A 0.00143 M concentration of MnO4^- is not a reasonable solution .
<h3>Number of moles of carbonate</h3>
The ions left in solution are Na^+ and NO3^-
Number of moles of calcium nitrate = 100/1000 L × 1 = 0.1 moles
Since;
1 mole of sodium carbonate reacts with 1 mole of calcium nitrate then 0.1 moles of sodium carbonate were used.
<h3>Conductivity of filtrate</h3>
The claim of the student that the concentration of sodium carbonate is too low is wrong because the value was calculated from concentration and volume of calcium nitrate and not using the precipitate. If the filtrate is tested for conductivity, it will be found to conduct electricity because it contains sodium and NO3 ions.
2) In the reaction as shown, the MnO4^- ion was reduced.
The initial volume is 3.4 mL while the final volume is 29.6 mL.
Number of moles of MnO4^- ion = (29.6 mL - 3.4 mL)/1000 × 0.0235 M = 0.0006157 moles
<h3>The calculations are performed as follows</h3>
- If 2 moles of MnO4^- reacted with 5 moles of acid
0.0006157 moles of MnO4^- reacted with 0.0006157 moles × 5 moles/ 2 moles
= 0.0015 moles
- In this case, number of moles of acid = 0.139 g/90 g/mol = 0.0015 moles
Number of moles of MnO4^- = 0.00143 M × (29.6 mL - 3.4 mL)/1000
= 0.000037 moles
- If 2 moles of MnO4^- reacts with 5 moles of acid
0.000037 moles of MnO4^- reacts with 0.000037 moles × 5 moles/ 2 moles
= 0.000093 moles
- Hence, this is not a reasonable amount of solution.
Learn more about MnO4^- : brainly.com/question/10887629
<span>CO2 (carbon) is the main product that results from burning paper. The paper is the reactant
When paper burns in a fire, the reactants are mostly carbon (the main substance in the paper) and oxygen (from the air).Co2
If paper is loaded with CaCO3 such as cigarette paper, CaCO3 decompose in to CaO and CO2</span>
Answer:
- 178 ºC
Explanation:
The ideal gas law states that :
PV = nRT,
where P is the pressure, V is the volume, n is number of moles , R is the gas constant and T is the absolute temperature.
For the initial conditions :
P₁ V₁ = n₁ R T₁ (1)
and for the final conditions:
P₂V₂= n₂ R T₂ where n₂ = n₁/2 then P₂ V₂ = n₁/2 T₂ (2)
Assuming V₂ = V₁ and dividing (2) by Eqn (1) :
P₂ V₂ = n₁/2 R T₂ / ( n₁ R T₁) then P₂ / P₁ = 1/2 T₂ / T₁
4.10 atm / 25.7 atm = 1/2 T₂ / 298 K ⇒ T₂ = 0.16 x 298 x 2 = 95.1 K
T₂ = 95 - 273 = - 178 º C
If the cyclist rode at an average speed of 10mph for 15 miles...
we can solve by dividing the distance by the speed to get time using the equation...
Δspeed = Δdistance / Δtime
Δtime = Δdistance / Δspeed
Δtime = 15 miles / 10 mph = 1.5 hours
