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garik1379 [7]
3 years ago
11

How do you calculate the density of a gas, based on its temperature and pressure?

Chemistry
1 answer:
stiks02 [169]3 years ago
5 0
We can rearrange the ideal gas equation:
PV = nRT, where n is the number of moles equivalent to:
n = mass / Mr
PV = mRT/Mr
m/V = PMr/RT
density = PMr / RT; where Mr and R are constant.
You might be interested in
Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.
zepelin [54]

Answer:

balanced equation mole ratio 5 2 mol NO/1 mol O2

10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

3 0
2 years ago
Which of the following changes will always be true for a spontaneous reaction? (3 points)
Alika [10]
Answer is: <span>- delta G.
</span>The change in Gibbs free energy (ΔG), at constant temperature and pressure, is: <span>ΔG=ΔH−TΔS.
</span>ΔH<span> is the change in enthalpy.
</span>ΔS is change in entropy.
T is temperature of the system.
When ΔG is negative, a reaction (<span>occurs without the addition of external energy)</span><span> will be spontaneous (</span>exergonic).

3 0
3 years ago
What is oxidized in a galvanic cell with aluminum and gold electrodes ?
k0ka [10]

Answer:

Aluminum metal

Explanation:

In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.

First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:

Al^{3+}+3e^-\rightarrow Al; E^o=-1.66 V

Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V

Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.

Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.

Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):Al\rightarrow Al^{3+}+3e^-; E^o=1.66 V\\Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V

Notice that the overall cell potential upon summing is:

E_{cell}=1.66 V + 1.50 V=3.16 V

Meaning we obey the law of galvanic cells.

Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.

6 0
2 years ago
Which of the following would most likely result in an increase in reaction rate? placing the reactants on a hotplate
puteri [66]

Answer:

placing the reactants on a hot plate

Explanation:

If the temperature goes up, the reaction rate will increase. Because the particle will move faster and makes the kinetic energy larger.

4 0
3 years ago
3. Which part of a cell is not found in the cytoplasm?
kvv77 [185]

Answer:

vacuole

Explanation:

Brainliest please

6 0
2 years ago
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