Answer:
<h2>Heterogeneous</h2>
Explanation:
<h3><em>Milk </em><em>seems</em><em> to</em><em> be</em><em> </em><em>homogeneous</em><em> mixture</em><em> </em><em>but </em><em>actually</em><em> </em><em>milk </em><em>is </em><em>a </em><em>heterogeneous</em><em> </em><em>mixture</em><em> </em><em>and </em><em>a </em><em>colloid</em><em> </em><em>solution</em><em>.</em></h3>
Answer:
Henry Moseley
Explanation:
Dmitry Mendeleef and Lothar Meyer proposed a periodic table based on the atomic mass.
They stated a periodic law expressed as "chemical properties of elements are a periodic function of their atomic weights".
But, Henry Moseley in 1900s re-stated periodic law by changing the basis of the law from atomic weight to atomic number.
The present periodic law is stated as "the properties of elements are a periodic function of their atomic number".
The
equation for the photosynthesis reaction in which carbon dioxide and water
react to form glucose is .
The hear reaction is the difference between the bond dissociation energies in
the products and the bond dissociation energies of the reactants
The
reactant molecules have 12 C = O, 12 H - O bonds while the product molecules
have 5 C - C, 7 C – O, 5 H – O, and 6 O = O bonds. The average bond
dissociation energies for the bonds involved in the reaction are 191 for C = O,
112 for H – O, 83 C –C, 99 C – H, 86 C – O, 119 O = O.
Substitute
the average bond dissociation energies in the equation for and
calculate as follows
=
[12 (C=O) + 12 (H-O)] – [5(C-C) + 7(C-H) + 7 (C-O) + 5(H-O) + 6(O=O)]
=
[12x191 kcal/mol + 12x112 kcal//mol] – [5x83 kcal/mol + 7x99 kcal/mol + 7x86
kcal/mol + 5x112 kcal/mol + 6x119 kcal/mol]
=
3636 kcal/mol – 2984 kcal/mol = 652 kcal/mol x 4.184 Kj/1kcal = 2.73x10^3 kJ/mol
So,
enthalpy change for the reaction is 652 kcal/mol or 2.73x10^3 kJ/mol
<span> </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)
Instability of an atoms nucleus can result from an excess of either neutrons or protons . So neutrons and protons .
