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jasenka [17]
4 years ago
13

. If a burning candle is covered by a large glass beaker, the flame will go out. How does the beaker affect the conditions descr

ibed by the fire triangle?
A. It removes oxygen from the flame.
B. It limits the heat available for the flame to continue to burn under the glass beaker.
C. It prevents more oxygen from reaching the flame once the existing oxygen is used up.

D. It limits the fuel available to the flame
Chemistry
1 answer:
Dima020 [189]4 years ago
3 0
It is a it removes oxygen 
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A(1,2),B(1,4),C(6,2),
KIM [24]

Answer:

????

Explanation:

where's the question

4 0
3 years ago
What assumptions do we make in order to use the Henderson-Hasselbalch equation? a. Both the weak acid and its conjugate base are
zepelin [54]

Answer:

The final and initial concentration of the acid and it's conjugate base are approximately equal, that is we use the weak acid approximation.

Explanation:

The Henderson-Hasselbalch is used to calculate the pH of a buffer solution. It depends on the weak acid approximation.

Since the weak acid ionizes only to a small extent, then we can say that [HA] ≈ [HA]i

Where [HA] = final concentration of the acid and [HA]i = initial concentration of the acid.

It also follows that [A^-] ≈ [A^-]i where [A^-] and[A^-]i refer to final and initial concentrations of the conjugate base hence the answer above.

7 0
3 years ago
A sample of oxygen gas is compressed from 30.6 L to 1.8 L at constant temperature pressure of 1.8 atm. Calculate the amount of e
ad-work [718]

Answer:

the change in the internal energy of the system is 3,752.67 J

Explanation:

Given;

initial volume of the gas, V₁ = 30.6 L

final volume of the gas, V₂ = 1.8 L

constant pressure of the gas, P = 1.8 atm

Energy released by the system, Q = 1.5 kJ = 1,500 J

Apply pressure-volume work equation, to determine the work done on the gas;

w = -PΔV

w = -P(V₂ - V₁)

w = - 1.8 atm(1.8 L - 30.6 L)

w = 51.84 L.atm

w = 51.84 L.atm x 101.325 J/L.atm

w = 5,252.67 J

The change in the internal energy of the system is calculated as;

ΔU = Q + w

Since the heat is given out, Q = - 1,500 J

ΔU = -1,500 J  +  5,252.67 J

ΔU = 3,752.67 J

Therefore, the change in the internal energy of the system is 3,752.67 J

3 0
3 years ago
The following reaction is at equilibrium in a sealed container.
ale4655 [162]

Answer:

D.Lowering the temperature is the best option.

Explanation:

The value of equilibrium constants aren't changed with change in the pressure or concentrations of reactants and products in equilibrium. The only thing that changes the value of equilibrium constant is a change of temperature.

In the reaction below for example;

A + B <==>C+D

If you have moved the position of the equilibrium to the right (and so increased the amount of C and D), why hasn't the equilibrium constant increased?

Let's assume that the equilibrium constant mustn't change if you decrease the concentration of C - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does?

If you decrease the concentration or pressure of C, the top of the Kc expression gets smaller. That would change the value of Kc. In order for that not to happen, the concentrations of C and D will have to increase again, and those of A and B must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before.

3 0
3 years ago
3. A 4.00 gram sample of solid gold was heated from 274K to 314K. How much energy was involved?
MrMuchimi
Q = mcΔT = (4.00 g)(0.129 J/g•°C)(40.85 °C - 0.85 °C)
Q = 20.6 J of energy was involved (more specifically, 20.6 J of heat energy was absorbed from the surroundings by the sample of solid gold).
3 0
3 years ago
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