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Natalka [10]
3 years ago
10

N 2H 4 + H202 N2 + H 20​

Chemistry
1 answer:
Degger [83]3 years ago
7 0

Answer:

N2 + H20

Explanation:

chemical equation balancer

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What must be part of a quantitative observation?
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Quantitative observations include numerical data. Ex: 32 degrees, 10 inches, etc.

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What is the relationship between producers and consumers in a food chain a producers eat consumers b consumers eat producers c p
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Answer:

b consumers eat producers

Explanation:

In a food chain, the consumers are living organisms that eat organisms from a different population. Examples are herbivores like goats.

Producers make their own food in the food chain. Examples are plants.

The relationship between consumer and producers is hence that consumers consume / eat producers. Goats eat grass.

The correct option is;

b consumers eat producers

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What is the orbital notation of oxygen?
AlekseyPX

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Oxygen has 8 electrons. The electron configuration can be written as 1s22s22p4. To draw the orbital diagram, begin with the following observations: the first two electrons will pair up in the 1s orbital; the next two electrons will pair up in the 2s orbital.

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Hard water contains calcium and magnesium, which undergo oxidation and
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3 years ago
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For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell
frutty [35]

Answer:

\large\boxed{\text{0.512 V}}

Explanation:

We must use the Nernst equation

E = E^{\circ} - \dfrac{RT}{zF}lnQ

The equation for the cell reaction is is

2Cl⁻(0.384 mol·L⁻¹) + 2Co³⁺(0.324 mol·L⁻¹) ⇌ Cl₂(5.80 atm) + 2Co²⁺(0.158 mol/L)

Data:

E° = 0.483 V

R = 8.314 J·K⁻¹mol⁻¹

T = 25 °C

n = 2

F = 96 485 C/mol

 

Calculation:  

T = 25 + 273.15 = 298.15 K

Q = \dfrac{\text{[Cl}^{-}]^{2}[\text{Co}^{3+}]^{2}}{p_{\text{Cl}_{2}}^{2}\text{[Co}^{3+}]^{2}} = \dfrac{0.384^{2} \times 0.324^{2}}{5.80 \times 0.158^{2}} =0.1069\\\\E = 0.483 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(0.1069)\\\\=0.483 -0.01285 \times (-2.236) = 0.483 + 0.02872 = \textbf{0.512 V}\\\text{The cell potential is } \large\boxed{\textbf{0.512 V}}

7 0
4 years ago
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