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Bezzdna [24]
2 years ago
13

Pleaaaaase hellllp asaaaap

Chemistry
1 answer:
rewona [7]2 years ago
8 0

Answer:

solve example number 2 ....

Explanation:

solve examole number 2 .....

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Where are atoms located? (check all that apply)*
Jobisdone [24]
I think it’s 2 hope that helped
4 0
3 years ago
Please help! Which Zeros are significant in the measurement .0506010 kg?
Vitek1552 [10]
It would be a,b,c as the answer. hope this helps!
7 0
2 years ago
Analyze the bonds of CCl4. What is the shape of the CCl4 molecule? Is it symmetrical? Does this mean that the CCl4 molecule is p
Crank

C-Cl bond is polar.


As Cl is more electronegative than C,  

Cl pulls the shared electrons towards its side making the C-Cl bond polar.


Polar meaning C gets a partial positive charge and Cl gets a partial negative charge. Thus C-Cl bond becomes polar.


CCl4 molecule is tetrahedral.

And its shape is symmetrical.


That means its similar on all corners or all planes and it can be cut into equal halves from any plane.


C-Cl bond is polar.But the C-Cl polairty gets cancelled out as its pulled equally from all sides.


Since the molecule is symmetrical CCl4 is non polar.  


3 0
3 years ago
Sugar is dissolved in water. Which is the solute? sugar neither both water
lina2011 [118]

Answer:

Sugar

Explanation:

Defintion of a solute-dissolved in the solvent, sugar dissolves in water and is therefore the solute

3 0
3 years ago
Read 2 more answers
A 20.0 g piece of aluminum at 5.00 C is dropped into 20.2 g of water at 90.00 C. The final temperature is 75.00 C. Use the First
bekas [8.4K]

Answer:

The specific heat of aluminium is 0.906 J/g°C

Explanation:

Step 1: data given

Mass of aluminium = 20.0 grams

Temperature = 5.00 °C

Mass of water = 20.2 grams

Temperature of water = 90.00 °C

The final temperature = 75.00 °C

Specific heat of water = 4.184 J/g°C

Step 2: calculate the specific heat of aluminium

heat won = heat lost

Qaluminium = -Qwater

Q = m*c* ΔT

m(aluminium * c(aluminium) *ΔT(aluminium = -m(water) * c(water) *ΔT(water)

⇒with m(aluminium) = mass of aluminium = 20.0 grams

⇒with c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒with ΔT(aluminium) = the change of temperature = T2 - T1 = 75.00 °C - 5.00 °C = 70.00 °C

⇒with m(water) = the mass of water = 20.2 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = T2 - T1 = 75.00°C - 90.00 °C = -15.00 °C

20.0 * c(aluminium) * 70.00 = -20.2 * 4.184 * -15.00

c(aluminium) = 0.906 J/g°C

The specific heat of aluminium is 0.906 J/g°C

7 0
3 years ago
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