To name the compound written as cucl2 you would write\

The following data was collected for the formation of ammonia (NH3) based on the following overall reaction: N2 + 3H2 = 2NH3 N2

notsponge [240]

Answer : The unit for the rate constant in the rate law for the formation of ammonia is, $M^{-2}min^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$N_2+3H_2\rightarrow 2NH_3$

Rate law expression for the reaction:

$\text{Rate}=k[N_2]^a[H_2]^b$

where,

a = order with respect to $N_2$

b = order with respect to $H_2$

Expression for rate law for first observation:

$0.0021=k(0.10)^a(0.10)^b$ ....(1)

Expression for rate law for second observation:

$0.0084=k(0.10)^a(0.20)^b$ ....(2)

Expression for rate law for third observation:

$0.0672=k(0.20)^a(0.40)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{0.0084}{0.0021}=\frac{k(0.10)^a(0.20)^b}{k(0.10)^a(0.10)^b}\\\\4=2^b\\b=2$

Dividing 3 by 1 and also put value of b, we get:

$\frac{0.0672}{0.0021}=\frac{k(0.20)^a(0.40)^2}{k(0.10)^a(0.10)^2}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[N_2]^a[H_2]^b$

$\text{Rate}=k[N_2]^1[H_2]^2$

Now, calculating the value of 'k' by using any expression.

$0.0021=k(0.10)^1(0.10)^2$

$k=2.1M^{-2}min^{-1}$

The value of the rate constant 'k' for this reaction is $2.1M^{-2}min^{-1}$

That means, the unit for the rate constant in the rate law for the formation of ammonia is, $M^{-2}min^{-1}$

7 0

3 years ago

What have scientists been able to use to create the Geologic Time Scale?

dem82 [27]

The half-life of the carbon-14 isotope is used in dating fossils in a process called radiocarbon dating.

hope this is adequate.

3 0

3 years ago

A sample of gas has an initial volume of 3.88 l at a pressure of 707 mmhg . part a if the volume of the gas is increased to 5.43

JulsSmile [24]

Using boyles law
p1v1=p2v2
v1=3.88
v2=5.43
p1=707
p2=?
707x3.88=p2x5.43
2743.16/5.43=p2
p2=505mmHg

4 0

3 years ago

A solution of sodium hydroxide was titrated against a solution of sulfuric acid. How many moles of sodium hydroxide would react

jeka57 [31]

Answer:

2 mole of Sodium hydroxide reacts with 1 mole of Sulfuric acid

Explanation:

Write down the equation in the beginning with reactants and products:

NaOH + H₂SO₄ → Na₂SO₄ + H₂0

Now try to balance it. Try with Na first:

2NaOH + H₂SO₄ → Na₂SO₄ + H₂0

Na atoms are balanced. There are 6 Oxygen atoms on the right and 5 on the left. Balance by increasing the H₂O moles:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0

Check if H atoms are also balanced. They are. That means our final reaction is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0

2 Moles of NaOH reacts with 1 mole of H₂SO₄

5 0

3 years ago

In a synthesis reaction you start with 1.7L of Hydrogen how many liters of water will be produced?

jolli1 [7]

15.3 litres of water will be produced if we take 1.7 litres of Hydrogen

Explanation:

Let's take a look over synthesis reaction;

$H_{2}+ O_{2}$ $H_{2}O$

Balancing the chemical reaction;

$2H_{2} +O_{2}$ $2H_{2}O$

Thus, 2 moles of hydrogen molecules are required to form 2 moles of water molecules.

Equating the molarity;

$\frac{1.7*1}{2*2}$ = $\frac{x*1}{2*18}$

(Since, the molecular mass of hyd and water is 2 and 18 respectively)

x= $\frac{1.7*2*18}{2*2}$

x= 15.3 litres.

Thus,15.3 L of water will be produced if we take 1.7 litres of Hydrogen in a synthesis reaction.

6 0

3 years ago