10. Write the formula for the compounds: copper (1) chloride and copper (II) chloride

Chemistry

1 answer:

Natali [406]2 years ago

3 0

Answer:

Copper(II) chloride (CuCl2) reacts with several metals to produce copper metal or copper(I) chloride (CuCl) with oxidation of the other metal.

Explanation:

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An automobile gasoline tank holds 25.0 gal when full. how many pounds of gasoline will it hold if the gasoline has a density of

lora16 [44]

The amount (in pounds) the gasoline tank will hold when full is 153.76 lb

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

<h3>How to convert gallon to milliliters </h3>

1 gallon = 3785.412 mL

Therefore,

25 gallon = 25 × 3785.412

25 gallon = 94635.3 mL

<h3>How to determine the mass </h3>

Density = 0.737 g/mL
Volume = 94635.3 mL
Mass =?

Mass = Density × Volume

Mass = 0.737 × 94635.3

Mass = 69746.2161 g

<h3>How to convert grams to pounds </h3>

453.592 g = 1 lb

Therefore,

69746.2161 g = 69746.2161 / 453.592

69746.2161 g = 153.76 lb

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1 year ago

How many mols are present in the a sample of silver nitrate which has<br> 5.3x10^24 molecules.

Westkost [7]

Answer: 5.3 x 10^24 formula units of silver nitrate is equivalent to 8.8 moles of silver nitrate. Silver nitrate is an ionic compound, therefore, its representative particle is called a "formula unit" instead of molecule. For every mole of a substance, we know that there are 6.022 x 10^23 representative units of that substance. The amount of particles in one mole of substance is called Avogadro's number.

Further Explanation:

We can convert from number of representative particles to moles using the formula:

$\boxed {no. \ of \ moles \ = \ ( given \ no. \ of \ particles) \ (\frac{1 \ mole}{\ 6.022 \ x 10^{23} particles})}$

For this problem, we can calculate the number of moles by plugging in the given values to the equation above,

$no. \ of \ moles \ = (5.3 \ x \ 10^{24} \ formula \ units \ AgNO_{3}) \ (\frac{1 \ mole \ AgNO_{3}}{6.022 \ x 10^{23} \ formula \ units AgNO_{3}}) \\\\\boxed {no. \ of moles \ AgNO_{3} \ = \ 8.8 \ moles}$