Considering the ideal gas law, the volume of gas produced at 25.0 °C and 1.50 atm is 184.899 L.
<h3>Definition of ideal gas</h3>
An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
<h3>Ideal gas law</h3>
An ideal gas is characterized by absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:
P×V = n×R×T
<h3>Volume of gas</h3>
In this case, you know:
- P= 1.50 atm
- V= ?
- n= 500 g×
= 11.36 moles, being 44 
the molar mass of CO₂ - R= 0.082
- T= 25 C= 298 K (being 0 C=273 K)
Replacing in the ideal gas law:
1.50 atm×V = 11.36 moles×0.082 × 298 K
Solving:
V= (11.36 moles×0.082 × 298 K) ÷ 1.50 atm
<u><em>V= 184.899 L</em></u>
Finally, the volume of gas produced at 25.0 °C and 1.50 atm is 184.899 L.
Learn more about the ideal gas law:
<u>brainly.com/question/4147359?referrer=searchResults</u>
Answer:
1223.38 mmHg
Explanation:
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 
Also,
Moles = mass (m) / Molar mass (M)
Density (d) = Mass (m) / Volume (V)
So, the ideal gas equation can be written as:
Given that:-
d = 1.80 g/L
Temperature = 32 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (32 + 273.15) K = 305.15 K
Molar mass of nitrogen gas = 28 g/mol
Applying the equation as:
P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K
⇒P = 1223.38 mmHg
<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>
A)
NH⁴⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃0⁺<span>(aq)
- acid </span>a species that able to donate (H+): NH⁴⁺
- base a species that is able to accept a proton (H+): H₂O
- conjugate base a species formed when acid donates a proton (H+): NH₃
- conjugate acid a species formed by a base accepts a proton (H+): H₃0⁺
b)
CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq)
- base a species that is able to accept a proton (H+): CN⁻
- acid a species that able to donate (H+): H₂O
- conjugate acid a species formed by a base accepts a proton (H+): HCN
- conjugate base a species formed when acid donates a proton (H+): OH⁻
