The new pressure of the gas is calculated to be 40.7 kPa. Using the principle that P1 • V1 = P2 • V2, we can set 98.8 kPa (P1) multiplied by 21.7 mL (V1) equal to P2 (unknown pressure) multiplied by 52.7 mL (V2). To isolate P2, we rearrange the equation to P2 = (98.8 kPa • 21.7 mL) / 52.7 mL, resulting in P2 equal to 40.7 kPa.
Answer:
0.31%
Explanation:
For the chemical reaction:
I₂ + 2 S₂O₃²⁻ → 2 I⁻ + S₄O₆²⁻
0.043 L multiplied by 0.117 M of sodium thiosulfate gives 5.031x10⁻³ moles of S₂O₃²⁻
5.031x10⁻³ moles of S₂O₃²⁻ produces:
ClO⁻⁻ + 2 H⁺ + 2 I⁻ → I₂ + Cl⁻ + H2O
2.5156x10⁻³ moles of I₂ equates to moles of NaClO
2.5156x10⁻³ moles of NaClO times 
yields 0.187 g of NaClO
Thus, the mass percentage composition is:
= 0.31%
I hope this helps!
First, we need to identify the half-reaction for magnesium. It can be represented as:
Mg2+ + 2e- = Mg
Next, we will determine the overall charge generated during the electrolysis using the information derived from the half-reaction. The calculation follows:
4.50 kg Mg (1000 g / 1 kg) (1 mol / 24.305 g) (2 mol e- / 1 mol Mg) (96500 C / 1 mol e-) = 35733388.2 C
The provided EMF is given in voltage. Since 1 V equals J/C, 5 V translates to 5 J/C.
Therefore, 35733388.2 C (5 J/C) = 178666941 J
Finally, 178666941 J (1 kW-h / 3.6x10^6 J) = 49.63 kW-h
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The calculation of moles of chromium (III) nitrate produced is done as follows. First, you write the reaction equation: 3 Pb(NO3)2 + 2 Cr = 2 Cr(NO3)3 + 3 Pb. Then, by using the mole ratio from Pb(NO3)2 to Cr(NO3)3, which is 3 to 2, you can find the moles of Cr(NO3)3. Thus, for 0.85 moles of lead (IV) nitrate, it equates to 0.85 x 2 / 3 = 0.57 moles.
