2-) ion: 0.100 M
- Zinc (Zn2+) ion: 0.300 M
- Chloride (Cl-) ion: 0.600 M
- Barium (Ba2+) ion: 0.200 M
- Cyanide (CN-) ion: 0.400 M

6.7 g of LiNO3 from a 20 %(m/m) LiNO3 solution.

Answers

20%(m/m)

20 g  LiNO₃------------- 100 g solution
6.7 g LiNO₃ ------------ ?

m = 6.7 * 100 / 20

m = 670 / 20

m = 33.5 g 

hope this helps!

Background Info:The standard enthalpy of formation (ΔH∘f) is the enthalpy change that occurs when exactly1 mol of a compound is formed from its constituent elements under standard conditions. The standard conditions are 1 atm pressure, a temperature of 25 ∘C , and all the species present at a concentration of 1 M . A "standard enthalpies of formation table" containing ΔH∘f values might look something like this:SubstanceΔH∘fH(g)218 kJ/molH2(g)0 kJ/molBa(s)0 kJ/molBa2+(aq)−538.4 kJ/molC(g)71 kJ/molC(s)0 kJ/molN(g)473 kJ/molO2(g)0 kJ/molO(g)249 kJ/molS2(g)129 kJ/mol

Question:
What is the balanced chemical equation for the reaction used to calculate ΔH∘f of BaCO3(s)?If fractional coefficients are required, enter them as a fraction (i.e. 1/3). Indicate the physical states using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for aqueous solution.
EXPRESS ANSWER AS A CHEMICAL EQUATION. Please explain for me too!!!

Answers

You will get at first carbon dioxide by burning C:
C+O_2=CO_2
Burn hydrogen to obtain water:
H_2+(1)/(2)O_2=H_2O
Combine them:
CO_2+H_2O=H_2CO_3
Now react it with Ba:
Ba+H_2CO_3=BaCO_3+H_2
To sum up, the reaction is Ba+C+(3)/(2)O_2=BaCO_3, using hydrogen as a catalyst.

Answer:

You will get at first carbon dioxide by burning C:

C+O_2=CO_2

Burn hydrogen to obtain water:

H_2+\frac{1}{2}O_2=H_2O

Combine them:

CO_2+H_2O=H_2CO_3

Now react it with Ba:

Ba+H_2CO_3=BaCO_3+H_2

To sum up, the reaction is Ba+C+\frac{3}{2}O_2=BaCO_3, using hydrogen as a catalyst.

Explanation:

In a very violent reaction called a thermite reaction, aluminum metal reacts with iron (III) oxide to form iron metal and aluminum oxide according to the following equation:

Answers

Here's the equation:
Fe2 O3 + 2Al → 2Fe + Al2 O3 

Here's the question.
What mass of Al will react with 150g of Fe2 O3? 

In every 2 moles Al you need 1 mole Fe2O3 

moles = mass / molar mass 
moles Fe2O3 = 150 g / 159.69 g/mol 
= 0.9393 moles 

moles Al needed = 2 x moles Fe2O3 
= 2 x 0.9393 mol 
= 1.879 moles Al needed 

mass = molar mass x moles 
mass Al = 26.98 g/mol x 1.879 mol 
= 50.69 g 
= 51 g (2 sig figs) 

So the mass of Al that will react with 150g of Fe2 O3 is 51 grams.

Look up the density of n-butyl chloride (1-chlorobutane). Assume that this alkyl halide was prepared instead of the bromide. Decide whether the alkyl chloride would appear as the upper or the lower phase at each stage of the separation procedure: after the reflux, after the addition of water, and after the addition of sodium bicarbonate.

Answers

Answer:

See explanation below

Explanation:

First, you need to know the density of each compound in order to know this.

The density of 1-chlorobutane is 0.88 g/mL,

The density of water is 1 g/mL

The density of sodium bicarbonate is 2.2 g/cm3.

therefore, the one that has a greater density will always go at the lower phase.

In this case, after the reflux, it will stay in the lower phase, basically because you don't have another solvent with a greater density than the butane.

After adding water, it will be in the upper phase, water has a greater density.

After adding bicarbonate, it will be in the upper phase too.
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