Is 334mm bigger than 1 km

How to determine oxidation numbers

nadezda [96]

Answer:

1. The oxidation number of a free element is always 0.

2. The oxidation number of a monatomic ion equals the charge of the ion.

3. The oxidation number of H is +1, but it is -1 in when combined with less electronegative elements.

4. The oxidation number of O in compounds is usually -2, but it is -1 in peroxides.

Explanation:

7 0

4 years ago

Consider the following reaction: CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) The initial rate of the reaction is measured at several di

Ira Lisetskai [31]

Answer : The correct rate law for the reaction is,

$\text{Rate}=k[CHCl_3][Cl_2]^{1/2}$

Explanation :

Rate law : It 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:

$CHCl_3(g)+Cl_2(g)\rightarrow CCl_2(g)+HCl(g)$

Rate law expression for the reaction:

$\text{Rate}=k[CHCl_3]^a[Cl_2]^b$

where,

a = order with respect to $CHCl_3$

b = order with respect to $Cl_2$

Expression for rate law for first observation:

$0.0035=k(0.010)^a(0.010)^b$ ....(1)

Expression for rate law for second observation:

$0.0069=k(0.020)^a(0.010)^b$ ....(2)

Expression for rate law for third observation:

$0.0098=k(0.020)^a(0.020)^b$ ....(3)

Expression for rate law for fourth observation:

$0.027=k(0.040)^a(0.040)^b$ ....(4)

Dividing 1 from 2, we get:

$\frac{0.0069}{0.0035}=\frac{k(0.020)^a(0.010)^b}{k(0.010)^a(0.010)^b}\\\\2=2^a\\a=1$

Dividing 2 from 3, we get:

$\frac{0.0098}{0.0069}=\frac{k(0.020)^a(0.020)^b}{k(0.020)^a(0.010)^b}\\\\1.42=2^b\\b=\frac{1}{2}$

Calculation used :

$1.42=2^b\\\log (1.42)=b\log 2\\\log (\frac{1.42}{2})=b\\b=0.5=\frac{1}{2}$

Thus, the rate law becomes:

$\text{Rate}=k[CHCl_3]^1[Cl_2]^{1/2}$

7 0

3 years ago

Which of the following methods would be the best method for separating oil from water?

vazorg [7]

Oil and water can separate by utilizing a separating funnel, oil and water are totally insoluble in one another. The water structures lower while the oil forms the upper layer.

8 0

3 years ago

Rank the following atoms in order of increasing electronegativity.a. Se, O, S b. P, Na, Cl c. Cl, S, F d. O, P, N

Papessa [141]

Answer:

a)- Se, S, O

b)- Na, P, Cl

c)- S, Cl, F

d)- P, N, O

Explanation:

We can solve this problem by looking at the Periodic Table. In a group, electronegativity increases from bottom to top; whereas in periods it increases from left to right.

Thus, we order the elements from lower to higher electronegativity as follows:

a. Se, O, S ⇒ order: Se, S, O

Because they are all in the same group. Se is near the bottom, followed by S and O is at the top.

b. P, Na, Cl ⇒ order: Na, P, Cl

They are in the same period. Na is at the left, followed by P and Cl is nearest the right.

c. Cl, S, F ⇒ order: S, Cl, F

P and Cl are in the same period, and P is at the left, so it has the lowest electronegativity. F is in the same group of Cl, but at the top. F has the highest electronegativity.

d. O, P, N ⇒ order: P, N, O

N and P are in the same group, but P is at the bottom so it has the lower electronegativity. N and O are in the same period, but O is at the right, so it is the most electronegative.

8 0

3 years ago

Solid ammonium chloride, NH4Cl, is formed by the reaction of gaseous ammonia, NH3, and hydrogen chloride, HCl. NH3(g)+HCl(g)⟶NH4

Mashutka [201]

9.41 atm is the pressure in atmospheres of the gas remaining in the flask

<h3>What is the pressure in atmospheres?</h3>

The equation NH3(g) + HCl(g) ==> NH4Cl(s) is balanced.

Divide the moles of each reactant by its coefficient in the balanced equation, and the limiting reagent is identified as the one whose value is less. With the issue we now have...

6.44 g NH3 times 1 mol NH3/17 g equals 0.3688 moles of NH3 ( 1 = 0.3688)

HCl: 6.44 g of HCl times one mole of HCl every 36.5 g equals 0.1764 moles ( 1 = 0.1764). CONTROLLING REAGENT

NH4Cl will this reaction produce in grams

0.1764 moles of HCl multiplied by one mole of NH4Cl per mole of HCl results in 9.44 g of NH4Cl (3 sig. figs.)

the gas pressure, measured in atmospheres, that is still in the flask

NH3(g) plus HCl(g) results in NH4Cl (s)

0.3688......0.1764............0..........

Initial

-0.1764....-0.1764........+0.1764...Change

Equilibrium: 0.1924.......0...............+0.1924

There are 0.1924 moles of NH3 and no other gases in the flask. This is at a temperature of 25 °C (+273 = 298 °K) in a volume of 0.5 L. After that, we may determine the pressure by using the ideal gas law (P).

PV = nRT

P = nRT/V = 0.1924 mol, 0.0821 latm/mol, and 298 Kmol / 0.5 L

P = 9.41 atm

9.41 atm is the pressure in atmospheres of the gas remaining in the flask

To learn more about balanced equation refer to:

brainly.com/question/11904811

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7 0

1 year ago