Balanced equation for the reaction between ammonia and sulphuric acid is:
Explanation:
2NH3+H2SO4→(NH4)2SO4
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Answer:
Lines of latitude run from east to west.
Explanation:
Answer:
G<0, spontanteous
H<0, from equation
S>0, gas to solid
Explanation:
The small bags of silica gel you often see in a new shoe box are placed there to control humidity. Despite its name, silica gel is a solid. It is a chemically inert, highly porous, amorphous form of SiO2. Water vapor readily adsorbs onto the surface of silica gel, so it acts as a desiccant. Despite not knowing mechanistic details of the adsorption of water onto silica gel, from the information provided, you should be able to make an educated guess about the thermodynamic characteristics of the process. Predict the signs of ΔG, ΔH, and ΔS.
G<0, spontanteous
H<0, from equation
S>0, gas to solid
Answer:
50
Explanation:
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
Mᵣ: 30.01 32.00 46.01
2NO + O₂ ⟶ 2NO₂
Mass/g: 80.00 16.00
2. Calculate the moles of each reactant

3. Calculate the moles of NO₂ we can obtain from each reactant
From NO:
The molar ratio is 2 mol NO₂:2 mol NO

From O₂:
The molar ratio is 2 mol NO₂:1 mol O₂

4. Identify the limiting and excess reactants
The limiting reactant is O₂ because it gives the smaller amount of NO₂.
The excess reactant is NO.
5. Mass of excess reactant
(a) Moles of NO reacted
The molar ratio is 2 mol NO:1 mol O₂

(b) Mass of NO reacted

(c) Mass of NO remaining
Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO
42.34 g of water could be warmed from 21.4°C to 43.4°C by the pellet dropped inside it
Heat loss by the pellet is equal to the Heat gained by the water.
….(1)
where,
is the heat gained by water
is the heat loss by pellet
= mCΔT
where m = mass of water
C = specific heat capacity of water = 4.184 J/g-°C
ΔT = Increase in temperature
ΔT for water = 43.4 - 21.4 = 22°C
= m × 4.184 × 22 …. (2)
Now
=
×ΔT
where
= Heat capacity of pellet = 56J/°C
Δ T for pellet = 43.4 - 113 =- 69.6°C
= 56 × -69.6 = -3897.6 J
From equation (1) and (2)
-m× 4.184 × 22 =-3897.6
m= 42.34 g
Hence, 42.34 g of water could be warmed from 21.4 degrees Celsius to 43.4 degrees Celsius by the pellet dropped inside it.
Learn more about specific heat here brainly.com/question/16559442
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