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My name is Ann [436]
2 years ago
14

A 14.3-cm3 sample of tin has a mass of 0.104 kg.

Chemistry
1 answer:
VladimirAG [237]2 years ago
7 0
This is a true statement if it is density you are looking for... Density problem.....

Density is the ratio of the mass of an object to its volume.
D = m / V
D = 104g / 14.3 cm³ = 7.27 g/cm³ .............. to three significant digits

The conventions for the units of density is that grams per cubic centimeter (g/cm³) are usually used for solids, but will work for anything. Grams per milliliter (g/mL) are usually used for liquids and grams per liter (g/L) are for gases. Therefore, by convention, the units for tin (a solid) should be in grams per cubic centimeter.

Since 1 mL is equivalent to 1 cm³, then the density could be expressed as 7.27 g/mL.

The accepted value for the density of tin is 7.31 g/cm³
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As Jesse's hybrid accelerates to pass a car on the highway, he notices that his gas mileage drops from 40 miles per gallon to 15
e-lub [12.9K]
The most likely explanation for this observation is C. his car has turned on more pistons to provide the extra energy needed to accelerate.

When cruising, hybrid cars are able to employ electrical energy to drive the car. Moreover, even if a vehicle is not a hybrid, a greater amount of fuel is consumed when one accelerates because the vehicle has to generate a force larger than the force of air resistance in order for it to accelerate. This increased demand of force reduces the vehicle's fuel economy.
7 0
2 years ago
Read 2 more answers
The volume of a single strontium atom is 4.15×10-23 cm3. What is the volume of a strontium atom in microliters
Ivenika [448]

Answer:-  4.15*10^-^2^0\mu L

Solution:- It is a volume unit conversion problem where we are asked to convert the volume from cm^3 to microliters.

We know that:

1cm^3 = 1 mL

1mL=10^-^3L

and, 1L=10^6\mu L

Let's use these conversions factors for the desired conversion using dimensional as:

4.15*10^-^2^3cm^3(\frac{1mL}{1cm^3})(\frac{10^-^3L}{1mL})(\frac{10^6\mu L}{1L})

= 4.15*10^-^2^0\mu L

So, the answer is  4.15*10^-^2^0\mu L .

7 0
2 years ago
An atom in the ground state contains a total of 5 electrons 5 protons and 5 neutrons. Which Lewis electron-dot diagram represent
Luba_88 [7]
I don't know if you didn't gave a picture choice or if i didn't get the picture.
But lets call this atom A. Electron dot formula doesn't require Neutron and Protons, its main concern is valance elections.

So atom A has 5 electrons which means 2,3 it has 3 valance electrons. Its dot formula will become
:A.

I hope this helped.
7 0
2 years ago
Hydrogen was collected over water using the approach in the manual. The water temperature was 220C and the measured pressure ins
olasank [31]

Answer:

Pressure of hydrogen gas = 695.2 mmHg

Explanation:

Given:

Water temperature = 22°C

Pressure inside the tube = 715 mmHg

Find:

Pressure of hydrogen gas

Computation:

Using vapor pressure of water table

Water pressure at 22°C = 19.8 mmHg

Pressure inside the tube = Pressure of hydrogen gas + Water pressure at 22°C

715 = Pressure of hydrogen gas + 19.8

Pressure of hydrogen gas = 715 - 19.8

Pressure of hydrogen gas = 695.2 mmHg

3 0
2 years ago
Calculate the daily aluminum production of a 150,000 [A] aluminum cell that operates at a faradaic efficiency of 89%. The cell r
Gala2k [10]

Explanation:

It is known that in one day there are 24 hours. Hence, number of seconds in 24 hours are as follows.

                             24 \times 3600 sec

Hence, total charge passed daily is calculated as follows.

                      150,000 \times 24 \times 3600 sec

And, number of Faraday of charge is as follows.

                    \frac{150,000 \times 24 \times 3600 sec}{96500}

                     = 134300.52 F

The oxidation state of aluminium in Al_{2}O_{3} is +3.

                       Al^{3+} + 3e^{-} \rightarrow Al(s)

So, if we have to produce 1 mole of Al(s) we need 3 Faraday of charge.

Therefore, from 134300.52 F the moles of Al obtained with 89% efficiency is calculated as follows.

                \frac{134300.52 F}{3} \times \frac{89}{100}

                   = 39842.487 mol

or,               = 3.9842 \times 10^{4} mol

Molar mass of Al = 27 g/mol

Therefore, mass in gram will be calculated as follows.

            Mass in grams = 3.9842 \times 10^{4} mol \times 27

                                     = 107.57 \times 10^{4} g

                                     = 1075.7 kg/day

Thus, we can conclude that the daily aluminum production of given aluminium is 1075.7 kg/day.

8 0
2 years ago
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