Answer:
Va = (MbVb)/Ma
Explanation:
Divide both sides by Ma and voila!
Answer:
177.1 L
Explanation:
The excersise can be solved, by the Ideal Gases Law.
P . V = n . R . T
In first step we need to determine the moles of gas:
We convert T° from, C° to K → 20°C + 273 = 293K
We convert P from mmHg to atm → 760 mmHg = 1atm
1Dm³ = 1L → 190L
We replace: 190 L . 1 atm = n . 0.082 . 293K
(190L.atm) / 0.082 . 293K = 7.91 moles.
We replace equation at STP conditions (1 atm and 273K)
V = (n . R .T) / P
V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L
We can also make a rule of three:
At STP conditions 1 mol of gas occupies 22.4L
Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L
Positively charged and the sodium ion would be a cation
The product of prime polynomials is equivalent to 36x3 – 15x2 – 6x is letter B which is 3x(3x – 2)(4x 1). Below is the solution.
3x(3x - 2) (4x + 1)
= 9x2 - 6x (4x + 1)
= 36x3 + 9x2 + - 24x2 - 6x
= 36x3 - 15x2 - 6x
Answer:

Explanation:
We can use the Ideal Gas Law and solve for T.
pV = nRT
Data
p = 1.25 atm
V = 25.0 L
n = 2.10 mol
R = 0.082 06 L·atm·K⁻¹mol⁻¹
Calculations
1. Temperature in kelvins

2. Temperature in degrees Celsius
