Answer:
1. Hidracidas a. MX
2 Acidas c. MHXO
3. Oxacidas b. MXO
4. Basicas d. M(OH)X
Explanation:
¡Hola!
En este caso, de acuerdo con el concepto de sal, la cual está generalmente dada por la presencia de al menos un metal y un no metal, es posible encontrar cuatro tipos de estas; hidrácidas, oxácidas, básicas y ácidas, en las que las primeras dos son neutras pero la segunda tiene presencia de oxígeno, la tercera tiene iones hidróxido adicionales y la cuarta iones hidrógeno de más.
Debido a la anterior, es posible relacionar cada pareja de la siguiente manera:
1. Hidracidas a. MX
2 Acidas c. MHXO
3. Oxacidas b. MXO
4. Basicas d. M(OH)XO
En las que M se refiere a un metal, X a un no metal, H a hidrógeno y O a oxígeno.
¡Saludos!
<span>1. </span>To solve this we assume
that the gas is an ideal gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant temperature and number of moles of the gas
the product of PV is equal to some constant. At another set of condition of
temperature, the constant is still the same. Calculations are as follows:
P1V1 =P2V2
V2 = P1 x V1 / P2
V2 = 104.1 x 478 / 88.2
<span> V2 =564.17 cm^3</span>
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.
Atomic mass Carbon (C ) = 12.01 a.m.u
12.01 g ---------- 6.02x10²³ atoms
1.50 g ----------- ??
1.50 x ( 6.02x10²³ ) / 12.01 =
7.51x10²² atoms of C
hope this helps!
