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Elenna [48]
3 years ago
9

What volume of 0.480 m koh is needed to react completely with 15.8 ml of 0.150 m h2so4? assume the reaction goes to completion?

Chemistry
1 answer:
LiRa [457]3 years ago
3 0
I think this answer i write a answer and take takae a photo send you

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NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.084 M in
kolbaska11 [484]
Answer is: pH of solution is 5,17.
Kb(NH₃) = 1,8·10⁻⁵.
c(NH₄Cl) = 0,084 M = 0,084 mol/L.
Chemical reaction: NH₄⁺ + H₂O → NH₃ + H₃O⁺.
Ka · Kb = 10⁻¹⁴.
Ka(NH₄⁺) = 10⁻¹⁴ ÷ 1,8·10⁻⁵.
Ka(NH₄⁺) = 5,55·10⁻¹⁰.
[H₃O⁺] = [NH₃]  = x.
Ka(NH₄⁺) = [H₃O⁺] · [NH₃] ÷ [NH₄⁺].
5,55·10⁻¹⁰ = x² ÷ (0,084 M - x).
Solve quadratic equation: x = [H₃O⁺] = 6,8·10⁻⁶ M.
pH = -log[H₃O⁺].
pH = -log(6,8·10⁻⁶ M) = 5,17.
7 0
3 years ago
What is a decomposition reaction?
77julia77 [94]

Answer:

D. Two reactants combine to form a single ew product.

Explanation:

Because, A decomposition reaction produces multiple products from a single reactant.

4 0
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What is the molar mass of Mg(NO3)2
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Answer: The molar mass of magnesium nitrate is 148.3148 g/mol.

Explanation:

4 0
3 years ago
Calcular el tanto por ciento en masa de una disolución que tiene 25 ml de un soluto con una densidad de 0.8 g / cm3, y 500 ml de
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Answer:

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8 0
3 years ago
A teapot with a surface area of 785 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic
lina2011 [118]

Answer:

Given info: The surface area of teapot plated with silver is 700 cm2700 cm2, the cell is powered by 12.0-V12.0-V, the resistance of the cell is 1.80 Ω1.80 Ω, thickness of silver layer is 0.133-mm0.133-mm and density of silver is 10.5×103 kg/m310.5×103 kg/m3.

Write the expression for the mass of the silver.

m=ρAdm=ρAd (1)

Here,

ρρ is the density of silver.

AA is the surface area.

dd is the thickness of the silver layer.

Substitute 10.5×103 kg/m310.5×103 kg/m3 for ρρ, 700 cm2700 cm2 for AA and 0.133-mm0.133-mm for dd in equation (1) to find mass of the silver.

m=(10.5×103 kg/m3)(700 cm2(10−21 cm)2)(0.133-mm(10−3 m1 mm))=0.0978 kg(103 g1 kg)=97.8 gm=(10.5×103 kg/m3)(700 cm2(10−21 cm)2)(0.133-mm(10−3 m1 mm))=0.0978 kg(103 g1 kg)=97.8 g

Thus, the mass of silver is 97.8 g97.8 g.

Write the expression for number of moles of silver.

n=mWan=mWa (2)

Here,

mm is the mass of silver.

WaWa is the atomic weight.

The atomic weight of silver is 107.8 g/mole107.8 g/mole

Substitute 97.8 g97.8 g for mm, and 107.8 g/mole107.8 g/mole for WaWa in equation (2) to find number of moles of silver.

n=97.8 g107.8 g/mole=0.907 moln=97.8 g107.8 g/mole=0.907 mol

Thus, the number of moles of silver is 0.907 mol0.907 mol.

7 0
4 years ago
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