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2 years ago

How did the mass change when the copper coin was made to look silver?

Chemistry

1 answer:

il63 [147K]2 years ago

7 0

We have that The the mass change when the copper coin was made to look silver is an increase in mass

Correct option C

It increased.

It is important to note that the copper coin after its cutting into shape will have a specific mass or weight and the silver coating solution will also have a net value of mass or weight

Therefore

The the mass change when the copper coin was made to look silver is an increase in mass

Correct option C

It increased.

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Natalia dipped two feathers in oil. Then she dipped one feather in cold water and the other feather in hot water. She swirled bo

Alika [10]

Answer:

The hot water was better for removing the oil.

Explanation:

You can see that because the mass went down more with the hot water. So, that means that more oil was taken out of the feather with hot water.

8 0

2 years ago

How many moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Cel

o-na [289]

1.137448506 mol moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Celsius.

<h3>What is an ideal gas equation?</h3>

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. The terms are: p = pressure, in pascals (Pa). V = volume, in $m^3$ .

We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol:

PV= nRT

Given data:

P=100.0 kPa =0.986923 atm

T=100 degree celcius= 100 + 273 =373 K

V=35.5 L

Substituting the values in the equation.

n= $\frac{\;0,98 \;atm \;X \;35,5 \;L }{\;0,082\;atm / \;K mol \;X \;373 K}$

n= 1.137448506 mol

Hence, 1.137448506 mol moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Celsius.

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7 0

2 years ago

You are given a 1.55 g mixture of calcium nitrate and calcium chloride. You dissolve this mixture in 20 mL of water and add an e

irina [24]

Answer:

13.4 (w/w)% of CaCl₂ in the mixture

Explanation:

All the Cl⁻ that comes from CaCl₂ (Calcium chloride) will be precipitate in presence of AgNO₃ as AgCl.

To solve this problem we must find the moles of AgCl = Moles of Cl⁻. As 2 moles of Cl⁻ are in 1 mole of CaCl₂ we can find the moles of CaCl₂ and its mass in order to find mass percent of calcium chloride in the original mixture.

<em>Moles AgCl - Molar mass: 143.32g/mol -:</em>

0.535g * (1mol / 143.32g) = 3.733x10⁻³ moles AgCl = Moles Cl⁻

<em>Moles CaCl₂:</em>

3.733x10⁻³ moles Cl⁻ * (1mol CaCl₂ / 2mol Cl⁻) = 1.866x10⁻³ moles CaCl₂

<em>Mass CaCl₂ -Molar mass: 110.98g/mol-:</em>

1.866x10⁻³ moles CaCl₂ * (110.98g/mol) = 0.207g of CaCl₂ in the mixture

That means mass percent of CaCl₂ is:

0.207g CaCl₂ / 1.55g * 100 =

<h3>13.4 (w/w)% of CaCl₂ in the mixture</h3>

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3 years ago

How many molecules are in 10.00 moles of h2o?