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Gnesinka [82]
3 years ago
13

Silver chloride, often used in silver plating, contains 75.27%Ag. Calculate the mass of silver chloride required to plate 195 mg

mg of pure silver.
Chemistry
1 answer:
kenny6666 [7]3 years ago
6 0

Answer:

259 mg is the mass of silver chloride required to plate 195 mg of pure silver.

Explanation:

Silver chloride contains 75.27 percent silver. It can be concluded that:-

For every 100 mg of silver chloride, 75.27 mg of silver we can get.

So, to produce 195 mg of silver, the amount of silver chloride required can be calculated as:-

Mass=\frac{100\times 195}{75.27}\ mg=259\ mg

<u>259 mg is the mass of silver chloride required to plate 195 mg of pure silver.</u>

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When water is solid, do molecules move?
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What results when two waves that are completely out of phase meet?
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An atom that has 13  protons and 15  neutrons  is isotope  of Aluminium (answer C)

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What is the engine piston displacement in liters of an engine whose displacement is listed as 490 in^3?
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Answer:

490 in^3 = 8.03 L

Explanation:

Given:

The engine displacement = 490 in^3

= 490 in³

To determine the engine piston displacement in liters L;

(NOTE: Both in^3 (in³) and L are units of volume). Hence, to find the engine piston displacement in liters (L), we will convert in^3 to liters (L)

First, we will convert in³ to cm³

Since 1 in = 2.54 cm

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If 1 in³ = 16.387 cm³

Then 490 in³ =  (490 in³ × 16.387 cm³) / 1 in³ = 8029.63 cm³

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Now will convert cm³ to dm³

(NOTE: 1 L = 1 dm³)

1 cm = 1 × 10⁻² m = 1 × 10⁻¹ dm

∴ 1 cm³ = 1 × 10⁻⁶ m³ = 1 × 10⁻³ dm³

If 1 cm³ = 1 × 10⁻³ dm³

Then, 8029.63 cm³ = (8029.63 cm³ × 1 × 10⁻³ dm³) / 1 cm³ = 8.02963 dm³

≅ 8.03 dm³

∴ 8029.63 cm³ = 8.03 dm³

Hence, 490 in³ = 8029.63 cm³ = 8.03 dm³

Since 1L = 1 dm³

∴ 8.03 dm³ = 8.03 L

Hence, 490 in³ = 8.03 L

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