Earth is the right distance from the sun. It’s protected from harmful solar radiation by its magnetic field and the atmosphere keeps it warm. Earth contains the right chemical ingredients to sustain life such as H2O (water) and C (carbon)
Adding (S2O3)2- would affect the reaction mechanism that involves this ion. From the reaction mechanism given above, the equilibrium of step 2 would be affected. Adding the stock solution of (S2O3)2- would shift the equilibrium to the right thus making more products of the said mechanism. Also, the reaction rate of this step would occur faster than the original rate. This is based on Le Chatelier's Prinicple which states that a corresponding change would happen to the equilibrium of a reaction when pressure, concentration of the substances or temperature is changed. So, that after the addition, a color change would appear immediately because I3- would be removed slowly from solution, and would therefore be able to react with starch.
Answer:
Part A is just T2 = 58.3 K
Part B ∆U = 10967.6 x C
You can work out C
Part C
Part D
Part E
Part F
Explanation:
P = n (RT/V)
V = (nR/P) T
P1V1 = P2V2
P1/T1 = P2/T2
V1/T1 = V2/T2
P = Pressure(atm)
n = Moles
T = Temperature(K)
V = Volume(L)
R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.
bar = 0.986923 atm
N = 14g/mol
N2 Molar Mass 28g
n = 3.5 mol N2
T1 = 350K
P1 = 1.5 bar = 1.4803845 atm
P2 = 0.25 bar = 0.24673075 atm
Heat Capacity at Constant Volume
Q = nCVΔT
Polyatomic gas: CV = 3R
P = n (RT/V)
0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))
V = (nR/P) T
V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K
V = (0.28721/1.4803845) x 350
V = 0.194 x 350
V = 67.9036 L
So V1 = 67.9036 L
P1V1 = P2V2
1.4803845 atm x 67.9036 L = 0.24673075 x V2
100.52343693 = 0.24673075 x V2
V2 = P1V1/P2
V2 = 100.52343693/0.24673075
V2 = 407.4216 L
P1/T1 = P2/T2
1.4803845 atm / 350 K = 0.24673075 atm / T2
0.00422967 = 0.24673075 /T2
T2 = 0.24673075/0.00422967
T2 = 58.3 K
∆U= nC
∆T
Polyatomic gas: C
= 3R
∆U= nC
∆T
∆U= 28g x C
x (350K - 58.3K)
∆U = 28C
x 291.7
∆U = 10967.6 x C
The density of the rectangular block in g/mL is 7.0.
<u>Given the following data:</u>
- Mass of block = 22.8 gra1.94 kg
- Length of block = 3.21 cm
- Height of block = 1.84 in.
To find the density of the block in g/mL:
First of all, we would determine the volume of the rectangular block by using the following formula:
×
× 
<u>Conversion:</u>
1 in = 2.54 cm
5.83 in = X cm
Cross-multiplying, we have:

×
× 
Volume = 277.16 cubic centimeters.
<u>Note</u>: Milliliter (mL) is the same as cubic centimeters.
1000 grams = 1 kg
Y grams = 1.94 kg
Cross-multiplying, we have:
Y = 1940 grams
Now, we can find the density:

<em>Density </em><em>= 7</em><em>.0 g/mL</em>
Therefore, the density of the rectangular block in g/mL is 7.0.
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