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Xelga [282]
3 years ago
8

A sample of a gas is placed inside a cylinder that is a fixed size. The cylinder is heated as additional gas is added to it. Wha

t will most likely happen to the pressure of the gas inside the cylinder?
Chemistry
2 answers:
pashok25 [27]3 years ago
7 0

Answer:

The pressure will increase.

Explanation:

The gas state has the ability by the distance of their particles, to compress or expand, in this case, the cylinder can support certain quantity of gas inside it.

But when more quantity of gas is added, the particles will increase, in this way, the energy at which they move will increase by the number of particles present, so the heat increases.

Also, since inside there is a lot of particles moving, the pressure that they exhort is bigger.

makkiz [27]3 years ago
4 0
The pressure of gas will increase because gaseous state is the final state and even if the heat added is evaporating some more gas is still added. It also depends on the temperature of heat added, if the temperature doesn't change the it's most likely for the pressure to be stable...
Hope it helps
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Why is a molecule of CO2 nonpolar even though the bonds between the carbon atom and the oxygen atoms are polar?
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The correct answer is option 1. Carbon dioxide is nonpolar because the shape of the molecule  is symmetrical. It is a linear molecule where the oxygen atoms are symmetrical on each end. CO2 molecule do not have a region of unequal sharing.
7 0
2 years ago
The orbital radii of four planets in our solar system is shown in the following table. Orbital Radii Planet Orbital Radii (AU) W
nalin [4]

Answer:

w and x

Explanation:

6 0
3 years ago
Use the molar bond enthalpy data in the table to estimate the value of Δ∘rxn
MakcuM [25]

Answer:

ΔH°rxn = - 433.1 KJ/mol

Explanation:

  • CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)

∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state

∴ ΔH°CCl4(g) = - 138.7 KJ/mol

∴ ΔH°HCl(g) = - 92.3 KJ/mol

∴ ΔH°CH4(g) = - 74.8 KJ/mol

⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)

⇒  ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol

⇒ ΔH°rxn = - 433.1 KJ/mol

4 0
3 years ago
Read 2 more answers
Can someone please help me answer these questions and I’ll give you brainlest!!!!!
pickupchik [31]

Explanation:

first of all open the menu

5 0
2 years ago
Will mark the brainiest later for correct answers! Please show work.
Elodia [21]

Answer:

According to avogadro's law, 1 mole of every substance contains avogadro's number 6.023\times 10^{23} of particles and weighs equal to its molecular mass.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadros number}}

a. moles in 14.08 g of C_{12}H_{22}O_{11} = \frac{14.08g}{342.3g/mol}=0.04113moles

molecules in 14.08 g of C_{12}H_{22}O_{11} = 0.04113\times 6.023\times 10^{23}=0.2477\times 10^{23}

b. moles in 17.75 g of NaCl = \frac{17.75g}{58.5g/mol}=0.3034moles

molecules in 17.75 g of NaCl = 0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}

formula units 17.75 g of NaCl = 0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}

c. moles in 20.06 g of  CuSO_4.5H_2O= \frac{20.06g}{249.68g/mol}=0.08034moles

formula units in 20.06 g of  CuSO_4.5H_2O= 0.08034\times 6.023\times 10^{23}=0.4839\times 10^{23}

7 0
3 years ago
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