The answer would be millimeters
so answer choice d
Answer:
26.9 L SO₂
Explanation:
Step 1: Write the balanced equation
S(s) + O₂(g) = SO₂(g)
Step 2: Establish the appropriate volume ratio
For gases at the same conditions, the volume ratio is equal to the molar ratio. The volume ratio of O₂(g) to SO₂(g) is 1:1.
Step 3: Calculate the liters of SO₂ produced from 26.9 L of O₂
We will use the previously established volume ratio.
26.9 L O₂ × 1 L SO₂/1 L O₂ = 26.9 L SO₂
Answer:
a. The original temperature of the gas is 2743K.
b. 20atm.
Explanation:
a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:
T₁n₁ = T₂n₂
<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>
<em />
<em>Replacing with values of the problem:</em>
T₁n₁ = T₂n₂
X*7.1g = (X+300)*6.4g
7.1X = 6.4X + 1920
0.7X = 1920
X = 2743K
<h3>The original temperature of the gas is 2743K</h3><h3 />
b. Using general gas law:
PV = nRT
<em>Where P is pressure (Our unknown)</em>
<em>V is volume = 2.24L</em>
<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>
R is gas constant = 0.082atmL/molK
And T is absolute temperature (2743K)
P*2.24L = 0.20mol*0.082atmL/molK*2743K
<h3>P = 20atm</h3>
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Answer:
This would support Dalton's postulates that proposed the atoms are indivisible because no small particles are involved.
Explanation:
Experiment using the gas discharge tube by J.J Thomson led to the discovery of cathode rays which are now known as electrons.
Primarily, Thomson's experiment led to the discovery of cathode rays, electrons, as subatomic particles.
If the size of the atoms observed at the cathode is the same as that of the rays,we can conclude that the particles of the rays are the simplest form of matter we can have. This would suggest that the atom is indeed the smallest indivisible particle of a matter according to Dalton.
Answer:
The
of the given reaction is -129.6 kJ
Explanation:
The given chemical reaction is as follows.
Enthalpy of each reactant and products are as follows.
In the given chemical reaction involved two C-H bonds in the reactant side and one C-C bond in the product side therefore, the enthalpy of formation will be the negative.
Therefore, The of the given reaction is -129.6 kJ
