Answer:
, 
, 
, 
Explanation:
The four most abundant minerals in the continental crust are, , , , .
The percentages are
is called quartz
is called corundum
is called calcium oxide or quick lime
is called magnesium oxide or magnesia
Answer:
44Kj
Explanation:
These are the equations for the reaction described in the question,
Vaporization which can be defined as transition of substance from liquid phase to vapor
H2(g)+ 1/2 O2(g) ------>H2O(g). Δ H
-241.8kj -------eqn(1)
H2(g)+ 1/2 O2(g) ------>H2O(l).
Δ H =285.8kj ---------eqn(2)
But from the second equation we can see that it moves from gas to liquid, we we rewrite the equation for vaporization of water as
H2O(l) ------>>H2O(g)---------------eqn(3)
But the equation from eqn(2) the eqn does go with vaporization so we can re- write as
H2O ------> H2(g)+ 1/2 O2(g)
Δ H= 285.8kj ---------------eqn(4)
To find Delta h of the vaporization of water at these conditions, we sum up eqn(1) and eqn(4)
Δ H=285.8kj +(-241.8kj)= 44kj
Ok so this is what we know :
2KClO3 -> 2KCl + 3O2 (Always check if equation is balanced - in this case it is)
4.26moles
So we know that we have 4.26 moles of oxygen (O2). Now lets look at the ratio between KClO3 and O2.
We see that the ratio is 2:3 meaning that we need 2KClO3 in order to produce 3O2.
Therefore divide 4.26 by 3 and then multiply by 2.
4.26/3 = 1.42
1.42 * 2 = 2.84
Now we know that the molarity of KClO3 is 2.84 moles.
Multiply by R.M.M to find how many grams of KClO3 we have.
R.M.M of KClO3
K- 39
Cl- 35.5
3O- 3 * 16 -> 48
---------------------------
<span>122.5
</span>2.84 * 122.5 = 347.9 grams therefore the answer is (a)
348 grams needed of KClO3 to produce 4.26 moles of O2.
Hope this helps :).
Answer:
Use a ratio of 0.44 mol lactate to 1 mol of lactic acid
Explanation:
John could prepare a lactate buffer.
He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\3.5 = 3.86 + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\\log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 3.5 - 3.86 = -0.36\\\\\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 10^{-0.36} = \mathbf{0.44}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C3.5%20%3D%203.86%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%203.5%20-%203.86%20%3D%20-0.36%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%2010%5E%7B-0.36%7D%20%3D%20%5Cmathbf%7B0.44%7D)
He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.
For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.
Answer:
The description including its given problem is outlined in the following section on the clarification.
Explanation:
The given values are:
RBCC = 0.12584 nm
RFCC = 0.12894 nm
The unit cell edge length (ABCC) as well as the atomic radius (RBcc) respectively connected as measures for BCC (α-phase) structure:
√3 ABCC = 4RBCC
⇒ ABCC = 
⇒ = 
⇒ = 
Likewise AFCC as well as RFCC are interconnected by
√2AFCC = 4RFCC
⇒ AFCC = 
⇒ = 
⇒ = 
Now,
The Change in Percent Volume,
= 
= 
= 
= 
= 
Note: percent = %
