Answer:
<u>1.4 M</u>
Explanation:
n(HBr)=3.115/81
so, 3.115/81=0.0385mol
according to the reaction, n(HBr)=n(KOH)=0.038 mol
C(KOH)=n/V=0.0385/0.02745
0.0385/0.02745 =1.4 M
in short the answer is 1.4 M (molarity)
Answer:
0.6749 M is the concentration of B after 50 minutes.
Explanation:
A → B
Half life of the reaction = 
Rate constant of the reaction = k
For first order reaction, half life and half life are related by:


Initial concentration of A = ![[A]_o=0.900 M](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.900%20M)
Final concentration of A after 50 minutes = ![[A]=?](https://tex.z-dn.net/?f=%5BA%5D%3D%3F)
t = 50 minute
![[A]=[A]_o\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_o%5Ctimes%20e%5E%7B-kt%7D)
![[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}](https://tex.z-dn.net/?f=%5BA%5D%3D0.900%20M%5Ctimes%20e%5E%7B-0.02772%20min%5E%7B-1%7D%5Ctimes%2050%20minutes%7D)
[A] = 0.2251 M
The concentration of A after 50 minutes = 0.2251 M
The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M
0.6749 M is the concentration of B after 50 minutes.
Answer:
a) 26.98
b) 3
c) 3* 1.01 = 3.03
d) 3 *16 = 48
e) 26.98 + 3.03 + 48 = 78.01
f) 6.023 * 10²³
Answer:
The classification and illustrations are attached in the drawing.
Explanation:
It is possible to identify the pure substance observing the figure, since it is the only one that has 2 joined atoms (purple and blue) which forms a single compound.
On the other hand, the homogeneous mixture is identified by noting that its atoms are more united with respect to the heterogeneous mixture, highlighting that in homogenous mixtures the atoms, elements or substances are not visible to the naked eye and are in a single phase, instead in the heterogeneous mixture if they can be differentiated.
Answer:
Si las condiciones para que el magma permanezca líquido no perduran, el magma se enfriará y solidificará en una roca ígnea. Una roca que se enfría en el interior de la Tierra se denomina intrusiva o plutónica y su enfriamiento será muy lento, produciendo una estructura cristalina de granos grueso.
Explanation: