Arteries carry blood away from the heart and veins carry blood back to the heart. Hope that helps (:
<span>Stoichiometry deals with the quantitative measurement of reactants and products in a chemical reaction. Let suppose you are given with following reaction;
A + 2 B </span>→ 3 C
According to this reaction 1 mole of A reacts with 2 moles of B to produce 3 moles of C. Now using the concept of mole one can easily measure the amount of reactants reacted and the amount of product formed, as...
1 Mole Exactly equals 6.022 × 10²³ particles
1 Mole of Gas (at STP) exactly occupies 22.4 L Volume
1 Mole of any compound exactly equals the molar mass in grams
Therefore, <span>Stoichiometry is very helpful in quantitative analysis.</span>
Answer:
a. 1810mL
Explanation:
When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:
where the temperatures must be measured in Kelvin
To convert from Celsius to Kelvin, add 273, or use the equation: 
For this problem, one must also recall that standard temperature is 0°C (or 273K).
So, ![T_1 = 273[K]](https://tex.z-dn.net/?f=T_1%20%3D%20273%5BK%5D)
, and ![T_2 = (49.4+273)[K]=322.4[K]](https://tex.z-dn.net/?f=T_2%20%3D%20%2849.4%2B273%29%5BK%5D%3D322.4%5BK%5D)
.
![\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5D%29%7D%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%29%7D)
![\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%29%7D%28322.4%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%20%29%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29%7D%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29)
![1810.04571428[mL]=V_2](https://tex.z-dn.net/?f=1810.04571428%5BmL%5D%3DV_2)
Adjusting for significant figures, this gives ![V_2=1810[mL]](https://tex.z-dn.net/?f=V_2%3D1810%5BmL%5D)
Answer:
Option B. 2096.1 K
Explanation:
Data obtained from the question include the following:
Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =.?
Entropy is related to enthalphy and temperature by the following equation:
Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)
ΔS = ΔH / T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
ΔS = ΔH / T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K
