Ask question

Ask question

All categories

3 years ago

9

5. Which choice best represents the relationship between wavelength and frequency?

1 answer:

Arte-miy333 [17]3 years ago

4 0

When frequency increases, wavelength increases

You might be interested in

How to remember all your answers for a test?

Troyanec [42]

rewrite them and say them out loud or in your head repeatedly

8 0

4 years ago

Which best describes a benefit of using laser instead of traditional cutting machines?

-BARSIC- [3]

Laser cutting provides a more precise , clean cut.

7 0

3 years ago

Are production of seeds asexual or sexual

nydimaria [60]

Asexual

I hope this helps

please mark this as brainiest!!

6 0

4 years ago

Which change will increase the yield of hydrazine?

irinina [24]

The reaction of hydrogen and nitrogen produces hydrazine. The chemical reaction is expressed as 2H2 + N2 = N2H4. To increase the production of hydrazine, one could increase the amount of the reactants. The reaction is endothermic which means it absorbs heat. Therefore, adding heat would also increase the hydrazine produced.

3 0

4 years ago

Consider a solution formed by mixing 50.0 mL of 0.100 M H2SO4, 30.0 mL of 0.1133 M HOCl, 25.0 mL of 0.200 M NaOH, 25.0 mL of 0.1

NARA [144]

Answer:

The pH of this solution is 12.1.

Explanation:

$H_2SO_4$

$H_2SO_4\rightarrow 2H^++SO_4^{2-}$

1 mole of sulfuric acid gives 2 moles of hydrogen ions, then 0.100 M of sulfuric acid will give :

$[H^+]=2\times [H_2SO_4]=2\times 0.100 M= 0.200 M$

Volume of sulfuric acid solution = 50.0 mL = 0.050 L ( 1 mL = 0.001 L)

Moles of hydrogen ions in sulfuric acid solution :a

$a=0.200 M\times 0.050 L=0.01 mol$

HOCl

The dissociation constant value of HOCl = $K_a=3.5\times 10^{-8}$

$HOCl\rightleftharpoons H^++OCl^-$

initially

0.1133 M 0 0

At equilibrium :

(0.1133-x) x x

The expression of dissociation constant is given as:

$K_a=\frac{[H^+][OCl^-]}{[HOCl]}$

$3.5\times 10^{-8}=\frac{x\times x}{(0.1133-x)}$

Solving for x:

x = $6.295\times 10^{-5} M$

$[H^+]=6.295\times 10^{-5} M$

Volume of HOCl solution = 30.0 mL = 0.030 L ( 1 mL = 0.001L)

Moles of hydrogen ions in HOCl solution = b

$b=6.295\times 10^{-5} M\times 0.030 L=1.8885\times 10^{-6} mol$

Total moles of hydrogen ions = a + b

= $0.01 mol +1.8885\times 10^{-6} mol=0.010002 mol$

$NaOH$

$NaOH\rightarrow Na^++OH^{-}$

1 mole of NaOH gives 1 mole of hydroxide ions, then 0.200 M of NaOH acid will give :

$[OH^-]=1\times [NaOH]=1\times 0.200 M= 0.200 M$

Volume of NaOH solution = 25.0 mL = 0.025 L ( 1 mL = 0.001 L)

Moles of hydroxide ions in NaOH solution : c

$c=0.200 M\times 0.025 L=0.005 mol$

$Ba(OH)_2$

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

1 mole of $Ba(OH)_2$ gives 2 mole of hydroxide ions, then 0.100 M of

$[OH^-]=2\times [Ba(OH)_2]=1\times 0.200 M= 0.200 M$

Volume of $Ba(OH)_2$ solution = 25.0 mL = 0.025 L ( 1 mL = 0.001 L)

Moles of hydroxide ions in $Ba(OH)_2$ solution : d

$d=0.200 M\times 0.025 L=0.005 mol$

$KOH$

$KOH\rightarrow K^++OH^{-}$

1 mole of KOH gives 1 mole of hydroxide ions, then 0.170 M of KOH will give :

$[OH^-]=1\times [KOH]=1\times 0.170 M= 0.170 M$

Volume of KOH solution = 10.0 mL = 0.010 L ( 1 mL = 0.001 L)

Moles of hydroxide ions in KOH solution : e

$e=0.170 M\times 0.010 L=0.0017 mol$

Total moles of hydroxide ions : c + d + e

$0.005 mol + 0.005 mol + 0.0017 mol = 0.0117 mol$

Total moles of hydrogen ions = 0.010002 mol

Total moles of hydroxide ions = 0.0117 mol

1 mole of hydrogen ion reacts with 1 mole of hydroxide ion to from a water with neutral pH.

Hydrogen ions < hydroxide ion

So, 0.0117 moles of hydroxide ion will neutralize 0.0117 moles of hydrogen ions.

Moles of hydroxide left after neutralization = 0.0117 mol - 0.010002 mol

= 0.001698 moles

Concentration of hydroxide ions left in the solution :

$[OH^-]=\frac{0.001698 mol}{0.050 L+0.030 L+0.025 L+0.025 L+0.010 L}=0.01213 M$

$pOH=-\log[OH^-]=-\log[0.01213 M]=1.9$

pH = 14 - pOH

pH = 14 - 1.9= 12.1

The pH of this solution is 12.1.

7 0

4 years ago