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3 years ago

Which best describes the energy conversion that happens when hydroelectric

Chemistry

2 answers:

Rama09 [41]3 years ago

8 0

Answer:

Explanation:

Kinetic energy to electrical energy

Sliva [168]3 years ago

4 0

Answer:

"A hydraulic turbine converts the energy of flowing water into mechanical energy. A hydroelectric generator converts this mechanical energy into electricity.

Explanation:

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20)

anyanavicka [17]

Answer:

Explanation:

7 0

3 years ago

Which state of matter is spread apart and not connected

Vilka [71]

Gas I’m pretty sure

5 0

3 years ago

Read 2 more answers

Given the reaction:

dezoksy [38]

Answer:

V = 22.34 L

Explanation:

Given data:

Volume of O₂ needed = ?

Temperature and pressure = standard

Number of molecules of water produced = 6.0× 10²³

Solution:

Chemical equation:

2H₂ + O₂ → 2H₂O

Number of moles of water:

1 mole contain 6.022× 10²³ molecules

6.0× 10²³ molecules × 1 mole / 6.022× 10²³ molecules

0.99 mole

Now we will compare the moles of oxygen and water.

H₂O : O₂

2 : 1

0.996 : 0.996

Volume of oxygen needed:

PV = nRT

V = nRT/P

V = 0.996 mol × 0.0821 atm.L/mol.K × 273.15 K / 1 atm

V = 22.34 L

3 0

3 years ago

Upper n subscript 2 (g) plus 3 upper H subscript 2 (g) double-headed arrow 2 upper N upper H subscript 3 (g). At equilibrium, th

artcher [175]

Answer:

The <u>equilibrium constant</u> is:

$k_c=0.0030M^{-2}$

Explanation:

The correct equation is:

N₂(g) + 3H₂(g) ⇄ 2NH₃(g)

Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.

The equation for the equilibrium constant is:

$k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}$

Substituting:

$k_c=\dfrac{(0.105M)^2}{(1.1M)\cdot (1.50M)^3}$

$k_c=0.0030M^{-2}$

6 0

3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago