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Lelechka [254]
2 years ago
11

Reduction occurs at the cathode in

Chemistry
2 answers:
mina [271]2 years ago
8 0

Explanation:

option 2) voltaic cells only

<em><u>I </u></em><em><u>guess </u></em><em><u>so </u></em><em><u>this </u></em><em><u>is</u></em><em><u> </u></em><em><u>the </u></em><em><u>answer</u></em><em><u> </u></em><em><u>or </u></em><em><u>not </u></em>

Marta_Voda [28]2 years ago
5 0

Answer: 3

Explanation:

3) both electrolytic cells and voltaic cells

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6) (a) Calculate the absorbance of the solution if its concentration is 0.0278 M and its molar extinction coefficient is 35.9 L/
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Answer:

6) (a) 0.499; (b) 31.7 %

7) 0.15

Explanation:

6) (a) Absorbance

Beer's Law is

A = \epsilon cl\\A = \text{35.9 L&\cdot$mol$^{-1}$cm$^{-1}$} $\times$ 0.0278 mol$\cdot$L$^{-1} \times $ 0.5 cm = \mathbf{0.499}

(b) Percent transmission

A = \log {\left (\dfrac{1}{T}}\right)}\\\\\%T = 100T\\\\T = \dfrac{\%T}{100}\\\\\dfrac{1}{T} = \dfrac{100 }{\%T}\\\\A = \log \left(\dfrac{100 }{\%T} \right ) = 2 - \log \%T\\\\0.499 = 2 - \log \%T\\\\\log \%T = 2 - 0.499 = 1.501\\\\\%T = 10^{1.501} = \mathbf{31.7}

7) Absorbance

A = \log \left (\dfrac{I_{0}}{I} \right ) = \log \left (\dfrac{I_{0}}{0.70I_{0}} \right ) = \log \left (\dfrac{1}{0.70} \right ) = -\log(0.70) = \mathbf{0.15}}

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3 years ago
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2 years ago
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What are the chemical symbols for the two elements found in iron oxide
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For Oxygen: O

These are the two chemical symbols for the two elements found in Iron Oxide.
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The dissolution of 0.200 l of sulfur dioxide at 19 °c and 745 mmhg in water yields 500.0 ml of aqueous sulfurous acid. The solut
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Answer:

Molarity=1.22\ M

Explanation:

Given:  

Pressure = 745 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 745 / 760 = 0.9803 atm

Temperature = 19 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (19 + 273.15) K = 292.15 K  

Volume = 0.200 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9803 atm × 0.200 L = n × 0.0821 L.atm/K.mol × 292.15 K  

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From the reaction shown below:-

H_2SO_3+2NaOH\rightarrow Na_2SO_3+2H_2O

1 mole of H_2SO_4 react with 2 moles of NaOH

0.008174 mole of H_2SO_4 react with 2*0.008174 moles of NaOH

Moles of NaOH = 0.016348 moles

Volume = 13.4 mL = 0.0134 L ( 1 mL = 0.001 L)

So,

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.016348}{0.0134}\ M

Molarity=1.22\ M

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