Calculate the mass of a 0.297-mL sample of a liquid with a density of 0.930 g/mL.

"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

How many moles of H2O are in 42.0 g H2O?

cricket20 [7]

(42/18.02)=2.3307mol H2O

5 0

3 years ago

Read 2 more answers

Which compound most likely contains polar covalent bonds?

marin [14]

I would say water; water is extremely polar, and this is why it can break one of the strongest bonds, ionic bonds. NaCl, as you probably know, is a salt, and dissolves in water. However, the ionic bond holding the Na+ and the Cl- is extremely strong; the boiling point of NaCl is at 1413 degrees celcius (water is at 100 degrees celcius). This means that it requires A LOT of energy to break the bond, but water is able to dissolve and break the bond very easily. It is very polar, so I would answer your question with water. And the bond connecting the H and the O is a covalent bond.

3 0

3 years ago

Consider the following reaction: 2Mg(s)+O2(g)-->2MgO(s) delta H=-1204kJ

Pachacha [2.7K]

Answer:

a. The reaction is exothermic.

b. -87,9 kJ

c. 9,60g of Mg(s)

d. 602kJ are absorbed

Explanation:

Based on the reaction:

2Mg(s) + O₂(g) → 2MgO(s) ΔH = -1204kJ

a. The reaction is exothermic. Because ΔH<0. That means the reaction produces heat when occurs

b. 3,55g of Mg(s) are:

3,55g Mg × ( 1mol / 24,305g) = 0,146 moles of Mg(s)

As 2 moles of Mg(s) produce -1204 kJ of heat:

0,146 moles of Mg(s) × ( -1204kJ / 2mol Mg) = -87,9 kJ

c. If -238 kJ of heat were transferred. The moles of Mg(s) that react must be:

-238kJ × ( 2mol Mg / -1204kJ) = 0,395 moles of Mg(s). In grams:

0,395 moles × ( 24,305g / 1mol Mg) = 9,60g of Mg(s)

d. The reverse reaction is:

2MgO(s) → 2Mg(s) + O₂(g) ΔH = +1204kJ

40,5g of MgO(s) are:

40,5g MgO × ( 1mol MgO / 40,3044g) = 1,00 moles of MgO(s)

As 2 moles of MgO absorbe 1204kJ of energy:

1,00 moles of MgO(s) × ( +1204 kJ / 2mol MgO) = 602kJ are absorbed

I hope it helps!

7 0

3 years ago

(01.05 LC)

wel

Answer:

it is A trust me I know thank you

7 0

1 year ago

Read 2 more answers