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3 years ago

What is the mass in grams of 6.69 mol W?

Chemistry

1 answer:

PilotLPTM [1.2K]3 years ago

7 0

Go to the website call convertunits.com you should get your answer there

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Sylvanite is a mineral that contains 28.0% gold by mass. How much sylvanite would you need to dig up to obtain 73.0 g of gold?

Scorpion4ik [409]

You would have to dig up 261 g of sylvanite.

Mass of sylvanite = 73.0 g Au × (100 g sylvanite/28.0 g Au) = <em>261 g</em> sylvanite.

3 0

3 years ago

An experimental spacecraft consumes a special fuel at a rate of 372 L/min. The density of the fuel is 0.730 g/mL and the standar

Karolina [17]

Explanation:

First, we will calculate fuel consumption is as follows.

$372 L/min \times 1000 ml/L \times 0.730 g/ml \times \frac{1}{60} min/s$

= 4526 g/s

Now, we will calculate the power as follows.

Power = Fuel consumption rate × -enthalpy of combustion

= $4526 g/s \times -26.5 kJ/g$

= $1.19 \times 10^{5}$ kW

Thus, we can conclude that maximum power (in units of kilowatts) that can be produced by this spacecraft is $1.19 \times 10^{5}$ kW.

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3 years ago

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sergejj [24]

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8 0

2 years ago

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REY [17]

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6 0

2 years ago

For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [

sveta [45]

Explanation:

The given data is as follows.

$E^{o}$ = 0.483, $[Co^{3+}]$ = 0.173 M,

$[Co^{2+}]$ = 0.433 M, $[Cl^{-}]$ = 0.306 M,

$P_{Cl_{2}}$ = 9.0 atm

According to the ideal gas equation, PV = nRT

or, P = $\frac{n}{V}RT$

Also, we know that

Density = $\frac{mass}{volume}$

So, P = MRT

and, M = $\frac{P}{RT}$

= $\frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}$

= $\frac{9.0}{24.436}$

= 0.368 mol/L

Now, we will calculate the cell potential as follows.

E = $E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}$

= $0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}$

= $0.483 - 0.02955 log \frac{0.0689}{0.0162}$

= $0.483 - 0.02955 \times 0.628$

= 0.483 - 0.0185

= 0.4645 V

Thus, we can conclude that the cell potential of given cell at $25^{o}C$ is 0.4645 V.

4 0

3 years ago