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IgorC [24]
3 years ago
15

Non example of electrical energy

Chemistry
1 answer:
valentinak56 [21]3 years ago
4 0

Answer:

Non-example – creating energy, losing energy. Energy sources - The places we get energy from for heat and electricity. Examples – coal, sun, wind. Non-example – socket, batteries. Energy transformation - The movement of energy where the energy changes forms.

Explanation:

Hope this helps :)

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Define the law of conservation of charge and provide an example.
iren2701 [21]

Answer:

Conservation of Charge is the principle that the total electric charge in an isolated system never changes. The net quantity of electric charge, the amount of positive charge minus the amount of negative charge in the universe, is always conserved.

7 0
3 years ago
Water can be formed in the following reaction:
Komok [63]

The  moles  of  oxygen  gas (O2) that is needed is    4  moles


 Explanation

2H2 +O2 → 2H2O

The  moles of O2  is determined using the mole  ratio  of H2:O2

that is from equation  above H2:O2  is 2:1  

If the moles of H2  is 8  moles therefore  the moles  of O2

 = 8 moles x 1/2  = 4 moles

3 0
2 years ago
During free fall which object would have a greater acceleration, an object with a mass of 240 kg or an object with a mass of 10
inna [77]

Answer:

Leroy Jeakins

Explanation:

Catus jack f me in the asss

5 0
2 years ago
Your mommy buys you a helium balloon at the circus. It has a volume of 2.95 liters. What is the mass
GrogVix [38]

Answer:

if my calculation are correct, it's 295 grams

Explanation:

because liters converted to grams is .1=100 so if you take 2.95 times 100, it equals 295

4 0
2 years ago
The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
3 years ago
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