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3 years ago

Which product of the ultraviolet decomposition of cfcs acts as the catalyst for ozone decomposition?

1 answer:

8 0

The answer is Chlorine atoms. This is the product of the ultraviolet decomposition of cfcs acts as the catalyst for ozone decomposition.

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Peter and Jessica are both forensic scientists. Peter visits crime scenes to pr

ratelena [41]

Answer: Peter is a Field Officer and Jessica is a Lab Officer

Explanation:

For Plato it should be, Peter is a "Field officer" because he's going into the fiend of the crime scene to find his evidence

And Jessica is a "Lab officer" because she stay's in the lab to analyze the evidence brought to her by Peter

8 0

3 years ago

On the basis of which property of the elements are they divided into metals and non metals?

kicyunya [14]

The answer is the elements in a periodic table. On the basis of the elements in the periodic table, they are divided into metals and non metals.

8 0

3 years ago

Meet /svc-jgyu-fii join

arsen [322]

Answer:

what are we doing there???

bye, have a nice day!

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3 years ago

20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc

bezimeni [28]

Answer:

Concentration of the barium ions = $[Ba^{2+}] = 0.4654 M$

Concentration of the chloride ions = $[Cl^{-}]=0.9308 M$

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride = 0.1355 M

$n=0.1355 M\times 0.04389 L=0.005947 mol$

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

$\frac{1}{2}\times 0.05947 mol=0.029735 mol$ barium hydroxide

Moles of barium hydroxide = 0.029735 mol

$Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)$

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

$=1\times 0.029735 mol= 0.029735 mol$

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L

Concentration of the barium ions = $[Ba^{2+}]$

$[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M$

$Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)$

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

$[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M$

8 0

3 years ago

A chemist determined by measurements that 0.030 moles of barium participated in a chemical reaction. Calculate the mass of bariu

schepotkina [342]

Answer: 4.1 g of barium precipitated.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}$

Given : moles of barium = 0.030

Molar mass of barium = 137 g/mol

$0.030=\frac{x}{137}$

x= 4.1 g

Thus there are 4.1 g of barium that precipitated.

3 0

3 years ago