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mario62 [17]
3 years ago
8

Differences between expansion of solid and liquid​

Chemistry
1 answer:
julia-pushkina [17]3 years ago
7 0
Yep yep looks about right
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Calculate the heat absorbed/released (in Joules) by the solution when the cold pack solid dissolved in water in your experiment.
skad [1K]

Answer:

56765

Explanation:

i did the math and this is the answer

8 0
3 years ago
Which one of the following compounds utilizes both ionic and covalent bonding?A) C6H12O6.B) CC32-.C) CO2.D) MgCl2.E) Al2(SO4)3.
JulsSmile [24]

Answer:

Al2(SO4)3

Explanation:

Looking at this carefully, we will discover that Al2(SO4)3 is composed of Al^3+ and SO4^2-.

The aluminum and sulphate ions are ionically bonded. However, the oxygen and sulphur in the sulphate ion are covalently bonded.

Hence, Al2(SO4)3 contains both ionic and covalent bond.

8 0
3 years ago
Which of the following describes scientists?
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3 0
3 years ago
8. Sulfur has a first ionization energy of 1000 kJ/mol. Photons of what frequency are required to ionize one mole of Sulfur?​
Lynna [10]

Answer:

the frequency of photons v = 1.509\times10^{39}Hz

Explanation:

Given:  first ionization energy of 1000 kJ/mol.

No. of moles of sulfur = 1 mole

\Delta E_1 = 1000KJ/mol

We know that plank's constant

h = 6.626\times10^{-34} Js

Let the frequency of photons be ν

Also we know that ΔE = hν

this implies ν = ΔE/h

= \frac{10^6J}{6.626\times10^{-34} Js}

v = 1.509\times10^{39}Hz

Hence, the frequency of photons v = 1.509\times10^{39}Hz

6 0
3 years ago
Is 13.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode?
Andrews [41]
Answer: 
 Zn =⇒ Zn+2(0.10) + 2e- (anode)
 Zn+2(?M) + 2e- === Zn(s) (cathode)
 
 Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn 
 E = E^o -0.0592 log Q; in this case E^o is zero. 
 E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2 
 23 mV x 1 volt/1000mv = 0.023 Volts 
 0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
 0.023 = -0.0296 { log 0.10 – log [Zn+2] }
 0.023 = -0.0296{ -1 - log[Zn+2] }
 0.023 = +0.0296 + 0.0296log[Zn+2]
 -0.0066 = 0.0296log[Zn+2]
 -0.22= log[Zn+2]
 [Zn+2] = 10^-0.22 = 0.603 Molar

6 0
3 years ago
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