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11Alexandr11 [23.1K]
2 years ago
8

What are 3 ways to seperate metal particles and salt

Chemistry
1 answer:
xenn [34]2 years ago
3 0
You can use a magnet for the metal particles the you will have the salt left
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In a synthesis reaction, one reactant contains 346 J of chemical energy, and one reactant contains 153 J of chemical energy. The
Eduardwww [97]

Answer:

64J of energy must have been released.

Explanation:

Step 1: Data given

One reactant contains 346 J of chemical energy, the other reactant contains 153 J of chemical energy.

The product contains 435 J of chemical energy.

Step 2:

Since the energy is conserved

Sum of energy of Reactants = Energy of Products

Sum of energy of Reactants = 346 J + 153 J = 499 J

The energy of the product = 435 J

435 < 499

This means energy must have been lost as heat.

Step 3: Calculate heat released

499 J - 435 J = 64 J

64J of energy must have been released.

4 0
3 years ago
I WILL GIVE BRAINLEST ANSWER THOSE THREE QUESTIONS PLZZZZ HELP ME
miskamm [114]
For changing potential energy; When you drop a book, gravitational potential energy is transformed into kinetic energy. Your car transforms the chemical potential energy stored in gasoline into the kinetic energy of the car's motion.

For Kinetic Energy; kinetic energy can be converted into electrical energy by a generator or into thermal energy by the brakes on a car.
3 0
3 years ago
Particles of matter how have both potential and kinetic energy is that true or false?
guapka [62]
False: No,any particles of matter do not have any potential or kinetic energy.

5 0
3 years ago
Does anyone know Coulomb’s law??
Ierofanga [76]

Answer:

the force between charged particles increases when they are in solid form

3 0
3 years ago
Does anyone know how to do this? the solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What i
g100num [7]
It's simple, just follow my steps.

1º - in 1 L we have 0.0100~g of BaCO_3

2º - let's find the number of moles.

\eta=\frac{m}{MM}

\eta=\frac{0.0100}{197.3}

\boxed{\boxed{\eta=5.07\times10^{-5}~mol}}

3º - The concentration will be

C=5.07\times10^{-5}~mol/L

But we have this reaction

BaCO_3\rightleftharpoons Ba^{2+}+CO_3^{2-}

This concentration will be the concentration of Ba^{2+}~~and~~CO_3^{2-}

K_{sp}=\frac{[Ba^{2+}][CO_3^{2-}]}{[BaCO_3]}

considering [BaCO_3]=1~mol/L

K_{sp}=[Ba^{2+}][CO_3^{2-}]

and

[Ba^{2+}]=[CO_3^{2-}]=5.07\times10^{-5}~mol/L

We can replace it

K_{sp}=(5.07\times10^{-5})*(5.07\times10^{-5})

K_{sp}\approx25.70\times10^{-10}

Therefore the K_{sp} is:

\boxed{\boxed{\boxed{K_{sp}\approx2.57\times10^{-11}}}}
4 0
3 years ago
Read 2 more answers
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