18.The average or small size stars become a ________

A cylinder with a movable piston originally has a volume of 2805 mL and is filled with nitrogen to a pressure of 4.00

Sergio039 [100]

This problem is providing the initial volume and pressure of nitrogen in a piston-cylinder system and asks for the final pressure it will have when the volume increases. At the end, the answer turns out to be 2.90 atm.

<h3>Boyle's law</h3>

In chemistry, gas laws are used so as to understand the volume-pressure-temperature-moles behavior in ideal gases and relate different pairs of variables.

In this case, we focus on the Boyle's law as an inversely proportional relationship between both pressure and volume at constant both temperature and moles:

$P_1V_1=P_2V_2$

Thus, we solve for the final pressure by dividing both sides by V2:

$P_2=\frac{P_1V_1}{V_2}$

Hence, we plug in both the initial pressure and volume and final volume in order to calculate the final pressure:

$P_2=\frac{2805mL*4.00atm}{3864mL}\\ \\P_2=2.90atm$

Learn more about ideal gases: brainly.com/question/8711877

6 0

2 years ago

At which electrode does oxidation occur in a

Trava [24]

The right answer for the question that is being asked and shown above is that: "(2) the cathode in a voltaic cell and the anode in an electrolytic cell." At the status of electrode does oxidation occur in a voltaic cell and in an electrolytic cell is that the cathode in a voltaic cell and the anode in <span>an electrolytic cell</span>

5 0

3 years ago

Read 2 more answers

The enthalpy change for converting 1.00 mol of ice at -25.0 Â°c to water at 70.0 Â°c is __________ kj. the specific heats of ice

Margarita [4]

Answer is: 6,16 kJ.
1) changing temperature of ice from -25°C to 0°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 2 J/g·°C · 25°C
Q₁ = 900 J.
m(H₂O) = 1mol · 18 g/mol = 18 g.
C - <span>specific heat of ice.
</span>2) changing temperature of water from 0°C to 70°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 4,18 J/g·°C · 70°C
Q₁ = 5266,8 J.
C - specific heat of water.
Q = Q₁ + Q₂ = 900 J + 5266,8 J
Q = 6166,8 J = 6,16 kJ.

8 0

2 years ago

Which stratigraphic principle states the fact that sedimentary rocks are deposited in layers perpendicular to the direction of g

Anvisha [2.4K]

Answer:

omework Help. Steno's laws of stratigraphy describe the patterns in which rock layers are deposited. The four laws are the law of superposition, law of original horizontality, law of cross-cutting relationships, and law of lateral continuity.

Explanation:

5 0

3 years ago

Dimethyl sulfoxide [(ch3)2so], also called dmso, is an important solvent that penetrates the skin, enabling it to be used as a t

Bas_tet [7]

The molecular formula of dimethyl sulfoxide is $(CH_{3})_{2}SO$ . Molar mass of dimethyl sulfoxide is 78.13 g/mol. Calculate number of moles as follows:

$n=\frac{m}{M}=\frac{7.14\times 10^{3} g}{78.13 g/mol}=91.38 mol$

From the molecular formula, 1 mole of dimethyl sulfoxide contains 2 moles of Carbon, 6 moles of Hydrogen, 1 mole of Sulfur and 1 mole of oxygen.

Thus, 91.38 moles of dimethyl sulfoxide will have:

Carbon :

$n_{C}=2\times 91.38 moles=182.77 moles$

Hydrogen:

$n_{H}=6\times 91.38 moles=548.28 moles$

Sulfur:

$n_{S}=1\times 91.38 moles=91.38 moles$

Oxygen:

$n_{O}=1\times 91.38 moles=91.38 moles$

Since, 1 mole of an element equals to $6.023\times 10^{23}$ atoms thus, number of atoms can be calculated as:

Carbon:

$1 mole\rightarrow 6.023\times 10^{23} atoms$

Thus,

$182.77 moles\rightarrow 182.77\times 6.023\times 10^{23} atoms=1.10\times 10^{26} atoms$

Hydrogen:

$1 mole\rightarrow 6.023\times 10^{23} atoms$

Thus,

$548.28 moles\rightarrow 548.28\times 6.023\times 10^{23} atoms=3.30\times 10^{26} atoms$

Sulfur:

$1 mole\rightarrow 6.023\times 10^{23} atoms$

Thus,

$91.38 moles\rightarrow 91.38\times 6.023\times 10^{23} atoms=5.50\times 10^{25} atoms$

Oxygen:

$1 mole\rightarrow 6.023\times 10^{23} atoms$

Thus,

$91.38 moles\rightarrow 91.38\times 6.023\times 10^{23} atoms=5.50\times 10^{25} atoms$

Therefore, number of C, S, H and O atoms are $1.10\times 10^{26}, 5.50\times 10^{25}, 3.30\times 10^{26} and 5.50\times 10^{25} atoms$ respectively.

4 0

3 years ago

18.The average or small size stars become a _________ at the end of their life cycle.