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3 years ago

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Titanium metal requires a photon with a minimum energy of 6.94×10⁻¹⁹J to emit electrons. What is the wavelength of this light? E

xpress your answer using three significant figures.

1 answer:

Elena-2011 [213]3 years ago

5 0

Answer:2.86x10^-7m

Explanation:E=hc/^

E=6.94x10^-19J

c = 2.9979x10^8m/s

h= 6.626x10^-34Js

^ =( 6.626x10^-34)x( 2.9979x 10^8)/ 6.94x10^-19

= 2.86x10^-7m

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A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na

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Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f: after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

$Normality=molarity \times n-factor$

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point: $N_1V_1=N_2V_2$

Before equivalence point : $N_1V_1>N_2V_2$

After the equivalence point: $N_1V_1$

$N_1V_1=100\times1=100$

case a: 5.00 mL of 1.00 M NaOH

$N_2V_2=5\times1=5$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

$N_2V_2=100\times1=100$

$N_1V_1=N_2V_2$ hence it is equivalence point

case c: 10.0 mL of 1.00 M NaOH

$N_2V_2=10\times1=10$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

$N_2V_2=150\times1=150$

$N_1V_1$ hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

$N_2V_2=50\times1=50$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

$N_2V_2=200\times1=200$

$N_1V_1$ hence it is after the eqivalence point

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How would the freezing point of seawater compare with that of pure water?

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Which of the following pairs of elements could react to form an ionic compound? Check all that apply.

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Sodium thiosulfate, Na 2S 2O 3, is used as a "fixer" in black and white photography. Identify the reducing agent in the reaction

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Answer: Reducing agent in the given reaction is $S_{2}O^{2-}_{3}$ .

Explanation:

A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.

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Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.

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