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Olegator [25]
3 years ago
12

A shiny chunk of metal is found to have a mass of 37.28g. The metal is dropped into a graduated cylinder which contains 20.0 mL

of water. The volume of the cylinder rises to 34.0 mL. Determine the density of the metal. What might the metal be made of?

Chemistry
1 answer:
Amanda [17]3 years ago
6 0

Answer: The density of the material is 2.66 g/mL and it is likely this  is made of Aluminum

Explanation:

The first step to know the material of the chunk of metal is to calculate its density. The general formula for density is P (density) = \frac{m (mass)}{ v (volume)}. Moreover, in this case, it is known the mass is 37.28 g, but the volume is not directly provided. However, we know the water in the graduated cylinder had a volume of 20.0 mL and this increased to 34.0 mL when the chunk of metal is added, this means the volume of the metal is 14 mL (34.0 mL - 20.0 mL = 14 mL). Now let's calculate the density:

P = \frac{37.28g}{14.0mL}

P = 2.66 g/mL

This means the density of this metal is 2.66 g/mL, which can be rounded as 2. 7 g/mL, and according to the chart, this is the density of aluminum. Therefore, this material of this chunk is aluminum.

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If a 12-volt battery produces a current of 20 amps, what is the resistance?
mixas84 [53]

Answer:

0.6 Ω

Explanation:

From the question given above, the following data were obtained:

Voltage (V) = 12 V

Current (I) = 20 A

Resistance (R) =?

From Ohm's law,

V = IR

Where:

V => is the voltage

I => is the current

R => R is the resistance

With the above formula, we can obtain the resistance as follow:

Voltage (V) = 12 V

Current (I) = 20 A

Resistance (R) =?

V = IR

12 = 20 × R

Divide both side by 20

R = 12 / 20

R = 0.6 Ω

Thus the resistance is 0.6 Ω

4 0
3 years ago
The three main groups of elements are metals, nonmetals, and
Sedbober [7]

Answer:

Metalloids

Explanation:

.... .....................

4 0
2 years ago
Study the following unbalanced half-reaction. Which equation represents the balanced half-reaction? H2O2 ---> H2O
Aleonysh [2.5K]
O: 1*2 = 2*1 
<span>H 2 + 2 = 2*2 </span>

<span>answer C hope you get it right</span>
5 0
3 years ago
Read 2 more answers
The colligative molality of an unknown aqueous solution is 1.56 m.
yawa3891 [41]

Answer:

Vapor pressure of solution = 17.02 Torr

T° of boiling point for the solution is 100.79°C

T° of freezing point for the solution is -2.9°C

Explanation:

Let's state the colligative properties with their formulas

- <u>Vapor pressure lowering</u>

ΔP = P° . Xm . i

- <u>Boiling point elevation</u>

ΔT = Kb . m . i

-<u> Freezing point depressión</u>

ΔT = Kf . m . i

ΔP = Vapor pressure pure solvent (P°) - Vapor pressure solution

ΔT = T° boling solution - T° boiling pure solvent

ΔT = T° freezing pure solvent - T° freezing solution

i represents the Van't Hoff factor (ions dissolved in the solution). If we assume that the solute is non-volatile and the solution is ideal i = 1

Kf and Kb are cryoscopic and ebulloscopic constant, they are  specific to each solvent.

Vapor pressure works with mole fraction (Xm) and the only data we have is molality, so we consider 1.56 moles of solute and 1000 g of solvent mass.

Moles of solvent → solvent mass / molar mass of solvent

Moles of solvent → 1000 g / 18 g/mol = 55.5 moles

Mole fraction is moles of solute / Total moles (mol st + mol sv)

Mole fraction: 1.56 / (1.56 + 55.5) = 0.027

- Vapor pressure lowering

ΔP = P° . Xm . i

17.5 Torr - Vapor pressure of solution = 17.5 Torr . 0.027 . 1

Vapor pressure of solution = - (17.5 Torr . 0.027 . 1 - 17.5 Torr)

Vapor pressure of solution = 17.02 Torr

- Boiling point elevation

ΔT = Kb . m . i

T° boiling solution - 100° = 0.512 °C/ m . 1.56 m . 1

T°boiling solution = 0.512 °C/ m . 1.56 m . 1 + 100°C

T°boiling solution = 100.79°C

- Freezing point depression

ΔT = Kf . m . i

0°C - T° freezing solution = 1.86 °C/m . 1.56 m . 1

T° freezing solution = - (1.86 °C/m . 1.56 m)

T° freezing solution = -2.9°C

3 0
3 years ago
Acetylsalicylic acid (aspirin), HC9H7O4, is the most widely used pain reliever and fever reducer. Find the pH of 0.035 M aqueous
KatRina [158]

<u> </u> The pH of 0.035 M aqueous aspirin is 2.48

<u>Explanation:</u>

We are given:

Concentration of aspirin = 0.035 M

The chemical equation for the dissociation of aspirin (acetylsalicylic acid) follows:

               HC_9H_7O_4\rightleftharpoons H^++C_9H_7O_4^-

<u>Initial:</u>         0.035

<u>At eqllm:</u>    0.035-x        x         x

The expression of K_a for above equation follows:

K_a=\frac{[C_9H_7O_4^-][H^+]}{[HC_9H_7O_4]}

We are given:

K_a=3.6\times 10^{-4}

Putting values in above expression, we get:

3.6\times 10^{-4}=\frac{x\times x}{(0.035-x)}\\\\x=-0.0037,0.0033

Neglecting the value of x = -0.0037 because concentration cannot be negative

So, concentration of H^+ = x = 0.0033 M

  • To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

H^+ = 0.0033 M

Putting values in above equation, we get:

pH=-\log(0.0033)\\\\pH=2.48

Hence, the pH of 0.035 M aqueous aspirin is 2.48

7 0
3 years ago
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