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Svetach [21]
3 years ago
9

Which of the following pairs of elements is most likely to form an ionic compound?

Chemistry
1 answer:
den301095 [7]3 years ago
7 0

Answer:

analogue and digital media is a nightmare to be able

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Suppose that you melt 24 ml of ice. What is the volume of liquid water that results?
serious [3.7K]
The answer to this would be 22 mL
8 0
3 years ago
Please help with this question
tamaranim1 [39]
1.806x10^24
Written equation form(always start the equation off with what you know based off of the question!):

3mol(CCl4)•6.022x10^23/1mol = 1.806x10^24

Good luck!
3 0
3 years ago
What is ΔE in kJ for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same t
kenny6666 [7]

Answer: 20.7 kJ

Explanation:

According to first law of thermodynamics:

\Delta E=q+w

\Delta E=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done on the system=-P\Delta V  {Work is done on the system is positive as the final volume is lesser than initial volume}

w = 4.51 kcal = 4.51\times 4.184kJ=18.9kJ    (1kcal = 4.184kJ)

q = +1.79 kJ   {Heat absorbed by the system is positive}

\Delta E=+1.79+(18.9)=20.7kJ

Thus \Delta E for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same time is 20.7 kJ

4 0
3 years ago
The movement of one or more from one reactant to another is called
mrs_skeptik [129]

Answer: Redox Reaction

Explanation:

Redox reaction is the key chemical events in an oxidation-reduction also called Redox. It is the net movement of electrons from one reactants to another

7 0
3 years ago
How much pure acid is in 520 milliliters of a 13 ​% ​solution?
Liono4ka [1.6K]
It actually depends on the percentage of the concentration give. Percentages can be expressed as %mass/mass, %volume/volume or %mass/volume. To keep things simple, let's just assume that it is in %volume/volume. Thus, 13% of 520 mL is pure acid.

Volume of pure acid = 520*0.13 = 67.6 mL
4 0
3 years ago
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