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3 years ago

Describe 2 physical changes that could occur with a piece of paper

Chemistry

2 answers:

Crank3 years ago

6 0

Answer:

ima just take all your points and roll out tbh peace

Explanation:

n200080 [17]3 years ago

3 0

Answer:

Physical=Ripping

Chemical=Burning

Explanation:

I may be wrong. But I would say these are the answers.

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When a solute is dissolved in water, the solution's boiling temperature _______ and the freezing temperature _______?

ale4655 [162]

Increases and decreases

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3 years ago

One year is the amount of time it takes Earth to ____________. A. complete one rotation B. complete one revolution C. complete 5

jeka57 [31]

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3 years ago

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Name the particle discovered by James<br> Chadwick in 1932.

LenKa [72]

Answer:

He discovered neutrons in 1932

4 0

3 years ago

How many moles are represented by 118g of cobalt? Cobalt had an atomic mass of 58.93amu

skad [1K]

N=m/M
n=118/58.93=2
Answer: 2 moles

5 0

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