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2 years ago

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

1 answer:

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

You might be interested in

A slug takes 4.25 minutes to travel 11.2 centimeters. What is the speed of the slug in minutes per second?

nexus9112 [7]

Answer:

4.39 × $10^{-4}$ m/sec

Explanation:

7 0

2 years ago

You are studying a large tropical reptile that has a high and relatively stable body temperature. How would you determine whethe

Vesnalui [34]

Explanation:

Endothermic animals are also known as warm-blooded, they have the capacity to regulate their body temperature independent of the environment. They have mechanisms to compensate if heat loss exceeds heat generation (shivers) Or if heat generation exceeds the heat loss (panting, sweating).

On the other hand, ectothermal animals are known as cold blooded organisms and depend on external sources, like sunlight, to regulate their body temperature, reptiles are ectothermals.

To determine if the animal of interest is endo or ectothermal you’ll have to consider that is a reptile, you’ll also observe that it consumes less food and finally it’ll have more difficulties to adapt to sudden temperature changes.

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

8 0

2 years ago

If a large marshmallow has a volume of 2.75 in3 and density of 0.242 g/cm3, how much would it weigh in grams? 1 in3=16.39 cm3.

Anestetic [448]

• Volume of the marshmallow:

V = 2.75 in^3 (but, 1 in^3 = 16.39 cm^3)

V = 2.75 × 16.39 cm^3

V = 2.75 × 16.39 cm^3

V = 45.0725 cm^3

• Density:

d = 0.242 g/cm^3

• Mass:

m = d × V

m = (0.242 g/cm^3) × (45.0725 cm^3)

m = (0.242 g/cm^3) × (45.0725 cm^3)

m = 10.907545 g

m ≈ 10.9 g <——— this is the answer.

I hope this helps. =)

4 0

2 years ago

What is the molarity of a solution that contains 0.500 mole of kno3 dissolved in 0.500-liter of solution?

belka [17]

Answer : The molarity of solution is, 1.00 M

Explanation : Given,

Moles of $KNO_3$ = 0.500 mol

Volume of solution = 0.500 L

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Moles of }KNO_3}{\text{Volume of solution (in L)}}$

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

$\text{Molarity}=\frac{0.500mol}{0.500L}=1.00mole/L=1.00M$

Therefore, the molarity of solution is, 1.00 M

7 0

2 years ago

Q1) A vapor-compression refrigeration system operates on the cycle of Fig. 9.1. The refrigerant is 1,1,1,2-Tetrafluoroethane. Gi

hoa [83]

Answer:

i) 0.5071 (kg/s)

ii) -1407.1 kj/kg

iii) 204.05 Kw

iv) 5.881

v) 9.238

Explanation:

Given Data:

evaporation temperature ( T ) = 4°c = 277.15 K

Condensation Temperature ( T ) = 34°c = 307.15 K

n ( compressor efficiency ) = 0.76

refrigeration rate = 1200 kJ.s^-1

i) determine the circulation rate of the refrigerant

m = $\frac{Q}{H2 - H1}$ = $\frac{Q}{H2 - H4\\}$ ------- 1

Q = 1200 Kj.s^-1

H2 = entropy at step 2 = 2508.9 (kJ / kg ) ( gotten from Table F )

H4 = entropy at step 4 = 142.4 ( kJ/ kg )

back to equation 1

m ( circulation rate of refrigerant ) = 0.5071 (kg/s)

ii) heat transfer rate in the condenser

Q = m ( H4 - H3 )

= 0.5071 ( 142.4 - 2911.27 )

= -1407.1 kj/kg

where H3 = H2 + ΔH23 = 2911.27 (kj/kg) ( as calculated )

iii) power requirement

w = m * ΔH23

= 0.5071 (kg/s) * 402.37 (kj/kg) = 204.05 Kw

where: ΔH23 = $\frac{H'3 - H2 }{0.76}$ = $\frac{2814.7-2508.9}{0.76}$ = 402.37 (kj/kg)

iv) coefficient of performance of a cycle

W = Qc / w

= 1200 Kj.s^-1/ 204.05 kw

= 5.881

v) coefficient of performance of a Carnot refrigeration cycle

$w_{carnot} = \frac{T2}{T4 - T2}$

= 277.15 / ( 307.15 - 277.15 )

= 9.238

4 0

2 years ago