RbOH is a strong base that dissociates completely and HCl is a strong acid that too dissociates completely. the complete reaction between the acid and base is;
RbOH + HCl ---> RbCl + H₂O
stoichiometry of acid to base is 1:1
At neutralisation point
H⁺ mol = OH⁻ mol
mol = molarity x volume
if Ma - molarity of acid and Va - volume of acid reacted
Mb - molarity of base and Vb - volume of base reacted
Ma x Va = Mb x Vb
0.5 M x 52.8 mL = Mb x 60.0 mL
Mb = 0.44 M
molarity of base - 0.44 M
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
Answer:
1. It depends what type of method you are using. if it is Height x Width x Length then it will not work for an irregular shape because it has extra pieces that would not be included.
2. The second method would work for both regular and irregular shapes because you would have to know or find out the volume of the regular shape to get the volume for the irregular shape.
3. It also depends on what you are doing, if you are doing a regular shape then use the first method, if it's an irregular shape then use the second method, if you do the maths correctly both should give you an accurate answer for what you want to achieve.
4. No, because the sugar would dissolve.
5. No, on this case the displacement method would not work because of the weight difference
Explanation:
All the answers for you!
Answer:
40.94 g
Explanation:
Given data:
Mass of NO₂ = ?
Volume = 20.0 L
Pressure = 110.0 Pka
Temperature = 25°C
Solution:
Pressure = 110.0 KPa (110/101 = 1.1 atm)
Temperature = 25°C (25+273 = 298.15 K)
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
n = PV/RT
n = 1.1 atm × 20.0 L / 0.0821 atm.L/ mol.K ×298.15 K
n = 22 / 24.5 /mol
n= 0.89 mol
Mass of NO₂:
Mass = number of moles × molar mass
Mass = 0.89 mol × 46 g/mol
Mass = 40.94 g
The answer is either D or B. hope this helps.
