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OleMash [197]
2 years ago
5

What is the balanced equation if electrolysis?​

Chemistry
2 answers:
Travka [436]2 years ago
7 0

Reaction:

2 \:  H2O(l)  \: → 2 H2(g) + O2(g)

  • <u>Hydrogen molecules produced is thus twice the number of oxygen molecules. </u>

  • <u>Assuming equal temperature and pressure for both gases, the produced hydrogen gas has, therefore, twice the volume of the produced oxygen gas</u>
ra1l [238]2 years ago
5 0

Answer:

It's Al 3+ + 3e - → Al

Explanation:

Because three negatively charged electrons are needed to balance the three positive charges on the aluminum ion

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I need help with this question ​
VARVARA [1.3K]

Answer:

2.25×10¯³ mm.

Explanation:

From the question given above, we obtained the following information:

Diameter in micrometer = 2.25 μm

Diameter in millimetre (mm) =?

Next we shall convert 2.25 μm to metre (m). This can be obtained as follow:

1 μm = 1×10¯⁶ m

Therefore,

2.25 μm = 2.25 μm / 1 μm × 1×10¯⁶ m

2.25 μm = 2.25×10¯⁶ m

Finally, we shall convert 2.25×10¯⁶ m to millimetre (mm) as follow:

1 m = 1000 mm

Therefore,

2.25×10¯⁶ m = 2.25×10¯⁶ m /1 m × 1000 mm

2.25×10¯⁶ m = 2.25×10¯³ mm

Therefore, 2.25 μm is equivalent to 2.25×10¯³ mm.

6 0
3 years ago
A city continuously disposes of effluent from a wastewater treatment plant into a river. The minimum flow in the river is 130 m3
Vera_Pavlovna [14]

Answer:

2.54\ \text{mg/L}

Explanation:

C = Allowable concentration = 1.1 mg/L

Q_1 = Flow rate of river = 130\ \text{m}^/\text{s}

Q_2 = Discharge from plant = 37\ \text{m}^3/\text{s}

C_1 = Background concentration = 0.69 mg/L

C_2 = Maximum concentration that of the pollutant

The concentration of the mixture will be

C=\dfrac{Q_1C_1+Q_2C_2}{Q_1+Q_2}\\\Rightarrow C_2=\dfrac{C(Q_1+Q_2)-Q_1C_1}{Q_2}\\\Rightarrow C_2=\dfrac{1.1(130+37)-130\times 0.69}{37}\\\Rightarrow C_2=2.54\ \text{mg/L}

The maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant is 2.54\ \text{mg/L}.

6 0
2 years ago
Calculate the moles and grams of solute in each solution. D. 2.0 L of 0.30M Na2SO4. I already have A, B, and C. Thanks!
e-lub [12.9K]
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol


Find the molar mass
2Na = 23 * 2 = 46 grams
1S   = 32 * 1 = 32 grams
O4   = 16 * 4 = 64 grams
Total =            142 grams / mol

Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???

given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams. 

85.2 are in a 2 L solution that has a concentration of 0.6 mol/L


4 0
3 years ago
Read 2 more answers
15. List the substances A-E in order from most dense to least dense based on the facts provided.
Volgvan

Answer:

So 1st it is B then D then E then a then C

3 0
3 years ago
Given that an E2 reaction proceeds with anti periplanar stereochemistry, draw the products of each elimination. The alkyl halide
serg [7]

Answer:

See explanation below

Explanation:

The first two pictures show the reagents used in these reactions a) and b). As it was stated, An E2 reaction proceeds with an antiperiplanar stereochemistry, so in the case of reaction a) it fill form a product with the groups in opposite directions. In other words, a Trans product.

In the case of reaction b) we have the same reaction, with the difference that we have changed the CH3 and phenyl group of positions. This will cause that the reaction will proceed the same but the stereochemistry of the final product will be changed too. In this case, and according to the picture 3 attached, we can see that the product formed is a cis product. So we can conclude that the relation of product a) and b) is that they are isomers, the trans and cis isomers respectively. See picture below for mechanism and products

7 0
3 years ago
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