The diagram below represents reaction of reactant A and reactant B combining to form product C.

33.23 grams of a thin sheet of iron is completely oxidized in 7 days. How would you express the rate of conversion of the silver

tresset_1 [31]

Answer:

A. 0.0440 moles/day

Explanation:

First, let's figure out how many moles 33.23 grams of silver is. We do this by dividing the number of grams by the molar mass of silver, which is 107.87 g/mol:

33.23 g Ag ÷ 107.87 g/mol = 0.3081 mol Ag

Now, let's divide this by 7 to get the rate per day:

0.3081 mol Ag ÷ 7 days = 0.0440 mol/day

Thus, the answer is A.

7 0

3 years ago

26 Which process occurs in an operating voltaic cell?(1) Electrical energy is converted to chemical energy.

tia_tia [17]

Answer is: (2) Chemical energy is converted to electrical energy.

An electrochemical cell (voltaic or galvanic cell) is generating electrical energy from chemical reactions.

In galvanic cell, specie (for example zinc and zinc cations) from one half-cell, lose electrons (oxidation) and species from the other half-cell (for example copper and copper cations) gain electrons (reduction).

Oxidation on the zinc anode: Zn(s) → Zn²⁺(aq) + 2e⁻.

Reduction on the copper cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).

8 0

3 years ago

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Find the mass of one atom of uranium-235. Recall that the mass in atomic mass units is equal to the mass in grams of one mole of

stiv31 [10]

Answer:

$3.90*10^{-25}kg/atom$

Explanation:

The molar mass of uranium-235 is 235 g/mol. So one mole of uranium-235 has a mass of 235 g. Put differently 6.022×10^23 atoms of uranium-235 have a mass of 235 g. Knowing that, how can we use that to find the mass of one atom?

mass of one atom = $\frac{235 g}{1mol} *\frac{1 mol}{6.022*10^{23}atoms } *\frac{1kg}{1000g}= 3.90*10^{-25}kg/atom$

7 0

3 years ago

The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W

Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹

Explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$