Answer:
The outside temperature is -45.8°C
Explanation:
When a gas keeps on constant its moles and its pressure, we can assume that volume will be increased or decreased as the T° (absolute T° in K).
V1 / T1 = V2 / T2
2.95L/298K = 2.25L / T2
(2.95L/298K ) . T2 = 2.25L
T2 = 2.25L . 298K / 2.95L
T2 = 227.2K
T°K - 273 = T°C
227.2K - 273 = -45.8°C
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I think c cause if he is in a relationship that’s abusive he needs to speak up before it goes way worst.
Answer:
237.2 mL.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
(XMV) acid = (XMV) base.
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HCl = (XMV) NaOH.</em>
<em></em>
For HCl; X = 1, M = 0.5 M, V = ??? mL.
For NaOH, X = 1, M = 1.54 M, V = 77.0 mL.
<em>∴ V of HCl = (XMV) NaOH / (XV) HCl = (</em>1)(1.54 M)(77.0 mL) / (1)(0.5 M) = <em>237.2 mL.</em>
C. Decreasing the temperature
D. Raising the pressure
<h3>Further explanation</h3>
Given
Reaction
2SO₂+O₂⇔2SO₃+energy
Required
Changes to the formation of products
Solution
The formation of SO₃ is an exothermic reaction (releases heat)
If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)
While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient
in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2
As the temperature decreases, the equilibrium will shift towards the exothermic reaction, so the reaction shifts to the right towards SO₃( products-favored)
And increasing the pressure, then the reaction shifts to the right SO₃( products-favored)⇒the number of coefficients is greater
