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aleksandrvk [35]
2 years ago
9

How can increase the rate of the preparation copper i iodide?​

Chemistry
1 answer:
BARSIC [14]2 years ago
5 0

Answer:

Preparation. Copper(I) iodide can be prepared by heating iodine and copper in concentrated hydriodic acid, HI. In the laboratory however, copper(I) iodide is prepared by simply mixing an aqueous solution of potassium iodide and a soluble copper(II) salt such copper sulfate.

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Conclusion on friendship​
Black_prince [1.1K]

Explanation:

<h3>Conclusion. A good friendship is very difficult to come across. That is why we should appreciate this divine relationship that is based on understanding and feelings. ... A true friend is one of the most precious possessions that one can have in his life.</h3>

Hope it's helpful for Yuh bro!!

3 0
2 years ago
Read 2 more answers
Classify each of the four compounds as a conjugated, isolated, or cumulated diene. Compound A: Two alkenes are joined by a sigma
Lina20 [59]

Explanation:

Conjugated diene is the one that contains alternate double bonds in its structure. That means both the double bonds are separated by a single bond.

Cumulated diene is the one that contains two double bonds on a single atom. This means it has two double bonds continuously.

Isolated double-bonded compound has a single bond isolated by two to three single bonds.

Compound A: Two alkenes are joined by a sigma bond.

For example:

-CH_2=CH-CH=CH2-

It is a conjugated diene.

Compound B: Two alkenes are joined by a C H 2 group.

It is a cumulative diene.

Compound C: Two alkenes are joined by C H 2 C H 2.

Then it is an isolated alkene.

Compound D:  A cyclohexene has a double bond between carbons 1 and 2. Carbon 3 is an sp 2 carbon that is bonded to another s p 2 carbon with an alkyl substituent.

Hence, compound D is a conjugated diene.

8 0
3 years ago
A gas with a volume of 4.0L at a pressure of 205 kPa is allowed to expand to a volume of 12000 mL. What is the pressure in atmos
Leni [432]

Answer: The pressure in atmospheres is 0.674 in the container if the temperature remains constant.

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

P_1V_1=P_2V_2  

where,

P_1 = initial pressure of gas  = 205 kPa

P_2 = final pressure of gas = ?

V_1 = initial volume of gas  = 4.0 L

V_2 = final volume of gas = 12000 ml = 12 L    (1L=1000ml)  

205\times 4.0=P_2\times 12  

P_2=68.3kPa=0.674atm        (1kPa=0.0098atm)

Therefore, the pressure in atmospheres is 0.674 in the container if the temperature remains constant.

8 0
3 years ago
What is the net ionic equation of the reaction of MgSO4 with Pb(NO3)2?
NemiM [27]

Hey there!:

Write the molecular equation for the reaction of MgSO4 with Pb(NO3)2 :

MgSO4(aq) + Pb(NO3)2(aq) ---> Mg(NO3)2(aq) + PbSO4(s)

Write the total ionic equation  for the reaction :

Mg²⁺ (aq) + SO₄⁻² (aq) + Pb²⁺ (aq) + 2 NO₃⁻¹ (aq) + PbSO₄(s)

Therefore:

Cancel the spectator ions on both sides:

Pb²⁺ (aq) + SO₄⁻² (aq) ---> PbSO4(s)


Hope that helps!

4 0
3 years ago
What is the freezing point of a solution of 498mL of water (solute) dissolved in 2.50 L of ethanol (solvent), C2H5OH? The densit
jok3333 [9.3K]

Answer:

Freezing T° of solution is -142.4°C

Explanation:

This excersise is about colligative properties, in this case freezing point depression,

ΔT = Kf . m . i

Where ΔT = Freezing T° of solvent - Freezing T° of solution

Kf = Cryoscopic constant

m = mol/kg (molality)

i = Number of ions dissolved.

Water is not ionic, so i = 1

Let's find out m.

We determine mass of water, by density

498ml . 1 g/mL = 498 g

We convert the mass of water to moles → 498 g . 1mol/18g = 27.6 moles

We determine mass of solvent by density

2500 mL . 0.789 g/mL = 1972.5 g

Notice, we had to convert L to mL to cancel units.

1 cm³ = 1 mL

We convert the mass from g to kg → 1972.5 g . 1kg /1000g = 1.97kg

We determine m = mol/kg → 27.6mol / 1.97kg = 13.9 m

Kf for ethanol is: 1.99 °C/m

Freezing T° for ethanol is: -114.6°C

We replace at formula: - 114.6°C - Freezing T° solution = 1.99 °C/m . 13.9 m . 1

- 114.6°C - Freezing T° solution = 27.8 °C

- Freezing T° solution  = 27.8°C + 114.6°C

Freezing T° Solution = - 142.4 °C

7 0
2 years ago
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