<h3><u>Answer;</u></h3>
10.80 ° C
<h3><u>Explanation;</u></h3>
From the information given;
Initial temperature of water = 24.85°C
Final temperature of water = 35.65°C
Mass of water = 1000 g
The specific heat of water ,c = 4.184 J/g °C.
The heat capacity of the calorimeter = 695 J/ °C
Change in temperature ΔT = 35.65°C - 24.85°C
= 10.80°C
Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
Answer:
2.14 moles of H₂O₂ are required
Explanation:
Given data:
Number of moles of H₂O₂ required = ?
Number of moles of N₂H₄ available = 1.07 mol
Solution:
Chemical equation:
N₂H₄ + 2H₂O₂ → N₂ + 4H₂O
now we will compare the moles of H₂O₂ and N₂H₄
N₂H₄ : H₂O₂
1 : 2
1.07 : 2×1.07 = 2.14 mol
Answer:
1, 1, 2, 3
Explanation:
The numbers 1 and 8 both have 1 sig. fig.
The number 13 has 2 sig. figs.
The number 104 has 3 sig. figs.
Answer:
Na(s) + C(s, graphite) + 1/2 H₂(g) + 3/2 O₂(g) → NaHCO₃(s)
Explanation:
The standard formation reaction is the synthesis of 1 mole of a substance from its elements in their most stables forms under standard conditions. The balanced chemical equation is:
Na(s) + C(s, graphite) + 1/2 H₂(g) + 3/2 O₂(g) → NaHCO₃(s)
