1. 12 L = 12 dm³
2. 3.18 g
<h3>Further explanation</h3>
Given
1. Reaction
K₂CO₃+2HNO₃⇒ 2KNO₃+H₂O+CO₂
69 g K₂CO₃
2. 0.03 mol/L Na₂CO₃
Required
1. volume of CO₂
2. mass Na₂CO₃
Solution
1. mol K₂CO₃(MW=138 g/mol) :
= 69 : 138
= 0.5
mol ratio of K₂CO₃ : CO₂ = 1 : 1, so mol CO₂ = 0.5
Assume at RTP(25 C, 1 atm) 1 mol gas = 24 L, so volume CO₂ :
= 0.5 x 24 L
= 12 L
2. M Na₂CO₃ = 0.03 M
Volume = 1 L
mol Na₂CO₃ :
= M x V
= 0.03 x 1
= 0.03 moles
Mass Na₂CO₃(MW=106 g/mol) :
= mol x MW
= 0.03 x 106
= 3.18 g
Cars, aeroplanes, factories etc
Answer:
1.46g of PbCrO₄ are the theoretical yield
Explanation:
Theoretical yield is defined as the maximum amount of products that could be produced (Assuming a yield of 100%).
The reaction of Lead (II) nitrate with sodium chromate is:
Pb(NO₃)₂(aq) + Na₂CrO₄(aq) → PbCrO₄(s) + 2NaNO₃ (aq)
First, we need to find molar mass of each reactant in order to determine limiting reactant (As the reaction is 1:1, the reactant with the lower number of moles is the limiting reactant). The moles of the limiting reactant = moles of Lead (II) chromate (The precipitate):
<em>Moles Pb(NO₃)₂ -Molar mass: 331.21g/mol-</em>
1.50g * (1mol / 331.21g) = 4.53x10⁻³ moles Pb(NO₃)₂
<em>Moles Na₂CrO₄ -Molar mass: 161.98g/mol-</em>
1.75g * (1mol / 161.98g) = 0.0108 moles
Pb(NO₃)₂ is limiting reactant and moles of PbCrO₄ are 4.53x10⁻³ moles. The mass is:
4.53x10⁻³ moles PbCrO₄ * (323.19g / mol) =
<h3>1.46g of PbCrO₄ are the theoretical yield</h3>
Answer:
In alpha decay, shown in Fig. 3-3, the nucleus emits a 4He nucleus, an alpha particle. Alpha decay occurs most often in massive nuclei that have too large a proton to neutron ratio. An alpha particle, with its two protons and two neutrons, is a very stable configuration of particles. Alpha radiation reduces the ratio of protons to neutrons in the parent nucleus, bringing it to a more stable configuration. Many nuclei more massive than lead decay by this method.
Explanation:
The food raises the temperature of the calorimeter by 1.29 °C. The amount of calorie produced by the combustion would be: 1.29°C * <span> 6.15 kJ/°C= 7.9335 kJ.
If the food weight is </span>0.521g, then the food value would be:
7.9335 kJ/ 0.521g / (4.184kcal/kJ) =
(15.227 kJ/g) / (4.184kcal/kJ)= 3.64 kilo calorie/gram
