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Harlamova29_29 [7]
2 years ago
12

In the reaction between 15.8 g of NH3 and 22.7 g oxygen, water and nitrogen monoxide are

Chemistry
1 answer:
LenKa [72]2 years ago
4 0

Explanation:

limiting is NH3

NO:27/88g

H2O: 25/09g

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WILL GIVE BRANLIEST TO WHOEVER ANSWERS THIS QUESTION>>> How many valence electrons does NaHCo3 have?
nydimaria [60]

Answer:

NaHCO3

Na = 1

H = 1

C = 4

Explanation:

5 0
3 years ago
Hydrofluoric acid, hf, has a ka of 6.8 × 10−4. what are [h3o+], [f−], and [oh−] in 0.710 m hf?
STALIN [3.7K]

Answer:

[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.

Explanation:

  • For a weak acid like HF, the dissociation of HF will be:

<em>HF + H₂O ⇄ H₃O⁺ + F⁻.</em>

[H₃O⁺] = [F⁻].

<em>∵ [H₃O⁺] = √Ka.C,</em>

Ka = 6.8 x 10⁻⁴, C = 0.710 M.

∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.

<em>∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.</em>

<em></em>

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺]</em> = 10⁻¹⁴/(2.2 x 10⁻²) = <em>4.55 x 10⁻¹³.</em>

6 0
3 years ago
A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field
maksim [4K]

Answer:

the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  })

Substituting the values into the equation, it becomes

E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  }) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }) = 7.12 × 10⁵(1 - \frac{0.12}{0.1216}) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

7 0
3 years ago
Can someone answer dis:
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Answer:

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Explanation: hope this helps???

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