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Naya [18.7K]
3 years ago
10

A sample of hydrogen gas will behave most like an ideal gas under conditions of?

Chemistry
1 answer:
fenix001 [56]3 years ago
6 0
Conditions:
Low pressure and low temperature
Low pressure and high temperature
High pressure and low temperature
High pressure and high temperature
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Aluminum has a density of 2.7 g/cm3, how much space in cm3 would 81 grams of aluminum occupy? Show steps to answering this equat
Evgesh-ka [11]

Answer:

30 cm³

Explanation:

Step 1: Given data

  • Density of aluminum (ρ): 2.7 g/cm³
  • Mass of aluminum (m): 81 g
  • Volume occupied by aluminum (V): ?

Step 2: Calculate the volume occupied by aluminum

The density of aluminum is equal to its mass divided by its volume.

ρ = m/V

V = m/ρ

V = 81 g / 2.7 g/cm³

V = 30 cm³

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3 years ago
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Answer:

Imao ;/

Explanation:

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What factor would speed up a chemical reaction
vekshin1

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Adding Catalyst can Speed up a Chemical reaction

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What units do we use to measure mass?
MA_775_DIABLO [31]

Answer:

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3 years ago
PLEASE HELP!!!! WILL MARK BRAINLIEST!!!! A rigid steel container with a volume of 30 L is filled with oxygen to a pressure of 9.
GuDViN [60]

Answer:

12 atm

Explanation:

First, let us convert Celcius into Kelvin: 28.0 °C = 301.15 K and 129.0 °C = 402.15 K

For this question we must employ the Combined Gas Law: \frac{P_1V_1}{P_2V_2}=\frac{n_1RT_1}{n_2RT_2}, where P_1 is the initial pressure and P_2 is the new pressure.

We know that intitially, P=9 atm, V=30 L, and T=301.15K. From our problem, only temperature and pressure changes, while the number of moles, volume and the gas constant, R, stay the same, so they are irrelevant.

Thus, the filled out Combined Gas Law would be:

\frac{9 atm}{P_2}=\frac{301.15K}{402.15K}, where the volume, moles of gas, and R are cancelled out.

We can manipulate this equation to derive the new pressure. We find that

9atm≈0.74885P_2.  

This means that

P_2≈9/0.74885≈12 atm

4 0
2 years ago
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