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alina1380 [7]
3 years ago
6

Which equation shows conservation of charge

Chemistry
2 answers:
Semmy [17]3 years ago
8 0
I think it’s Reaction
Zina [86]3 years ago
3 0

Answer:

In essence, charge conservation is an accounting relationship between the amount of charge in a region and the flow of charge into and out of that region, given by a continuity equation between charge density ρ ( x ) {displaystyle rho (mathbf {x} )} and current density J ( x ) {displaystyle mathbf {J} (mathbf {x} )} .

Hope this helps!

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Ideal Gas Law 1
neonofarm [45]

Answer:

1•323

Explanation:

mass of carbon=12

mass of oxygen=2*16=32

no of moles = 3/44

PV=nRT

V=(0.06818*8.31*288)/123.3*10^3

4 0
3 years ago
I just need the balanced redox equation for the cell using the oxidation and reduction half reactions. PLEASE AND ILL MARK BRAIN
harkovskaia [24]

Answer: For the balanced redox equation, you should add the half reactions.

Like this.

Explanation:

Was it helpful? Or do you need different answer?

3 0
3 years ago
Which of the following has a positive enthalpy without a temperature change?
KiRa [710]
 i think the answer is cooling a liquid
6 0
4 years ago
Read 2 more answers
How many molecules are in 4.50 moles of H2O?
alisha [4.7K]
C can be the only correct answer - 6.023 x 10^23 is the amount of molecules in a mol of an element. 4.5 x 6.023 x 10^23 can not equal anything but C.

4.5 x 6.023 x 10^23 = 2.71035 x 10^24
8 0
3 years ago
Read 2 more answers
A generic salt, AB, has a molar mass of 241 g/mol and a solubility of 2.60 g/L at 25 °C. What is the Ksp of this salt at 25 °C?
alekssr [168]
When the salt dissolves, it dissociates as follows;
AB --> A⁺ + B⁻
The molar solubility is amount of moles of the salt that can be dissolved in 1 L of solution.
solubility of the salt is - 2.60 g/L, to find molar solubility we have to multiply by the molar mass.
2.60 g/L /241 g/mol = 0.0108 mol/L
ksp = [A⁺][B⁻]
molar solubility of salt is equal to the molar solubility of cation and anion 
ksp = (0.0108)(0.0108)
ksp = 1.17 x 10⁻⁴
3 0
3 years ago
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