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Pachacha [2.7K]
3 years ago
11

Calcium metal reacts with a potassium chloride solution to form calcium chloride and potassium ions. Classify this reaction. Ca(

s) + 2KCl(aq) → CaCl2(s) + 2K+(aq)
Physics
1 answer:
Scilla [17]3 years ago
8 0
The given reaction above is an example of <em>single replacement reaction. </em>This is a certain type of reaction in which an element of a compound is replaced by a free element. In this case, chlorine (Cl) is replaced by calcium (Ca). 
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Why is magnesium the limiting reactant in this experiment
Keith_Richards [23]

Answer:

Explanation:

  • Magnesium is being oxidized by the oxygen in the air to magnesium oxide. This is a highly exothermic combustion reaction, giving off intense heat and light. The reaction of the combustion of magnesium in oxygen is given below:
  •                                     2Mg(s) + O2(g) → 2MgO(s)
  • The stoichiometric factor is 2 moles of magnesium are burned for every 1 mole of oxygen (2mol Mg/1mol O2). If the magnesium strip weighs 1 gram, then there is 0.04 mol of magnesium (1 gram divided by 24.3 grams/mol Mg) available in the reaction. The amount of oxygen required to completely react with the magnesium strip is:
  • 0.04 mol Mg x (1 mol O2 / 2 mol Mg) = 0.02 mol O2 x 16 g/mol O2 = 0.32 gram O2.
  • The magnesium will burn until consumed entirely. There is much more oxygen available in the atmosphere than needed to consume the magnesium. Thus the magnesium is the limiting reactant because it determines the amount of product formed.
4 0
3 years ago
Please need help with this
lutik1710 [3]

The first and third choices could both do it, but the first choice makes a much better, clear demonstration.

8 0
3 years ago
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What magnitude charge creates a 32.08 N/C electric field at a point 2.84 m away?
sammy [17]

Answer:

Q = 29.4 x 10⁻⁹ C.

Explanation:

Electric field due to a charge Q at distance d is given by coulomb law as follows

Electric field

E = k Q /d²

where for air k which is a constant is 9 x 10⁹

E=\frac{k\times Q}{d^2}

Given E = 32.08 , d = 2.84 m

Putting these values in the relation above, we have

[tex]32.08=\frac{9\times 10^9\times Q}{(2.84)^2}[/tex]

Q = 29.4 x 10⁻⁹ C.

5 0
2 years ago
A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 t. What minimum current in the rod is nee
slamgirl [31]

Answer:

i = 4.9 A

Explanation:

Force on a current carrying rod due to magnetic field is given as

F = iLB

here we know that

i =current in the rod

B = 0.10 T

L = 1.0 m

now magnetic force is balanced by the weight of the rod

so we will have

iLB = mg

i(1.0)(0.10) = 0.05 \times 9.8

i = 4.9 A

8 0
3 years ago
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Svetllana [295]

Answer:

DO NO KNOW AND I HOPE YOU CAN FIND IT

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Explanation:

6 0
3 years ago
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