Explanation:
Below is an attachment containing the solution.
Answer:
V = 1.434 L
Explanation:
Given data:
Mass of argon = 4.24 g
Temperature = 58.2 °C
Pressure = 1528 torr
Volume = ?
Solution:
58.2 °C = 58.2 + 273 = 331.2 K
1528/760= 2.01 atm
<em>Number of moles:</em>
Number of moles = mass/molar mass
Number of moles = 4.24 g / 39.948 g/mol
Number of moles = 0.106 mol
<em>Volume:</em>
PV = nRT
V = nRT/P
V = 0.106 mol ×0.0821. atm. L. mol⁻¹. K⁻¹ × 331.2K/ 2.01 atm
V = 2.88 atm L/ 2.01 atm
V = 1.434 L
Answer:
4Ba(CO3) -> 4BaO2 + 2CO2
Explanation:
I looked at the oxygens to balance this. Ba(CO3) normally has 3 oxygens. BaO2 and CO2 have 4 oxygens total. The common multiple of 3 & 4 is 12. So there should be 12 oxygens on both sides. Then I just found the coefficients that would give 12 oxygens on both sides and can balance the rest of the atoms.
Answer: I2 is the Oxidant; while the 2S2O3(-2) is the reductant.
Explanation:
An Oxidant is any substance that oxidizes, or receives electrons from, another; in so doing, it becomes reduced in oxidation number.
A Reductant thus exactly the opposite.
Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:
I2 --> 2I-.
The oxidation number reduced from 0 to -1.
In contrast, the oxidation number of 2S2O3(-2) increases from -4 to -2.
Thus, I2 is the Oxidant; while the 2S2O3(-2) is the reductant.
The mass percent of sulfurous acid in the new solution : 38.9%
<h3>Further explanation</h3>
<em>In a container you have 800 g of a 35% by mass solution of sulfurous acid, from which 80 ml of water evaporates. What is the mass percent of sulfurous acid in the new solution? data: density of water is 1g / ml.</em>
<em />
solution 1
composition :
solution 2(new solution)
composition :
- Total mass of new solution after water evaporated
- %mass of acid in a new solution
