What is the total charge of an iron nucleus

What is the volume that you should dilute 30mL of a 12.0M H2SO4 solution to obtain a 0.160M H2SO4 solution?

shutvik [7]

To solve this problem, the dilution equation (M1 x V1 = M2 X V2) must be used. The given values in the problem are M1= 12.0 M, V1= 30 mL, and M2= 0.160 M. To solve for V2,
V2=M1xV1/M2
V2= (12x30)/0.16
V2= 2,250 mL.
The correct answer is 2.25 L.

8 0

2 years ago

"Compounds A and B react to give a single product, C. Write the rate law for each of the following cases and determine the units

olasank [31]

Answer:

Part a: Units of k is $M^{-2}s^{-1}$ where reaction is first order in A and second order in B

Part b: Units of k is $M^{-1}s^{-1}$ where reaction is first order in A and second order overall.

Part c: Units of k is $M^{-1}s^{-1}$ where reaction is independent of the concentration of A and second order overall.

Part d: Units of k is $M^{-3}s^{-1}$ where reaction reaction is second order in both A and B.

Explanation:

As the reaction is given as

$A+B \rightarrow C$

where as the rate is given as

$r=k[A]^x[B]^y$

where x is the order wrt A and y is the order wrt B.

Part a:

x=1 and y=2 now the reaction rate equation is given as

$r=k[A]^1[B]^2$

Now the units are given as

$r=k[A]^1[B]^2\\M/s =k[M]^1[M]^2\\M/s =k[M]^{1+2}\\M/s =k[M]^{3}\\M^{1-3}/s =k\\M^{-2}s^{-1} =k$

The units of k is $M^{-2}s^{-1}$

Part b:

x=1 and o=2

x+y=o

1+y=2

y=2-1

y=1

Now the reaction rate equation is given as

$r=k[A]^1[B]^1$

Now the units are given as

$r=k[A]^1[B]^1\\M/s =k[M]^1[M]^1\\M/s =k[M]^{1+1}\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k$

The units of k is $M^{-1}s^{-1}$

Part c:

x=0 and o=2

x+y=o

0+y=2

y=2

Now the reaction rate equation is given as

$r=k[A]^0[B]^2$

Now the units are given as

$r=k[B]^2\\M/s =k[M]^2\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k$

The units of k is $M^{-1}s^{-1}$

Part d:

x=2 and y=2

Now the reaction rate equation is given as

$r=k[A]^2[B]^2$

Now the units are given as

$r=k[A]^2[B]^2\\M/s =k[M]^2[M]^2\\M/s =k[M]^{2+2}\\M/s =k[M]^{4}\\M^{1-4}/s =k\\M^{-3}s^{-1} =k$

The units of k is $M^{-3}s^{-1}$

7 0

3 years ago

Complete the three words. The oxidation state tells you how many electrons an atom has g....................... , l.............

fgiga [73]

Answer:Gained, Lost , Shared

Explanation:

The oxidation state tells you how many electrons an atom has GAINED.................. , LOST....................... , or SHARED........................ , in forming a compound.

Oxidation state is defined as the the total number of electrons that an atom gains or loses when forming a chemical bond with another atom.

----To form an ionic bond for example in NaCl, Na, with 11 electrons and one valence electron in its outermost shell donates or lose that valence electron to Chlorine with 17 electron and 7 in its outermost shell. Therefore Sodium, Na acquires the +1 oxidaton state to become stable and Chlorine acquires the -1 oxidation state to become stable forming the NaCl compound.

To form a covalent compound, There must be sharing of electrons between atoms.For example, in PCl3, The phosphorous atom with atomic number 15 shares its three unpaired electrons with the single valence electrons of three chlorine atoms. making the four molecules to attain stability with Phosphorous having +3 and the chlorine atoms having -1 oxidation states

8 0

2 years ago

4NH3 (g)+502(g)->6H20g)+4NOg) What is the enthalpy change for this reaction?

Feliz [49]

Enthalpy is energy of bonds broken - energy of bonds formed. Here, the NH3 and O2 are broken and H2O and NO are formed. So the energy to break the NH3 bonds is 3 times the amount of energy it takes to break a N-H single bond (because there are three of them in a NH3 molecule) and then multiplied by 4 because there are four particles.

So the energy of the bonds broken is 12x the energy to break a N-H single bond plus 5x the amount of energy to break an O—O double bond (you don’t multiply this by anything because in each O2 molecule there is only one bond).

The energy of the bonds formed is 6*2 = 12 Times the amount of energy for a O-H single bond plus 4 times the amount of energy required to break a N—O double bond.

Subtract energy of bonds broken - energy of bonds formed and this is the change in enthalpy.

To know what type of bond it is, draw the Lewis structure.

7 0

3 years ago

In a metallic bond, electrons

kobusy [5.1K]

The answer would be c

8 0

2 years ago