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Tom [10]
3 years ago
7

The forensic technician at a crime scene has just prepared a luminol stock solution by adding 15.0 gg of luminol into a total vo

lume of 75.0 mLmL of H2OH2O. What is the molarity of the stock solution of luminol? Express your answer with the appropriate units.
Chemistry
1 answer:
Firdavs [7]3 years ago
6 0

Answer:

1.13 M

Explanation:

Given data

  • Mass of luminol (solute): 15.0 g
  • Volume of solution = volume of water = 75.0 mL = 0.0750 L
  • Molar mass of luminol: 177.16 g/mol

The molarity of the stock solution of luminol is:

M = mass of solute / molar mass of solute × liters of solution

M = 15.0 g / 177.16 g/mol × 0.0750 L

M = 1.13 M

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What is the specific heat capacity of an unknown metal if 75.00 g of the metal absorbs 418.6J of heat and the temperature rises
EleoNora [17]

Answer:

The specific heat capacity of the unknown metal is 0.223 \frac{J}{g*C}

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • Q= 418.6 J
  • c= ?  
  • m= 75 g
  • ΔT= 25 C

Replacing:

418.6 J= c* 75 g* 25 C

Solving:

c=\frac{418.6 J}{75 g*25 C}

c= 0.223 \frac{J}{g*C}

<u><em>The specific heat capacity of the unknown metal is 0.223 </em></u>\frac{J}{g*C}<u><em></em></u>

<u><em> </em></u>

<u><em></em></u>

3 0
2 years ago
A galvanic cell consists of a Cu(s)|Cu2+(aq) half-cell and a Cd(s)|Cd2+(aq) half-cell connected by a salt bridge. Oxidation occu
Kryger [21]

Answer:

Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)

Explanation:

A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.

The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.

Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)

6 0
4 years ago
What effect does the anion of an ionic compound have on the appearance of the solution?
choli [55]

Answer: B. The anion affects the color of the solution more than the intensity of the color.

Explanation:

An ionic bond is gotten when an electron is transferred from a metal atom to a non-metal one. It should be noted that the ionic bonds simply has an anion and a cation.

An anion is formed when a valence election is gained by a non metal while a cation is formed when the metal ion misplaces a valence electron.

The effect of the anion of an ionic compound on the appearance of the solution is that the anion affects the color of the solution more than the intensity of the color.

6 0
3 years ago
If the half life of iridium-182 is 15 years, how much of a 3 gram sample is left after 2 half-lives?
padilas [110]

Answer:

D. 0.75 grams

Explanation:

The data given on the iridium 182 are;

The half life of the iridium 182, t_{(1/2)} = 15 years

The mass of the sample of iridium, N₀ = 3 grams

The amount left, N(t) after two half lives is given as follows;

N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}} }

For two half lives, t = 2 × t_{(1/2)}

∴ t = 2 × 15 = 30

\dfrac{t}{t_{(1/2)}} = \dfrac{30}{15} = 2

\therefore N(t) = 3 \times\left (\dfrac{1}{2} \right )^2 = 0.75

∴ The amount left, N(t) = 0.75 grams

4 0
3 years ago
Calculate the mass of sucrose needed to prepare a 2000 grams of 2.5% sucrose solution.
meriva

Answer:

50 g Sucrose

Explanation:

Step 1: Given data

  • Mass of solution: 2000 g
  • Concentration of the solution: 2.5%

Step 2: Calculate the mass of sucrose needed to prepare the solution

The concentration of the solution is 2.5%, that is, there are 2.5 g of sucrose (solute) every 100 g of solution. The mass of sucrose needed to prepare 2000 g of solution is:

2000 g Solution × 2.5 g Sucrose/100 g Solution = 50 g Sucrose

5 0
3 years ago
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