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3 years ago

0.57 centimeter to meter

Chemistry

1 answer:

uysha [10]3 years ago

6 0

If 100cm is 1 metre
Then 0.57 cm =

If less more divides,
Hence 100÷0.57 × 1

= 0.0057 Metres.

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Iron(II) can be oxidized by an acidic K2Cr2O7 solution according to the net ionic equation: Cr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe

cestrela7 [59]

Answer:

The molar concentration of Fe²⁺ in the original solution is 1.33 molar

Explanation:

Moles of K₂Cr₂O₇ = Molarity x Volume (lit)

= 0.025 x 35.5 x 10⁻³

= 0.0008875

Cr₂O₇²⁻ + 6 Fe²⁺ + 14 H⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O

From equation

1 mole K₂Cr₂O₇ used for the oxidation of 6 moles Fe²

0.0008875 mole K₂Cr₂O₇ used for the oxidation of = $\frac{6 X 0.0008875}{1}$ = 0.005325 mole of Fe²

Molarity = $\frac{No. of moles of solute}{Volume of solution(lit)}$

Molar concentration of Fe² = $\frac{0.005325 X 1000}{25}$ = 1.33 molar

So molar concentration of Fe²⁺ in the original solution = 1.33 molar

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3 years ago

A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita

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Answer:

0.21 g

Explanation:

The equation of the reaction is;

NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)

Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles

Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles

Since the reaction is 1:1, NaCl is the limiting reactant.

1 mole of NaCl yields 1 mole of AgCl

0.00147 moles of NaCl yields 0.00147 moles of AgCl

Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol

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2 years ago

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The concentration of Ca2+ ions is half that of the Cl- ions.

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If 27.3% of a sample of silver-112 decays in 1.52 hours, what is the half-life (in hours to 3 decimal places)?

ICE Princess25 [194]

<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.

<u>Explanation:</u>

All radioactive decay processes undergoes first order reaction.

To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

$k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}$

where,

k = rate constant = ?

t = time taken = 1.52 hrs

$[A_o]$ = Initial concentration of reactant = 100 g

[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g

Putting values in above equation, we get:

$k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}$

To calculate the half life period of first order reaction, we use the equation:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life period of first order reaction = ?

k = rate constant = $0.2098hr^{-1}$

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs$

Hence, the half life of the sample of silver-112 is 3.303 hours.

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3 years ago