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Naddik [55]
3 years ago
12

How many moles in 65.5 grams of chlorine

Chemistry
1 answer:
jekas [21]3 years ago
4 0
I believe it would be 2322.1714999999604
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Please help! ill give you 90 points!
Luba_88 [7]

Answer:

1.d = 2.70 g/mL

2.d = 13.6 g/mL

4.d = 1896 g / 212.52 cm3 = 8.9 g/cm3

Explanation:If this helped subscribe to Amiredagoat Yt

3 0
3 years ago
What is the concentration of barium ions in a 1.00 M BaCl2 solution?
Luden [163]
Concentration of Ba2+ is 1.00 mol/dm3 or 1M
5 0
3 years ago
The azide ion, n−3, is a symmetrical ion, all of whose contributing structures have formal charges. draw three important contrib
GREYUIT [131]

The azide ion N3- can actually be represented by 3 resonance structures.

(check attached image for the structures)

 

<span>Among the three, the Structure 1 is the most important one. While structure 3 almost makes no contribution due to the positive charges located on the adjacent atoms and the overall higher formal charge.</span>

3 0
4 years ago
Read 2 more answers
How many moles of carbon are in the sample?
jeka94

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5 0
3 years ago
A 125G sample of water was heated to 100.0°C and then I borrow platinum 20.0°C is dropped into the beaker the temperature of the
Sophie [7]

Answer:

mass of platinum = 2526.12 g

Explanation:

Given data:

Mass of water = 125 g

Initial temperature of water= 100.0°C

Initial temperature of Pt = 20.0°C

Final temperature = 235°C

Specific heat of Pt = 0.13 j/g°C

Specific heat of water = 4.184 j/g°C

Mass of platinum = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

Q(w) = Q(Pt)

m.c.  (T2 - T1)   =   m.c.   (T2 - T1)

125 g × 4.184 j/g°C ×  (235°C - 100.0°C)  = m × 0.13 j/g°C ×  (235°C - 20°C)

125 g × 4.184 j/g°C × 135°C  = m × 0.13 j/g°C × 215°C

70605 j = m×27.95 j/g

m = 70605 j /27.95 j/g

m = 2526.12 g

5 0
4 years ago
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