Answer:
1.852 g of CO2 were produced in the chemical reaction
Explanation:
The problem is pretty simple. We can write down the chemical reaction that is involved to help us understand better what is going on in the process:
CaCO3 (aq) + HAc (aq) → CaAc (aq) + CO2 (g) + H2O (l)
Let's think this through: we have a tablet that has an active compound (CaCO3) and an inert substance that weighs 0.833 g. When we add 58.072 g of an acid solution (represented in the equation as HAc because we are not told specifically which acid is being added), CO2 is formed and released from solution as gas leaving us with an aqueous solution that weighs 57.053 g.
Having said that, we know that the only mass lost during the reaction is due to the formation of CO2 gas. Therefore, we sum the reactants (the tablet + the acid solution) and subtract the mass of the remnant solution. This value will indicate us the amount of CO2 formed:
0.833 g of the Tablet + 58.072 g from the Acid solution = 58.905 g
58.905 g of reactants - 57.053 g of remnant solution = 1.852 g of produced CO2.
Surplus cause the other ones aren’t in english
Q=mc(deltaT)
Q is the amount of energy which you are looking for
M is the mass which you can find
C is the specific heat of water which is 4.18 J/gC
DeltaT is the change in temperature which you can find.
To find the mass, first you must know that the density of water is 1g/mL, meaning that 200 mL has a mass of 200 g. This means that to find the total mass (m in the equation) all you need to do is add the mass of water and NaOH.
200 g + 2.535 g=202.535 g.
To find deltaT you would need to take the final temperature minus the initial temperature.
27.8C-24.2C=3.6C
Then these values can be substituted into the equation:
q=(202.635g)(4.18J/gC)(3.6C)
Q=3049.25 J
Technically this should be rounded off to 1 significant figure (200 mL only had 1), but ignoring signficiant figure rules this should be correct. Also, sometimes other units like calories or kJ may be asked for, meaning that a conversion or alternate c value would be used.
Answer:
ΔT = 20.06 °C
Explanation:
The equation used for this problem is as follow,
Q = m Cp ΔT ----- (1)
Where;
Q = Heat = 1.17 kJ = 1170 J
m = mass = 24.1 g
Cp = Specific Heat Capacity = 2.42 J.g⁻¹.°C⁻¹
ΔT = Change in Temperature = <u>??</u>
Solving eq. 1 for ΔT,
ΔT = Q / m Cp
Putting values,
ΔT = 1170 J / 24.1 g × 2.42 J.g⁻¹.°C⁻¹
ΔT = 20.06 °C
Answer:
Moles of NO₂ = 0.158
Explanation:
SO 2 ( g ) + NO 2 ( g ) ⇄ SO 3 ( g ) + NO ( g )
According to the law of mass equation
= ![\frac{[SO_{3} ][NO]}{[SO_{2}][NO_{2} ]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSO_%7B3%7D%20%5D%5BNO%5D%7D%7B%5BSO_%7B2%7D%5D%5BNO_%7B2%7D%20%20%5D%7D)
⇒ 3.10 = ![\frac{(1.00)(1.00)}{(2.30) [NO_{2} ]}](https://tex.z-dn.net/?f=%5Cfrac%7B%281.00%29%281.00%29%7D%7B%282.30%29%20%5BNO_%7B2%7D%20%5D%7D)
At equilibrium [SO₃] = [NO]
⇒ [NO₂] = 
⇒ [NO₂] = 0.158
So. number of moles of NO₂ at equilibrium added = 0.158
