Part 1: What is a black dwarf? Part 2: What is the fate of Sol?

As the volume of a gas at constant temperature did decreased,the pressure increases because

frosja888 [35]

Answer:

boyles law

Explanation:

volume is inversely proportional to pressure

7 0

2 years ago

The isomers butane and methyl propane have

madreJ [45]

Explanation:

“The isomers butane and methyl propane have the same molecular formula and different properties”, this is because structural isomers usually have different properties to their parent.

3 0

3 years ago

In the PhET simulation window, click the radio button labeled Mystery in the Blocks menu on the right-hand side of the screen. U

Yuliya22 [10]

Answer:

B, D, E, C, A

Explanation:

We have 5 blocks with their respective masses and volumes.

Block Mass Volume

A 65.14 kg 103.38 L

B 0.64 kg 100.64 L

C 4.08 kg 104.08 L

D 3.10 kg 103.10 L

E 3.53 kg 101.00 L

The density (ρ) is an intensive property resulting from dividing the mass (m) by the volume (V), that is, ρ = m / V

ρA = 65.14 kg / 103.38 L = 0.6301 kg/L

ρB = 0.64 kg / 100.64 L = 0.0064 kg/L

ρC = 4.08 kg / 104.08 L = 0.0392 kg/L

ρD = 3.10 kg / 103.10 L = 0.0301 kg/L

ρE = 3.53 kg / 101.00 L = 0.0350 kg/L

The order from least dense to most dense is B, D, E, C, A

4 0

3 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

4 0

3 years ago

How much sucrose (g) do you need to weight in order to prepare 19.16 g of a 13.1 % (weight percent) solution?

Nonamiya [84]

<span>2.51 grams
You want to prepare 19.16 g of some solution which will have 13.1% of it's mass being sucrose. So we just need to perform some simple multiplication: 19.16g * 0.131 = 2.50996g
Rounding to 3 significant figures gives 2.51 g.</span>

3 0

3 years ago