The chemical formula for sodium and sulfur is Na2S I believe...
2.2311 moles of gas are there in a 50. 0 l container at 22. 0 °c and 825 torrs.
<h3>What is an ideal gas?</h3>
An Ideal gas is a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
Assuming the gas is ideal, we can solve this problem by using the following equation:
PV = nRT
Where:
P = 825 torr ⇒ 825 / 760 = 1.08 atm
V = 50 L
n = ?
R = 0.082 atm·L·mol⁻¹·K⁻¹
T = 22 °C ⇒ 22 + 273.16 = 295.16 K
We input the data:
1.08 atm x 50 L = n x 0.082 atm·L·mol⁻¹·K⁻¹ x 295.16 K
And solve for n:
24.20312
n = 2.2311 mol
Hence, 2.2311 moles of gas are there in a 50. 0 L container at 22. 0 °c and 825 torrs.
Learn more about ideal gas here:
brainly.com/question/23580857
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Answer:
<h3>In a solid, molecules are packed together, and it keeps its shape. ... Matter is the "stuff" of the universe, the atoms, molecules and ions that make up all physical substances. In a solid, these particles are packed closely together and are not free to move about within the substance.</h3>
Answer:
Molarity of the packet is 0.5M
Explanation:
In the reaction of acetic acid with NaOH:
CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺
<em>1 mole of acetic acid reacts with 1 mole of NaOH.</em>
<em />
When you are titrating the acid with NaOH, you reach equivalence point when moles of acid = moles of NaOH.
Moles of NaOH are:
3.0mL = 3.0x10⁻³L ₓ (0.1 mol / L) =<em> 3.0x10⁻⁴ moles</em> of NaOH = moles of CH₃COOH.
Now, you find the moles of acetic acid in the hot sauce packet. But molarity is the ratio between moles of the acid and liters of solution.
As you don't know the volume of your packet, <em>you can assume its density as 1g/mL. </em>Thus, volume of 0.6g of hot sauce is 0.6mL = 6x10⁻⁴L.
And molarity of the packet is:
3.0x10⁻⁴ moles acetic acid / 6x10⁻⁴L =
<h3>0.5M</h3>
Enzymes affect the rate of the reaction in both the forward and reverse directions; the reaction proceeds faster because less energy is required for molecules to react when they collide. Thus, the rate constant (k) increases. Figure 3: Lowering the Activation Energy of a Reaction by a Catalyst.
<em>Best of luck,</em>
<em>-Squeak</em>