Answer:
n = 2.58 mol
Explanation:
Given data:
Number of moles of argon = ?
Volume occupy = 58 L
Temperature = 273.15 K
Pressure = 1 atm
Solution:
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
1 atm × 58 L = n × 0.0821 atm.L/ mol.K × 273.15 K
58 atm.L = n × 22.43 atm.L/ mol.
n = 58 atm.L / 22.43 atm.L/ mol
n = 2.58 mol
Answer:
0.125. work- divide the volume value by 1000
873. work- multiply the length value by 100
98100. work- Conversion factor: 1 kg = 100000 cg
1) Centigram = Kilogram * 100000
2) Centigram = 0.981 * 100000
3) Centigram = 98100
285.65. work- 12.5°C + 273.15 = 285.65K
446.85. work- 720K − 273.15 = 446.85°C
346.25. work- 73.1°C + 273.15 = 346.25K
Explanation:
i hope this helps:) brainliest plss??
The complete balanced chemical
equation is:
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
In statement form: 4mol NH3 reacts with 5 mol O2 to produce 6
mol H2O
First let us find for the limiting reactant:
>molar mass NH3 = 17 g/mol
moles NH3 = 54/17 = 3.18 mol NH3
This will react with 3.18*5/4 = 3.97 mol O2
>molar mass O2 = 32g/mol
moles O2 = 54/32 = 1.69 mol O2
We have insufficient O2 therefore this is the limiting
reactant
From the balanced equation:
For every 5.0 mol O2, we get 6.0 mol H2O, therefore
moles H2O formed = 1.69
mol O2 * 6/5 = 2.025 mol
Molar mass H2O = 18g/mol
<span>mass H2O formed = 2.025*18 = 36.45 grams H2O produced</span>
Answer: A and D, I believe
Explanation:
