Answer:
The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L
Explanation:
Given data:
Number of moles of HF = 6.62×10⁻³ mol
Volume of HF in litter at STP = ?
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
Standard temperature = 273 K
Standard pressure = 1 atm
Now we will put the values in formula.
1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K
V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K / 1 atm
V = 148.38×10⁻³ L
Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.
Answer:
100 j
Explanation:
bascially u j subtract 50 from 150 because the system lost 50 j but theres 150 in its surroundings
The metric prefixes which is not one multiple of ten when moving forward or backward between prefixes is: none of the above
Option c is the correct answer
<h3>What is metric prefix?</h3>
A metric prefix simply refers to a unit prefix that precedes a basic unit of measure to indicate a multiple or sub-multiple of the unit.
So therefore, the metric prefixes which is not one multiple of ten when moving forward or backward between prefixes is: none of the above
Complete question:
Which of the following metric prefixes are not one multiple of ten when moving forward or backward between prefixes?
a. micro
b.nano
c. none of the above
Learn more about metric prefix:
brainly.com/question/2050691
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The final gas pressure : 175.53 atm
<h3>Further explanation</h3>
Maybe the complete question is like this :
A ridged steel tank filled with 62.7 l of nitrogen gas at 85.0 atm and 19 °C is heated to 330 °C while the volume remains constant. what is the final gas pressure?
The volume remains constant⇒Gay Lussac's Law
<em>When the volume is not changed, the gas pressure in the tube is proportional to its absolute temperature </em>
P₁=85 atm
T₁=19+273=292 K
T₂=330+273=603 K
