What happens if the metal you throw in is MORE REACTIVE than the<br> metal ion in solution?

Which element is oxidized in the reaction below? fe(co)5 (l) + 2hi (g) → fe(co)4i2 (s) + co (g) + h2 (g)

Hunter-Best [27]

The element which is oxidized in the reaction below
that is

Fe(Co)5 (l) + 2 Hi (g)= Fe (CO)4 i2(s) +Co (g)+H2 (g) is

Fe this is because fe in Fe(CO)5 compound move from oxidation state of Positive five (5^+) to oxidation state of positive six ( 6^+) in Fe(CO)4i2 compound. fe is oxidized by Hi while Hi itself is reduced by Fe,therefore Fe act as a reducing agent

7 0

3 years ago

URGENT HELP! I dont understand anything of this and this is for tomorrow

Tatiana [17]

Basically, the graph shows a linear (line) relationship with mass and volume. The slope will equal the density of the substance because mass/volume is the equation for density. Your paper is a bit unclear, but I can understand this much, Hope this helped a little bit :)

3 0

3 years ago

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium

lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

5 0

3 years ago

For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow

lora16 [44]

Answer:

Rate constant of the reaction is $3.3\times 10^{-3} M^{-2} s^{-1}$ .

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

$R=k[A]^a[B]^b[C]^c$

Rate(R) of the reaction in trail 1 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.30 M$

$R=9.0\times 10^{-5} M/s$

$9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c$ ...[1]

Rate(R) of the reaction in trail 2 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.90 M$

$R=2.7\times 10^{-4} M/s$

$2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c$ ...[2]

Rate(R) of the reaction in trail 3 ,when :

$[A]=0.60 M,[B]=0.30 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c$ ...[3]

Rate(R) of the reaction in trail 4 ,when :

$[A]=0.60 M,[B]=0.60 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c$ ...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

$R=k[A]^2[B]^0[C]^1$

Rate constant of the reaction = k

$9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1$

$k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}$

$k=3.3\times 10^{-3} M^{-2} s^{-1}$

7 0

3 years ago

Question 11 (1 point)

erastova [34]

Answer:

remove product

Explanation:

Removing the product will always shift the equilibrium to the right. This is based on the Le Chatelier's principle which states that "if any of the conditions of a system in equilibrium is changed, the system will adjust itself in order to annul the effect of the change".

If a system at equilibrium is disturbed, by changing the concentration of one of the substances all the concentrations will change until a new equilibrium point is reached.
Removing the product will increase the concentration of the species on the left hand side, the equilibrium will shift to the right.

5 0

3 years ago

What happens if the metal you throw in is MORE REACTIVE than the metal ion in solution?