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INDIVIDUAL WORK-15
For students all specialities
MODUL 2
Among the given elements has smallest electronegativity: | |
a | Cesium |
b | Fluorine |
c | Silicon |
d | Sodium |
f | Iodine |
0,49 g of hydroxide of twovalent metal reacted with sulfuric acid, as a result was formed 0,8 g of salt. Formula of the hydroxide is: | |
a | Co(OH)2 |
b | Mg(OH)2 |
c | Cu(OH)2 |
d | Fe(OH)2 |
f | Ca(OH)2 |
In a subgroup of alkaline metals cesium has smallest electronegativity, as it has: | |||
a | Valent electrons comparetively most remote from nucleus | ||
b | Greatest number of valent electrons compared with other elements | ||
c | Greatest atomic mass | ||
d | Greatest number of neutrons in the nucleus | ||
| |||
How many valent electrons has an element if its electronic structure looks like 1 s 22 s 22 p 63 s 23 p 4: | |
a | |
b | |
c | |
d | |
f |
In a highest oxide of an element a ratio between atoms of oxygen and the element makes 1:1. How many valent р -orbitals are filled with electrons in the atom of this element: | |
a | |
b | |
c | |
d |
On a basis of what laws has been made a sequence of elements: carbon, nitrogen, oxygen, fluorine: | |
a | The elements are located in ascending order of atomic mass |
b | Elements of one group |
c | All belong to metals |
d | All belong to nonmetals |
f | Elements of one period |
From the given compounds the formation of hydrogen bonds is possible for: | |
a | Н2О |
b | СН3ОН |
c | CF4 |
d | НІ |
f | HCl |
Find the charge of complex ion, charge, covalency and coordination number of the central atom in potassium tetrahydroxozincate: | |
a | 2–, 2+, 2, 2 |
b | 2+, 2+, 4, 4 |
c | 2–, 2+, 2, 4 |
d | 2–, 2+, 4, 2 |
f | 2–, 2+, 4, 4 |
Concentration of ions Cl– in 0,01 M solution of the complex compound [Co(NH3)4Cl2]Cl3 (first instability constant is К1 = 4×10–5) is: | |
a | mol/l |
b | mol/l |
c | mol/l |
d | 0,03 mol/l |
f | 0,01 mol/l |
Enthalpy of methane formation, found by using of the following thermochemical equations Н2(g)+1/2О2(g) = Н2О(liq); DН = –285,84 kJ С(solid) + О2(g)= СО2(g); DН = –393,51 kJ СН4(g) + 2О2(g) = 2Н2О(liq) + СО2(g); DН = –890,31 kJ, is equal to: | |
a | –213,5 kJ |
b | +159,76 kJ |
c | –679,35 kJ |
d | –74,88 kJ |
f | +56,95 kJ |
For realization of reaction have taken a certain quantity of substances. The reaction ran during 30 minutes, and the measurement of its rate was carried out every 5 minutes. How will be changed the reaction rate during its course: | |
a | Will pass through a maximum |
b | Decrease |
c | Will pass through a minimum |
d | Will not change |
f | Will increase |
In a closed vessel runs homogeneous reaction 2CO + O2 = 2CO2. The initial concentrations of reactants were equal c(CO) = 0,5 mol/l, c(O2) = 0,2 mol/l. Find equilibrium constant of reaction and equilibrium concentrations of all components, if at some temperature in the reaction had entered 20 % of carbon (ІІ) oxide: | |
a | Кр= 0,416, [CO] = 0,4 mol/l, [O2] = 0,15 mol/l, [CO2] = 0,1 mol/l |
b | Кр= 0,516, [CO] = 0,5 mol/l, [O2] = 0,3 mol/l, [2CO2] = 0,2 mol/l |
c | Кр= 0,616, [CO] = 0,6 mol/l, [O2] = 0,4 mol/l, [2CO2] = 0,3 mol/l |
d | Кр= 0,316, [CO] = 0,3 mol/l, [O2] = 0,2 mol/l, [2CO2] = 0,15 mol/l |
f | Кр= 0,208, [CO] = 0,2 mol/l, [O2] = 0,075 mol/l, [2CO2] = 0,05 mol/l |
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