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Butoxors [25]
3 years ago
5

Jodi reacts 6.087 grams of cobalt (mw=58.933) with a solution containing excess gold (iii) chloride and recovers 13.572 g of sol

id gold (mw=196.967). Which reaction, 1 or 2, most likely occurred in their experiment?
Chemistry
1 answer:
Makovka662 [10]3 years ago
8 0

The correct answer is reaction 2 most likely occurred in their experiment.

So the given two reactions are as follows:

1) Au³⁺(aq) + Co(s) --> Au(s) + Co³⁺(aq)

2) 2Au³⁺(aq) + 3Co(s) --> 2Au(s) + 3Co³⁺(aq)

Mass of Co = 6.087 g

Molar mass of Co = 58.933 g/mol

Moles of Co =\frac{6.087 g}{58.933 g/mol} = 0.1032

Mass of Au = 13.572 g

Molar mass of Au = 196.967 g/mol

Moles of Au = \frac{13.572 g}{196.967 g/mol} = 0.069

Now based on reaction 1) the molar ratio between Co and Au is 1:1

So as per this reaction 0.1032 moles of Co would produce 0.1032 mole of Au

Again based on reaction 2) the molar ratio between Co and Au is 3:2

So as per this reaction 0.1032 moles of Co would produce \frac{2 x 0.1032}{3} mole of Au

or, As per the second reaction 0.1032 moles of Co would produce 0.0688 moles of Au

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A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt
Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt

integrating factor

=e^{\int\limits (1/40) \, dt}

integrating factor =e^{(1/40)t}

multiply on both sides by  =e^{(1/40)t}

dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t

integrate on both sides

\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
6 0
3 years ago
Which of the following is true when 1 M KCl is diluted to semi-molar KCl?
eimsori [14]

The solution before dilution and after dilution contains same number of moles, and water is added for dilution.

Option B

<h3><u>Explanation:</u></h3>

Suppose before dilution, the solution contains x moles of KCl in Y liter of water. Now as the concentration got halved, then the solution contains x moles of KCl in 2Y kiters of solution. So the number of moles of KCl in the solution remained constant.

Again, as the solution is diluted to half of the concentration, water must have been added with the solution to make it dilute.

4 0
3 years ago
A) release of toxic gas
faust18 [17]
In my opinion the answer is D
5 0
3 years ago
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How many electrons will a neutral atom of carbon have if it’s nucleus has 6 protons and 8 neutrons?
Alexeev081 [22]

Answer:

6

Explanation:

number of protons equal number of electrons for the atom to be stable

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3 years ago
Balancing chemical equations
mojhsa [17]

<span>2H2 + O2 → 2H2O</span>

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<span>okay???</span>

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