The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Water could be made to boil lower than its normal boiling point of 100 degrees Celsius at 92 degrees Celsius by lowering the atmosphere or external pressure.
The boiling point is the temperature at which the vapor pressure of liquid equals the external or atmosphere pressure.
So, if you decrease the external pressure the temperature needed for the liquid reach the lower external pressure is also lower..
You can accomplish by taking the water to higher levels in the Earth (atmosphere pressure at high altitudes is lower than at sea level) or by creating vaccum.
The three concerns that the residents of this area might be:
1) <span>The cost of not moving forward is extremely high, so they opposed the plan, as they think it would affect US economy
2) </span><span>Nuclear waste disposal capability is an environmental imperative, so their environment would be polluted by very radioactive materials.
3) </span>Demand for new nuclear plants also demands disposal capability which supports national security but again, their site will be no longer for them. But unfortunately, <span>Extensive studies consistently show Yucca Mountain to be a sound site for nuclear waste disposal so the plan can't be abolished.
Hope this helps!</span>
Answer:
The answer is 1, 2, 3 for the first question and 1,2,3 again for the second question.
Explanation:
The true statements about complex IV are:
1) This complex catalyzes the reduction of oxygen to water.
2) Despite consuming protons, this complex pumps H+ across the membrane at a rate of 4 per molecule of O2 reduced.
3) The protons flow across a gradient to the inner membrane space.
The true statemennts about complex IV are:
1) This complex catalyzes the reduction of oxygen to water.
2) Despite consuming protons, this complex pumps H+ across the membrane at a rate of 4 per molecule of O2 reduced.
3) The protons flow across a gradient to the inner membrane space.
Answer:
5.47 x 10^23 atoms in Ag (sliver)
Explanation:
0.909 mol of Ag x 6.02 x 10^23 atoms of Ag/ 1 mol of Ag
