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!
Answer:
The Earth's surface is constantly changing through forces in nature. The daily processes of precipitation, wind and land movement result in changes to landforms over a long period of time. Driving forces include erosion, volcanoes and earthquakes. People also contribute to changes in the appearance of land.
Answer:
Why is copper used for most electrical wiring? All metals have some amount of resistivity to electrical currents, which is why they require a power source to push the current through. The lower the level of resistivity, the more electrical conductivity a metal has
Answer:
12 Ethene gas, CH is completely burned in excess oxygen to form carbon dioxide and water
The equation for this exothermic reaction is shown.
CH 30, - 200, 2H,0
The table shows the bond energies involved in the reaction
bond
bond energy
(kJ/moly
614
413
C-C
CHH
0 0
СО
495
799
O-H
467
What is the total energy change in this reaction?
A-954 kJ/mol
B-1010 kJ/mol
C-1313 kJ/mol
D-1359 kJ/mol
Explanation:
thats all you said
Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol
