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Mkey [24]
3 years ago
8

Copper(II) chloride hydrate, also known as copper chloride hydrate, is the inorganic compound with the chemical formula CuCl2 .

x(H2O), where x can range from 1 to 7. A 2.060 g sample of copper (II) chloride hydrate was dissolved in 50.0 mL of deionized water. The concentration of copper (II) ion in the solution was determined spectrophotometrically using the equation from the calibration curve, was 0.200 M. Determine the formula of the copper (II) chloride hydrate.
Chemistry
1 answer:
aksik [14]3 years ago
4 0

Answer:

CuCl₂·4H₂O

Explanation:

First we <u>calculate the moles of copper (II) ion in the solution</u>, using the <em>given volume and concentration</em>:

  • 50.0 mL ⇒ 50.0/1000 = 0.05 L
  • 0.05 L * 0.200 M = 0.01 mol Cu⁺²

For each Cu⁺² mol there's also one CuCl₂ mol.

We <u>convert 0.01 CuCl₂ moles into grams</u>, using its <em>molar mass</em>:

  • 0.01 CuCl₂ mol * 134.45 g/mol = 1.344 g

That means that out of the 2.060 g of the sample, 1.344 g are of CuCl₂. This means that there are (2.060 - 1.344) 0.716 g of water.

We <u>convert those 0.716 g of water to moles</u>:

  • 0.716 g ÷ 18 g/mol = 0.04 mol H₂O

There are four times as many H₂O moles as there are CuCl₂ moles. This means the formula is CuCl₂·4H₂O

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In the important industrial process for producing ammonia (the Haber Process), the overall reaction is: N2(g) + 3H2(g) → 2NH3(g)
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3 0
4 years ago
A mylar balloon is filled with helium gas to a pressure of 107 kPa when the temperature is 22°C. If the temperature changes to 4
brilliants [131]

Answer:

Explanation:

To solve this problem, we have to remember the ideal gas theory. We going to use the Gay Lussac Law.

The equation is:

PiTo = PoTi

Where:

<em>Pi =initial pressure</em>

<em>Po = final pressure</em>

<em>Ti = initial temperature</em>

<em>To = final temperature</em>

Let's use this equation and replace the values:

<em>Pi = 107 kPa</em>

<em>Ti = 22°C</em>

<em>To = 45 °C </em>

<em>Po = ? </em>

<em />

You can use the units as the problem gave you.

Using the Gay Lussac law:

107 kPa x 45 °C = Po x 22°C

Po = (4,815 kPa°C / 22°C) = 218.86 kPa

Po = 218 kPa

5 0
3 years ago
A 1.40 L sample of O2 at 645 Torr and 25 °C, and a 0.751 L sample of N2 at 1.13 atm and 25 °C, are both transferred to the same
anyanavicka [17]

Answer:

  • P(O₂) = 0.595 atm
  • P(N₂) = 0.424 atm
  • Total Pressure = 1.019 atm

Explanation:

To solve this problem we use PV=nRT for both gases in their containers, in order to <u>calculate the moles of each one</u>:

  • O₂:

645 Torr ⇒ 645 /760 = 0.85 atm

25°C ⇒ 25 + 273.16 = 298.16 K

0.85 atm * 1.40 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0487 mol O₂

  • N₂:

1.13 atm * 0.751 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0347 mol N₂

Now we can <u>calculate the partial pressure for each gas in the new container</u>, because the number of moles did not change:

  • O₂:

P(O₂) * 2.00 L = 0.0487 mol O₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(O₂) = 0.595 atm

  • N₂:

P(N₂) * 2.00 L = 0.0347 mol N₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(N₂) = 0.424 atm

Finally we add the partial pressures of all gases to <u>calculate the total pressure</u>:

  • Pt = 0.595 atm+ 0.424 atm = 1.019 atm
6 0
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