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Elan Coil [88]
2 years ago
9

Please help!!!!!!!!!!!!!!!

Chemistry
1 answer:
svlad2 [7]2 years ago
3 0
The 2nd one is the right answer

Hope this helped
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explain what happens when anhydrous calcium chloride are exposed to the atmosphere for about two days​
PtichkaEL [24]

Answer:

Explanation:

Calcium chloride is <u>deliquescent.</u> If exposed to air, it will absorb sufficient water from the air to allow it to dissolve. After a short while, instead of a white lump, you will have a pool of clear liquid.

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3 years ago
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1. Liquid
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2 years ago
How much heat is added if 0.0318g of water is increased in temperature by 0.364 degrees C?
SSSSS [86.1K]

Answer:

0.04838J

Explanation:

Heat is a form of energy that is transferred from one body to another as the result of a difference in temperature between the bodies , here heat is added to the water as a result of temperature change of 0.364 degreesC

Given:change in temperature=0.364

Mass of water=0.0318g

But we need specific heat capacity of water which is

4.2 J/g°C

Then we can calculate How much heat is added by using below formula

Energy = Mass * specific heat capacity *(change in temperature)

energy =0.0318g* 4.18g*0.364

=0.04838J

8 0
3 years ago
PLEASE HELP!! How could using either more or less of a chemical in a scientific experiment cause a change in results?
charle [14.2K]
The chemical could have more or less of a reaction to the other chemicals in the experiment
4 0
3 years ago
The ΔHcomb value for anethole is -5539 kJ/mol. Assume 0.840 g of anethole is combusted in a calorimeter whose heat capacity (Cal
bonufazy [111]

Answer:

Final temperature of calorimeter is 25.36^{0}\textrm{C}

Explanation:

Molar mass of anethole = 148.2 g/mol

So, 0.840 g of anethole = \frac{0.840}{148.2}moles of anethole = 0.00567 moles of anethole

1 mol of anethole releases 5539 kJ of heat upon combustion

So, 0.00567 moles of anethole release (5539\times 0.00567)kJ of heat or 31.41 kJ of heat

6.60 kJ of heat increases 1^{0}\textrm{C} temperature of calorimeter.

So, 31.41 kJ of heat increases (\frac{1}{6.60}\times 31.41)^{0}\textrm{C} or 4.76^{0}\textrm{C} temperature of calorimeter

So, the final temperature of calorimeter = (20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}

3 0
3 years ago
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