All categories

3 years ago

How can wind be used as a source of power?

Chemistry

2 answers:

Andre45 [30]3 years ago

7 0

The motion of the wind can turn wind turbines

kicyunya [14]3 years ago

5 0

Answer:

Drilling in the earth can provide new sources of wind

Explanation:

You might be interested in

The reaction 2NO(g) + O2(g) → 2NO2(g) is a synthesis reaction. True False

Trava [24]

true

not sure though

but I think

7 0

3 years ago

Read 2 more answers

To prepare for a laboratory?

yKpoI14uk [10]

Answer:

You have to prepare for the lab (Materials, work, paper etc.)

Set up the lab know where the lab will be taking place

Read thru the experiment before doing the lab

Make a hypothesis

Write down notes, observations, measures anything important to help with the lab!

6 0

3 years ago

In ironmaking, iron metal can be separated from iron ore (Fe2O3) by heating the ore in a blast furnace in the presence of coke,

mel-nik [20]

The limiting reactant is iron ore, the theoretical yield of iron metal is 701.344 kg, and the theoretical yield of carbon dioxide is 413.292 kg.

<h3>Stoichiometric problem</h3>

From the equation of the reaction:

$2 Fe_2O_3(s) + 3 C(s) --- > 4 Fe(s) + 3 CO_2(g)$

The mole ratio of iron ore to carbon is 2:3.

Mole of 1000 kg of iron ore = 1000000/159.69

= 6,262 moles

Mole of 120 kg carbon = 120000/12

= 10,000 moles

Thus, it appears that the carbon is in excess while the iron ore is limited in availability.

The mole ratio of the iron ore and the iron produced is 1:2. Thus, the equivalent number of moles of iron produced will be:

6,262 x 2 = 12,524 moles

Mass of 12,524 moles of iron = 12,524 x 56

= 701,344 g or 701.344 kg

Thus, the theoretical yield of iron is 701.344 kg.

The mole ratio of the iron ore and the carbon dioxide produced is 2:3. The equivalent mole of carbon dioxide produced will be:

6,262 x 3/2 = 9,393 moles

Mass of 9,393 moles carbon dioxide = 9,393 x 44

= 413,292 or 413.292 kg

The theoretical yield of carbon dioxide is, therefore, 413.292 kg.

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

3 0

1 year ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

Answer: The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

Explanation:

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

Reduction half reaction: $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

Net reaction: $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

What is the specific heat of a substance if 690 J of heat are required to raise the temperature of a 100 g

tangare [24]

Answer:

Explanation:

8 0

3 years ago