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VashaNatasha [74]
3 years ago
13

Which half-reaction equation represents the

Chemistry
2 answers:
erica [24]3 years ago
8 0
The reduction reaction is the gain of electrons while oxidation reaction is the loss of electrons. For potassium ion(K+), the reaction should be K+ + e- ==> K. So the answer is (1).
GalinKa [24]3 years ago
4 0

Answer : The correct option is, (1) K^++e^-\rightarrow K

Explanation :

Oxidation-reduction reaction : It is a type of reaction where the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is a reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

As per question, the half-reaction equation of reduction of a potassium ion will be :

K^++e^-\rightarrow K

In this reaction, the oxidation state of potassium changes from (+1) to (0) that means the oxidation state decreases. So, this reaction shows the half-reaction equation of reduction of a potassium ion.

Therefore, the correct option is, (1) K^++e^-\rightarrow K

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Assuming complete distribution, what is the molarity of 10 milligrams of lisinopril in 100 liters?
SVEN [57.7K]

Answer:

2.47x10^{-7}   M

Explanation:

Lisoprisil's molecular mass is 405.488g/mol, we'll use this fact to calculate molarity, which units are mol/L, and we proceed to the calculus:

  • First, we'll unify unities, the 10 milligrams of lisinopril we'll transform into grams.

10mg*\frac{1g}{1000mg}=0.01g

  • Now that we have the same unities we'll calculate molarity using the molecular mass, the grams of lisinopril and the liters in which these grams are, let's consider that our final unities have to be mol/L.

\frac{1mol}{405.488g}*\frac{0.01g}{100L}=2.47x10^{-7}   M

I hope you find this information useful and interesting! Good luck!

3 0
3 years ago
Be sure to use proper significant figures. The molecular weight of NaCl is 58.44 grams/mole. To prepare a 0.50 M solution: You w
Misha Larkins [42]
Your answer would be,
Molarity = moles of solute/volume of solution we needed, 29.22(g)(mol) of NaCI

= 29.22(g)/58.44(g)(mol^-1)(1)/1(L)
= 0.500(mol)(L^-1)



Hope that helps!!!
5 0
3 years ago
Read 2 more answers
Water acts as a<br> solvent<br> solute<br> solution<br> mixture
Serhud [2]
Answer: universal Solvent

Explanation: water can dissolve many substances than any other solvent. Because of this, water is sometimes called a UNIVERSAL SOLVENT.
5 0
3 years ago
If the distance between a point charge and a neutral atom and is multiplied by a factor of 5, by what factor does the force on t
dexar [7]

Given :

The distance between a point charge and a neutral atom and is multiplied by a factor of 5.

To Find :

By what factor does the force on the neutral atom by the point charge change.

Solution :

We know, electrostatic force between two object is directly proportional to product of charge and inversely proportional to distance between them.

Now, charge in neutral atom is 0 C.

So, the electrostatic force between two of them is also 0 N.

Therefore, by changing distance between the charge the forces did no change ( it remains zero).

3 0
3 years ago
When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

m=1.00 g/mL\times 100.0 mL = 100 g

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

q=m\times c\times (T_{final}-T_{initial})

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = 4.18 J/^oC

T_{final} = final temperature = 23.1^oC

T_{initial} = initial temperature = 28.9^oC

Now put all the given values in the above formula, we get:

q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0
3 years ago
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