I need to find the orbit notation 1. Ne - 1s 2s 2p 3s 3p 4s 3d 4p 5s

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

erastovalidia [21]

Answer : The initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

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:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

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

Putting values in equation 1, we get:

$6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1$

$\text{Rate}=3.4\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

8 0

2 years ago

Match the following chemical reactions:

S_A_V [24]

Answer:

Synthesis - 4

reversible- 2

exchange- 1

decomposition-3

Explanation:

In synthesis reaction two or more components combines to form a single product. example 2H2+O2⇒2H2O

In reversible reaction two reactants combine to form two products . The products then reacts and forms back the reactants. example N2 +3H2 ⇒2NH3

In exchange reaction there is an alternation of ions of reactants to form new products. AB+CD ⇒AC + BD

In decomposition reaction, molecules of a compound break down by the action of heat or light or catalyst. example CaCO3 ⇒CaO +CO2

3 0

2 years ago

How many alkali metals will combine with 1 oxygen family member

Daniel [21]

The alkali metals are so reactive that they are never found in nature in elemental form. Although some of their ores are abundant, isolating them from their ores is somewhat difficult. For these reasons, the group 1 elements were unknown until the early 19th century, when Sir Humphry Davy first prepared sodium (Na) and potassium (K) by passing an electric current through molten alkalis. (The ashes produced by the combustion of wood are largely composed of potassium and sodium carbonate.) Lithium (Li) was discovered 10 years later when the Swedish chemist Johan Arfwedson was studying the composition of a new Brazilian mineral. Cesium (Cs) and rubidium (Rb) were not discovered until the 1860s, when Robert Bunsen conducted a systematic search for new elements. Known to chemistry students as the inventor of the Bunsen burner, Bunsen’s spectroscopic studies of ores showed sky blue and deep red emission lines that he attributed to two new elements, Cs and Rb, respectively. Francium (Fr) is found in only trace amounts in nature, so our knowledge of its chemistry is limited. All the isotopes of Fr have very short half-lives, in contrast to the other elements in group 1.

7 0

3 years ago

Consider the reaction at 500 ° C 500°C . N 2 ( g ) + 3 H 2 ( g ) − ⇀ ↽ − 2 NH 3 ( g ) K c = 0.061 N2(g)+3H2(g)↽−−⇀2NH3(g)Kc=0.06

Sav [38]

Answer:

Q = 0.061 = Kc

Explanation:

Step 1: Data given

Temperature = 500 °C

Kc=0.061

1.14 mol/L N2

5.52 mol/L H2

3.42 mol/L NH3

Step 2: Calculate Q

Q=[products]/[reactants]=[NH3]²/ [N2][H2]³

If Qc=Kc then the reaction is at equilibrium.

If Qc<Kc then the reaction will shift right to reach equilibrium.

If Qc>Kc then the reaction will shift left to reach equilibrium.

Q = (3.42)² / (1.14 * 5.52³)

Q = 11.6964/191.744

Q = 0.061

Q = Kc the reaction is at equilibrium.

4 0

3 years ago

How many atoms are in 13.97 liters of water vapor at STP

Arisa [49]

Let's assume that water vapor has ideal gas behavior. 
Then we can use ideal gas formula,
PV = nRT

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.

P = 1 atm = 101325 Pa (standard pressure)
V = 13.97 L = 13.97 x 10⁻³ m³
n = ?
R = 8.314 J mol⁻¹ K⁻¹
T = 0 °C = 273 K (standard temperature)

By substitution,
101325 Pa x 13.97x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K
n = 0.624 mol

Hence, the moles of water vapor at STP is 0.624 mol.

According to the Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.

Hence, number of atoms in water vapor = 0.624 mol x 6.022 × 10²³ mol⁻¹
 = 3.758 x 10²³

5 0

3 years ago

I need to find the orbit notation 1. Ne - 1s 2s 2p 3s 3p 4s 3d 4p 5s​

I need to find the orbit notation 1. Ne - 1s 2s 2p 3s 3p 4s 3d 4p 5s