Aluminum is a substance because it is a particular type of matter that has particular and unique properties. aluminum is a unique element with a place on the periodic table
Answer:
Determining the Slope on a p-t Graph. It was learned earlier in Lesson 3 that the slope of the line on a position versus time graph is equal to the velocity of the object. ... If the object has a velocity of 0 m/s, then the slope of the line will be 0 m/s. The slope of the line on a position versus time graph tells it all.
Explanation:
#<em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em>o</em><em>n</em><em>l</em><em>e</em><em>a</em><em>r</em><em>n</em><em>i</em><em>n</em><em>g</em><em> </em>
Answer:
3.5 atm
Explanation:
As stated in the question pressure is required to counteract the natural tendency for water to dilute the more concentrated solution. The difference in concentrations will give us the answer using the osmotic pressure equation.
π = ( n/v) RT where n/v is the molarity (mol/L), R is the gas constant and T is the temperature.
The difference in osmotic pressure of the solutions is:
Δπ = Δ c RT where c is the difference in molar concentrations.
pressure required = Δπ = (0.190 - 0.048) M x 0.821 Latm/Kmol x 298 K
= 3.47 atm
Density= mass/volume
= 100/25
density = 4g/ml
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>
