<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 number of moles the value of PV/T is equal to some constant. At another
set of condition of temperature, the constant is still the same. Calculations
are as follows:</span>
P1V1/T1 = P2V2/T2
P2 = P1 (V1) (T2) / (T1) (V2)
P2 = 475 kPa (4 m^3) (277 K) / (290 K) (6.5 m^3)
P2 = 279.20 kPa
Therefore, the changes in the temperature and the volume lead to a change in the pressure of the system which is from 475 kPa to 279.20 kPa. So, there is a decrease in the pressure.
Redox
reactions are those in which the oxidation numbers of the elements involved are
changed.
Equation
1:
2Na(s)
+ Cl2(g) --> 2NaCl
The
oxidation numbers of Na and Cl in the reactant side are both zero because they
are in elemental form. In the product side, however, the oxidation numbers are
+1 and -1, respectively. Hence, this is an example of redox reaction.
Equation
2:
Cd(s)
+ Pb+2(aq) --> Cd2+(aq) + Pb(s)
The
oxidation numbers of Cd and Pb+2 in the reactant side are 0 and +2,
respectively. They are, however, +2 and 0 in the product side. Hence, this is
also a redox reaction.
Equation
3:
Pb(NO3)2(liq)
+ 2LiCl(aq) --> PbCl2(s) + 2LiNO3(aq)
The
oxidation numbers of the involved ions (both cations and anions) are not
changed. Hence, this is NOT an oxidation reaction.
Equation
4:
C(s)
+ O2(g) --> CO2(g)
Just
as the equation 1 and 2, the oxidation numbers of the reactants are not similar
to those in the product. Hence, this is an example of oxidation reaction.
Answer:
8
Explanation:
Oxidation:
Reduction:
We have to equalise the number of moles of electrons gained and lost in a redox reaction in order to get a balanced reaction.
Hence we have to multiply the oxidation reaction throughout by 6.
and adding the two half-reactions we obtain:
Still the total charge and number of oxygen is not balanced.
Since the reaction takes place in acidic conditions, we will add required number of H+ to the appropriate side to balance the charge and add half the amount of H2O to balance the hydrogen atoms.
We add 14 H+ on LHS and 7H2O on RHS to obtain:
Sum of coefficients of product cations = 6+2 = 8
Answer:
about 36.10 g
Explanation:
The proportion of interest is ...
mass/atoms = x/(1.810·10^24) = (12.0107 g)/(6.02214076·10^23)
Multiplying by 1.810·10^24, we find the mass of the sample to be ...
x = 36.0991 g
The mass of the sample is about 36.10 grams (to 4 s.f.).
Answer:
Diphosphorus pentoxide
Carbon dichloride
BCl3
N2H4
Explanation:
These are all covalent compounds. To name covalent compounds, you add prefixes to the beginning of their names depending on what the subscript is of each element. The prefixes are:
1: Mono
2: Di
3: Tri
4: Tetra
5: Penta
6: Hexa
7: Hepta
8: Octa
9: Nona
10: Deca
For example, since the first one is Phopsphorus with a 2 next to it, you add the prefix Di to it.
If the first element in the compound only has one, meaning no number next to it, you do not say mono. This is why we just say "Carbon" for the second one instead of "Monocarbon."
Finally, you always have to end the second element in the compound with "ide." So, "chlorine" becomes "chloride," "oxygen" becomes "oxide," and so on.
